tag:blogger.com,1999:blog-90589308657208227262018-07-03T07:29:45.311+01:00algebradabraAlgebran for digesting the notion of variable ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.comBlogger22125tag:blogger.com,1999:blog-9058930865720822726.post-74774684219750642942018-06-24T23:05:00.001+01:002018-06-29T08:18:37.538+01:00ALG 20Algebraderci! This post might be the last ALG post (at least for a while). Fittingly, perhaps, we devote this week's task to <span style="font-family: "arial" , "helvetica" , sans-serif;"><b>rules of single and double false position</b></span>, which are methods that go back about 2000 years and that surfaced in many parts of the world - China, Egypt, the Arab world, India and, eventually, Europe. It is not clear whether the methods spread from one region to another, or were discovered/invented spontaneously in several regions. They were generally used to solve practical problems and as such were known not just to mathematicians but to people such as merchants who would use them as algorithms, ie as rules that worked rather than as rules that needed to be justified and proved. The rules apply to situations that we would now represent with linear or affine equations, ie equations of the form <i>y </i>=<i> ax</i> and <i>y </i>=<i> ax </i>+<i> b</i> respectively. However the rules were devised many centuries before symbolic algebra was invented, and continued to be used for some centuries thereafter.<br />- <br /><b>MONDAY</b>: We start with a problem taken from the Eygyptan <i>Rhind Mathematical Papyrus</i> (RMP), which dates from about 1550 BC and was written by a 'copyist', Ahmose, who claims to have been copying work from one or several centuries earlier.<br />This particular problem is quite straightforward, and certainly doesn't need formal algebra, though that provides one way of solving it. We present it here to illustrate the <b>rule of single false position</b>, which is the way it is solved in the RMP.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-9tJXTaXAD_E/WzARL9Yig5I/AAAAAAAAAnI/I9IxFn8TTWw7awDFAprs4IRwjfBRxk_9wCLcBGAs/s1600/ALG-20A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-9tJXTaXAD_E/WzARL9Yig5I/AAAAAAAAAnI/I9IxFn8TTWw7awDFAprs4IRwjfBRxk_9wCLcBGAs/s1600/ALG-20A.jpg" /></a></div>The above problem is Problem 26 in the RMP. Problems 24, 25 and 27 are similar, though interestingly they would seem more complex to us than Problem 26, since their solutions are all fractional. In fact, the first three problems appear to get progressively easier! Something to think about, for the small-stepping devotees of 'intelligent practice'.<br />Here are the problems, expressed in modern form:<br />24: <i>x </i>+<i> x</i>/7 = 18<br />25: <i>x </i>+<i> x</i>/2 = 16<br />26: <i>x </i>+<i> x</i>/4 = 15<br />27: <i>x </i>+<i> x</i>/5 = 21.<br />-<br /><b>TUESDAY</b>: We show how a problem can be solved using the rule of <b>double false position</b> - or rather, we show the first two steps, and leave it to the reader to complete the method. We provide some help with the final step in Wednesday's version of the task.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-60sz-Lif61o/WzFuGrO3vZI/AAAAAAAAAnU/PRZyS928Oos4IW2p1rT98tX085vmDZh_ACLcBGAs/s1600/ALG-20B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-60sz-Lif61o/WzFuGrO3vZI/AAAAAAAAAnU/PRZyS928Oos4IW2p1rT98tX085vmDZh_ACLcBGAs/s1600/ALG-20B.jpg" /></a></div><b>Note</b>: We came across this Apples task in a very useful conference paper by <a href="https://www.researchgate.net/publication/242281258_Issues_in_the_Origin_and_Development_of_Hisab_al-Khata%27ayn_Calculation_by_Double_False_Position_Questions_Sur_l%27Origine_et_le_Developpement_de_Hisab_al-Khata%27ayn_Calcul_par_Double_Fausse_Position" target="_blank">Schwartz</a>, presented in 2004. Schwartz gives the source of the task as the <i>Liber Augmenti et Diminutionis</i>: "This book, probably from the 12th Century, is the translation of some now-lost earlier work attributed to one 'Abraham' and generally believed to have been written in Arabic or Hebrew" (ibid). The task can also be solved algebraically, though it wouldn't have been in its day. Interestingly, an algebraic approach turns out to be very cumbersome here, compared to the use of double false position.<br />-<br /><b>WEDNESDAY</b>: We revisit Tuesday's task but use a diagram to help make sense of the double false position rule.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-_PxqNvKY3IE/WzKlnTPSNTI/AAAAAAAAAns/FzZEhW3wF90SGilmr_mBuhRx9zxvb5buwCLcBGAs/s1600/ALG-20C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-_PxqNvKY3IE/WzKlnTPSNTI/AAAAAAAAAns/FzZEhW3wF90SGilmr_mBuhRx9zxvb5buwCLcBGAs/s1600/ALG-20C.jpg" /></a></div>In this example, the double false position rule can be written as (100×21 – 204×8)÷(21 – 8), which reduces to 36. An alternative version, which can perhaps be derived more directly from the diagram, is this:<br />100 – 8/(21–8) of (204–100)<br />= 100 – 8/13 of 104<br />= 100 – 64<br />= 36.<br /> The general form of the rule is (<i>x</i>₁<i>e</i>₂ – <i>x</i>₂<i>e</i>₁)÷(<i>e</i>₂ – <i>e</i>₁) which, presented rhetorically, would in ancient times probably have been treated simply as a practical algorithm to be learnt and applied.<br /><div class="separator" style="clear: both; text-align: center;"></div>-<br /><b>THURSDAY</b>: We apply the double false position rule to parts of a fish.... This problem was probably written by de la Grange. It appears in the 1822 translation into English of Euler's <i>Elements of Algebra</i>. The intention would have been to solve the task algebraically, using simultaneous equations. However, it also lends itself very nicely to the <i>double false position</i> approach.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-DeJYKu6bNAI/WzR-S2HA7bI/AAAAAAAAAn4/gt4FmJ0MimoxcgnUAkSC01Tr1ZCoCqA1ACLcBGAs/s1600/ALG-20D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-DeJYKu6bNAI/WzR-S2HA7bI/AAAAAAAAAn4/gt4FmJ0MimoxcgnUAkSC01Tr1ZCoCqA1ACLcBGAs/s1600/ALG-20D.jpg" /></a></div>Here, the weighted average of 11 and 31 can be written as (2×11 + 8×31)÷(8 + 2) = (22+248)÷10 = 27, so the head weights 27 lb. [In turn, the fish weighs 72 lb, as stated in the text.]<br />Here's a nice bar-model approach (thanks to Bernie Westacott). One can see how it corresponds to an algebraic approach using simultaneous equations.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-v2lTxhLJLP8/WzVajGNhHUI/AAAAAAAAAoQ/2WjGFxe5DBw6-2iJC4raa-S8klNj2KNiACLcBGAs/s1600/ALG-20D-bernie-westacott-fish.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1449" height="400" src="https://1.bp.blogspot.com/-v2lTxhLJLP8/WzVajGNhHUI/AAAAAAAAAoQ/2WjGFxe5DBw6-2iJC4raa-S8klNj2KNiACLcBGAs/s400/ALG-20D-bernie-westacott-fish.jpg" width="361" /></a></div><br />-<br /><b>FRIDAY</b>: Friday night is bath night and we take a (classic) bath.... This type of problem goes back centuries, though this one is only from 1936 (it is quoted in a paper co-authored by Arcavi). The problem is remarkably easy to solve using single false position. A greater challenge is to find an informal method - to start with, it might be helpful to simplify the task slightly, by ignoring the waste-pipe, ie by putting the plug in the bath. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-YhnpmH8GIiw/WzVWifCUvXI/AAAAAAAAAoE/8ZMBiqI5upkkPDjoVScAr4pRb5OyH-p4wCLcBGAs/s1600/ALG-20E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-YhnpmH8GIiw/WzVWifCUvXI/AAAAAAAAAoE/8ZMBiqI5upkkPDjoVScAr4pRb5OyH-p4wCLcBGAs/s1600/ALG-20E.jpg" /></a></div><span style="font-family: "arial" , "helvetica" , sans-serif;"><b>Valediction</b></span>: 20 sets of task, that's 20 weeks or almost 5 months of algebra tasks, 100 in all. Time to say Algebraderci (at least for a while). ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-57986251349605577052018-06-17T20:56:00.001+01:002018-06-21T23:54:51.657+01:00ALG 19This week we're looking at graphs that model 'real life' situations. The focus is on qualitative aspects of the graphs, though we do take a more analytic view at times. Pioneering work on school students' understanding of graphs was undertaken in the late 1970s and early 1980s by Daphne Kerslake, by Claude Janvier and by Malcolm Swan. I doubt whether one can better some of their outstanding tasks.<br />-<br /><b>MONDAY</b>: Mathematically, this task involves a nice context, though one that is perhaps not so familiar to today's school students - you might therefore have to tease out the crucial fact that audio tape has to travel at a <i>constant speed</i> (as it passes over the tape-head which picks up the sound). Most students will realise that as the left hand spool unwinds, the width (diameter) of the reel of tape decreases. This means the circumference also decreases and so each successive turn of the spool will feed out less and less tape. So if the tape passes the tape-head at a constant speed, the spool will rotate more and more quickly, and the width of the reel will decrease more and more rapidly.<br />This task should generate rich classroom discussion about these geometric aspects of the context and about the shape of the graph: the graph clearly slopes downwards, but is it a straight line?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-yvjbEZoqGq4/WyavZkCTwnI/AAAAAAAAAlw/c3Ox7UuBgu8HMkQi5m2x5OdGtJVjZtToQCLcBGAs/s1600/ALG-19A-tape.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-yvjbEZoqGq4/WyavZkCTwnI/AAAAAAAAAlw/c3Ox7UuBgu8HMkQi5m2x5OdGtJVjZtToQCLcBGAs/s1600/ALG-19A-tape.jpg" /></a></div>As the width of the reel of tape decreases more and more rapidly (over time), the <i>slope</i> of the graph gets progressively steeper (the <i>gradient</i> is negative throughout). This raises interesting questions about the slope near points A and B. Could it be horizontal at A and vertical at B? [Well, no, not quite...] And if we continue the curve, where might it cut the <i>x</i>-axis, and what would this mean in terms of the context?<br /><b>Note</b>: The curve through A and B is a parabola. It turns out that it passes roughly through the point (76, 0) and that the relationship between <i>M</i> minutes and <i>W</i> mm, is given by <i>M</i> ≂ 76 – (<i>W</i>²/30) [or, more precisely, <i>W</i> = 78.83 – (12<i>W</i>²/345)*, if we assume the given measurements are accurate].<br />*You might like to verify this formula (please correct me if I've got it wrong).<br />- <br /><b>Note 2</b>: Here is an alternative wording for the task. Would this be better?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-tPE3g-DpqpU/Wyes9itsGoI/AAAAAAAAAmI/uE7g3T0MB64g9gYm6ADW67pEYxb_JMrswCLcBGAs/s1600/ALG-19A-tape-plus.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://3.bp.blogspot.com/-tPE3g-DpqpU/Wyes9itsGoI/AAAAAAAAAmI/uE7g3T0MB64g9gYm6ADW67pEYxb_JMrswCLcBGAs/s400/ALG-19A-tape-plus.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>-<br /><b>TUESDAY</b>: In this task we assume that the pile of sand keeps its exact shape as it gets bigger. Students might initially think that if the time for which the sand flows is doubled (from 1 minute to 2 minutes), then the height of the pile will be doubled too. (It turns out that it takes 8 minutes rather than just 2, for the height to reach 2 metres - something we look at in Wednesday's task.)<br />However, it shouldn't take much discussion for students to realise that as the pile gets higher it gets wider too, so ever more sand is needed to raise the height by a given amount. This means that the graph will consist of a curve with an ever flatter positive slope. In turn this means that at the 2 minute mark, the height of the graph will be substantially less than 2 (m). However, at this juncture we wouldn't expect students to have a more precise sense of what that height will be. This is something we consider in a more analytic way in Wednesday's task.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-VIPCUiRH5ss/WygaMtinjcI/AAAAAAAAAmU/gdiQRhf-gW0JGukgnv56ZoABAd3-4XoKwCLcBGAs/s1600/ALG-19B-sand.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-VIPCUiRH5ss/WygaMtinjcI/AAAAAAAAAmU/gdiQRhf-gW0JGukgnv56ZoABAd3-4XoKwCLcBGAs/s1600/ALG-19B-sand.jpg" /></a></div>-<br /><b>WEDNESDAY</b>: We consider what the height of the pile of sand will be after 2 minutes, and use this information to draw a more accurate graph, which we then use to interpolate the height at 2 minutes.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-53D3-preO7g/WylmNEn-7LI/AAAAAAAAAmg/AmuZ4AIaqzo9QKm19V5yDOZIN3SfglM-ACLcBGAs/s1600/ALG-19C-sand.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-53D3-preO7g/WylmNEn-7LI/AAAAAAAAAmg/AmuZ4AIaqzo9QKm19V5yDOZIN3SfglM-ACLcBGAs/s1600/ALG-19C-sand.jpg" /></a></div>If we double the height of the pile of sand, from 1 m to 2 m, we also double the width (in two dimensions), so we have 2×2×2 = 8 times as much sand as before, so it will take 8 times as long to produce.<br />A good approximation for the value of the height at 2 minutes is 5/4. This is close to the cube-root of 2, since (5/4)-cubed is 125/64 = 1.953125.<br />-<br /><b>THURSDAY</b>: We step onto a travelator and think about some (nice, simple, straight line) distance-time graphs.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-ye9J2kQWltA/WyrUs-a6SnI/AAAAAAAAAmw/fN-vtXusoxwz-yyrxfmwYZf-JloWdX6mwCLcBGAs/s1600/ALG-19D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-ye9J2kQWltA/WyrUs-a6SnI/AAAAAAAAAmw/fN-vtXusoxwz-yyrxfmwYZf-JloWdX6mwCLcBGAs/s1600/ALG-19D.jpg" /></a></div>-<br /><b>FRIDAY</b>: We relate our distance-time graphs to speed-time graphs. This may not be easy - they look so very different!<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-yeSslBuU0T0/WywmVkhedaI/AAAAAAAAAm8/YI4VQcvZhmka7QZe5627cQmHw6Rltd1JACLcBGAs/s1600/ALG-19E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-yeSslBuU0T0/WywmVkhedaI/AAAAAAAAAm8/YI4VQcvZhmka7QZe5627cQmHw6Rltd1JACLcBGAs/s1600/ALG-19E.jpg" /></a></div>Before thinking about the speed-time graph for Mario, it is interesting to consider how students might arrive at the red line - ie the distance-time graph for Mario while he is walking on the travelator. One way would be to find a specific point [such as (10, 14)] and join it to (2, 2) with a straight line. Another way would be to consider the distance travelled each second on the travelator - namely 0.5 m due to the travelator, plus 1 m due to the fact that Mario is walking at 1 m per second relative to the surface of the travelator - so the red line has a slope of 1.5 (m per sec).<br />Marco's and Mario's speeds are constant for most of the time (except for the brief moments when their speeds change). This means their speed-time graphs will consist primarily of <i>horizontal</i> straight lines. As such, the graphs look very different from the distance-time graphs, where the increase in distance (from Lisa) over time is shown by the upward slope of the graphs. In contrast to this, on the speed-time graphs, the increasing distance is shown by the increase in area under the graphs as the time increases. This looks far less dramatic.ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-44089384600333548002018-06-11T00:06:00.001+01:002018-06-15T10:28:47.902+01:00ALG 18This week we look at the rules of arithmetic, as expressed through the use of <b>brackets</b>. But we start informally, with a 'real life' task, specifically a sour dough problem, which may not be to everyone's liking....<br />- <br /><b>MONDAY</b>: A lighthearted story about pizza slices.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-N5dUZGMhPxM/Wx2tbwSaUTI/AAAAAAAAAkc/QQ9Kog5eKLAdBTeulcmHIRa5EYIzRcehgCLcBGAs/s1600/ALG-18A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-N5dUZGMhPxM/Wx2tbwSaUTI/AAAAAAAAAkc/QQ9Kog5eKLAdBTeulcmHIRa5EYIzRcehgCLcBGAs/s1600/ALG-18A.jpg" /></a></div><b>TUESDAY</b>: Here we complete some numerical expressions involving brackets to describe a set of dots.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-T0LGC-1mqIc/Wx7z1JYOkNI/AAAAAAAAAko/FT53BHr9qQQ4F5_H9tM2N3eToOXdJpQagCLcBGAs/s1600/ALG-18B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-T0LGC-1mqIc/Wx7z1JYOkNI/AAAAAAAAAko/FT53BHr9qQQ4F5_H9tM2N3eToOXdJpQagCLcBGAs/s1600/ALG-18B.jpg" /></a></div>No pizzas.... No pseudo contexts...? Just dots. Simple.<br /><b>Note</b>: I have curtailed the string of expressions a bit. You might, for example, want to insert the expression 3×5×(20+?), and perhaps also 3×(5×20 + 5×?), between my second and third expressions.<br />-<br /><b>WEDNESDAY</b>: Here we look at expressions that give the total number of dots in a (recursive) dice pattern.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-Vs-FaPjW8yY/WyBHuGa1itI/AAAAAAAAAlA/eTLHoXkzfDMxYqiSUT1Ik1PG3oLiLrJwQCLcBGAs/s1600/ALG-18C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-Vs-FaPjW8yY/WyBHuGa1itI/AAAAAAAAAlA/eTLHoXkzfDMxYqiSUT1Ik1PG3oLiLrJwQCLcBGAs/s1600/ALG-18C.jpg" /></a></div><b>Note 1</b>: we have used brackets more frequently than is strictly necessary here, so that the order of operations is explicit and unambiguous, ie so that the expressions can be read without knowing the conventional order of operations. Of course, it is important to learn these conventions but it means that the current expressions should be accessible to a high proportion of students. Also, as the expressions refer to familiar, concrete elements that are easy to group and count, the work should help demystify the use of expressions involving brackets - students don't need to refer to rules like 'do the brackets first' to make sense of the notation.<br /><b>Note 2</b>: you might want to ask students to annotate the diagram, or to draw a revised diagram, to show how the dots are being structured by the various expressions. <br />-<br /><b>THURSDAY</b>: Here we use expressions to count dots in some partitioned/compound arrays.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-xKXQaZZRRb8/WyGX875gZ5I/AAAAAAAAAlM/kvhLyEatlKo0K1SL53iz8p6uHSPYP_aBQCLcBGAs/s1600/ALG-18D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-xKXQaZZRRb8/WyGX875gZ5I/AAAAAAAAAlM/kvhLyEatlKo0K1SL53iz8p6uHSPYP_aBQCLcBGAs/s1600/ALG-18D.jpg" /></a></div>Again, it might be helpful to annotate the sets of dots to show how they relate to the expressions.<br />-<br /><b>FRIDAY</b>: We look at some neat and some not so neat ways of performing the calculation 15 × 22.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-OH03QQpZ4uc/WyOGuaW_m5I/AAAAAAAAAlk/J25_zhlXp8sfOcqf1zC0NVrqj10Ut-lowCLcBGAs/s1600/ALG-18E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-OH03QQpZ4uc/WyOGuaW_m5I/AAAAAAAAAlk/J25_zhlXp8sfOcqf1zC0NVrqj10Ut-lowCLcBGAs/s1600/ALG-18E.jpg" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"></div>This slide is somewhat overloaded. It's really two tasks in one. And for the first task, it would be better to introduce the parts one at a time. Then, having worked through these exemplars, it should prove fruitful to devote time to students' methods of calculating 15 × 22, and to representing these as strings of expressions involving brackets. <br />The second task should help students take a more conscious, formal view of the distributive law and to think about when it applies.ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-34650755819283209272018-06-03T20:37:00.001+01:002018-06-07T23:54:09.993+01:00ALG 17She was just seventeen, You know what I mean, And the way she looked Was way beyond compare. How could I dance with another? Oooh!<i> When I saw her standing there</i>.<br />One of the great early Beatles tracks. With ALG 17, we're standing on the number line and seeing how our position is related to multiples, especially multiples of 2 and 3.<br />- <br /><b>MONDAY</b>: Where do multiples of 3 occur on the number line? How can we tell whether a multiple of 3 is also a multiple of 6? And of 9??<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-MXBj2wjc3ss/WxRCLfwr_4I/AAAAAAAAAjU/nGL1YQqQQrkgF4yiXQvXa0lkK2f5oKcLACLcBGAs/s1600/ALG-17A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-MXBj2wjc3ss/WxRCLfwr_4I/AAAAAAAAAjU/nGL1YQqQQrkgF4yiXQvXa0lkK2f5oKcLACLcBGAs/s1600/ALG-17A.jpg" /></a></div><b>TUESDAY</b>: This is a classic task that I was introduced to by Lulu Healy. You might want to generate some data and look for patterns, but it is also perfectly accessible via a generic (analytic) approach by thinking about the occurance of multiples in strings of consecutive numbers.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-eV4NL60wABQ/WxV9p7Zm9UI/AAAAAAAAAjg/HOS0_4mKr74mYhhSbhCYV4t18iWvAJFJwCLcBGAs/s1600/ALG-17B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-eV4NL60wABQ/WxV9p7Zm9UI/AAAAAAAAAjg/HOS0_4mKr74mYhhSbhCYV4t18iWvAJFJwCLcBGAs/s1600/ALG-17B.jpg" /></a></div><b> WEDNESDAY</b>: more multiple fun....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-5PFx8oQM8c8/WxfkjaExOII/AAAAAAAAAj4/OmrmIsCI-bAblOoXTNxeDxRHDyN2eYT5wCLcBGAs/s1600/ALG-17C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-5PFx8oQM8c8/WxfkjaExOII/AAAAAAAAAj4/OmrmIsCI-bAblOoXTNxeDxRHDyN2eYT5wCLcBGAs/s1600/ALG-17C.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>This task keeps catching me out! One moment I think it's OK, the next I think I've got it wrong and it doesn't work!<br />Part b) can be solved by simply going through the multiples of 7 and 8 in the standard times tables, which leads us to 63, 64. However, a key feature here is the relation of these numbers to 56 (= 7×8) .... Or one can make use of 'the difference of two squares': 8×8 is a multiple of 8, 8×8 – 1 = (8–1)(8+1) is a multiple of 7. [This generalises very nicely, eg to finding consecutive numbers that are multiples of 19 and 20 respectively.]<br />- <br /><b>THURSDAY</b>: A classic 'think of a number ....' task. The task has links with Tuesday's version of ALG 17 and one can get quite a long way by using an empirical, data-generating approach. However, its attraction lies in the fact that one can use some fairly routine algebra to throw light on the underlying structure. [This is all the more noteworthy, given that we often use algebra to ease our path to an answer while ignoring structure!] <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-mUjg797-2Ms/WxhhY1QN2_I/AAAAAAAAAkE/V-MMv0Pzb68ZvuA450L_A5G8W4HddNgGACLcBGAs/s1600/ALG-17D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-mUjg797-2Ms/WxhhY1QN2_I/AAAAAAAAAkE/V-MMv0Pzb68ZvuA450L_A5G8W4HddNgGACLcBGAs/s1600/ALG-17D.jpg" /></a></div><b>Note</b>: I came across this particular task in a 1995 paper by Alan Bell (<i>Purpose in school algebra</i>, JMB, 14, 41-73). Sadly, Alan died this year. (9 April 1929 - 5 April 2018)<br />-<br /><b>FRIDAY</b>: We take an explicit look at the structure of Thursday's task.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-awr1ECOeP7o/Wxm1kjzii-I/AAAAAAAAAkQ/Ph5nkhezoh44IuNmsAbGBW23gCIQU9fOgCLcBGAs/s1600/ALG-17E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-awr1ECOeP7o/Wxm1kjzii-I/AAAAAAAAAkQ/Ph5nkhezoh44IuNmsAbGBW23gCIQU9fOgCLcBGAs/s1600/ALG-17E.jpg" /></a></div>Here the geometric and/or symbolic representation may help us see that Thursday's task results in the product of the 'outer two' of three consecutive numbers. When the middle number is odd, the outer numbers will be even so both must be a multiple of 2 and one must also be a multiple of 4 (of course, they can be multiples of other numbers too). So their product is a multiple of 8.ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-12681300812563781692018-05-28T00:03:00.000+01:002018-06-01T09:18:48.471+01:00ALG 16This week we look at the structure of <b>arithmagons</b>. This involves algebraic thinking which is made more accessible by the use of algebraic symbolisation, and it is hoped that this might encourage students to use it, though in the absence of such symbolisation it involves algebraic thinking nonetheless.<br />-<br /><b>MONDAY</b>: A bank holiday and the beginning of school half term in the UK, but that's no reason to deprive ourselves of algebra. Here's our gentle introduction to arithmagons.<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-wLEG-o3zJVE/Wws3Iz1mPuI/AAAAAAAAAh4/f0taA10ifloKfnQwD4NN-MpWFESa7bAzgCLcBGAs/s1600/ALG-16A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-wLEG-o3zJVE/Wws3Iz1mPuI/AAAAAAAAAh4/f0taA10ifloKfnQwD4NN-MpWFESa7bAzgCLcBGAs/s1600/ALG-16A.jpg" /></a></div>Yes, the intro is rather subtle, but we don't want to over-excite our muddled friend Capt Scarlet... <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-qsBUjFaa8hE/Wws6mBCjSLI/AAAAAAAAAiE/wROHzHXhDksb6JevDdD50Ipy4FKY8ozPgCLcBGAs/s1600/various-artists-captain-scarlet-and-the-mysterons-part-1-century-21.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="745" data-original-width="750" height="198" src="https://2.bp.blogspot.com/-qsBUjFaa8hE/Wws6mBCjSLI/AAAAAAAAAiE/wROHzHXhDksb6JevDdD50Ipy4FKY8ozPgCLcBGAs/s200/various-artists-captain-scarlet-and-the-mysterons-part-1-century-21.jpg" width="200" /> </a></div><br />We can solve the question 'is there enough flour' by looking at the structure of the problem, which means <i>thinking algebraically</i>. We can take a <i>narrative</i> approach, in this kind of way:<br /><blockquote class="tr_bq">If we add the result of the two weighings, 130 + 160, this is counting the weight of bag A twice, so the total amount of flour is 290 g minus the weight of bag A. So the smaller bag A happens to be, the more flour we've got, and if A weighs 40 g or less, we will have at least 250 g.</blockquote>Or we can express this with <i>symbolic algebra,</i> in this kind of way:<br /><blockquote class="tr_bq"><i>a + b + a + c</i> = 130 + 160 = 290,<br />so the total number of grams, <i>a + b + c</i> = 290 – <i>a</i>,<br />so if <i>a </i>≤ 40, <i>a + b + c </i>≥ 250.</blockquote>-<br /><b>TUESDAY</b>: The arithmagon quietly makes an appearance, but we stick with the Peter pancakes story for now. [But it's not pancakes for ever.]<br />We now know the <i>combined weight for each pair of bags</i> and we look at an approach for hence finding the total weight of the bags - and the weight of any one bag. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-xd2MSOEz8mI/WwxcQzZx_NI/AAAAAAAAAiQ/Ivu5PApeudQI6kgIhSzQMjHPcYJui0lMgCLcBGAs/s1600/ALG-16B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-xd2MSOEz8mI/WwxcQzZx_NI/AAAAAAAAAiQ/Ivu5PApeudQI6kgIhSzQMjHPcYJui0lMgCLcBGAs/s1600/ALG-16B.jpg" /></a></div>NOTE: @ilarrosac (Ignacio Larrosa) has made the interesting observation on Twitter that another context for this problem is to find the radii of three touching circles, given the sum of pairs of radii. A nice thing about this is we can transform the arithmagon triangle into a triangle whose sides are proportional to these sums, in this kind of way (below). This also throws light on the situation where the value of <i>a</i>, <i>b</i> or <i>c</i> turns out to be negative.... as we shall see.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-bxs87OuLyGo/Ww2PrRphObI/AAAAAAAAAik/U9xc1huGDLMu-ywRhHm-JsQVAq0NdoGbACLcBGAs/s1600/ALG-16B-circles.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-bxs87OuLyGo/Ww2PrRphObI/AAAAAAAAAik/U9xc1huGDLMu-ywRhHm-JsQVAq0NdoGbACLcBGAs/s1600/ALG-16B-circles.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>- <br /><b>WEDNESDAY: </b>We can relax. The arithmogadon is here.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-CMEsUdKn_Ok/Ww3KFTLMw0I/AAAAAAAAAiw/OujXUIdlKl0wH6LoJ0kCBnMuP1ClU1IvgCLcBGAs/s1600/ALG-16C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-CMEsUdKn_Ok/Ww3KFTLMw0I/AAAAAAAAAiw/OujXUIdlKl0wH6LoJ0kCBnMuP1ClU1IvgCLcBGAs/s1600/ALG-16C.jpg" /></a></div>We are given the same information as in Tuesday's task, namely the value of the sum of each pairing of the three unknown numbers <i>a</i>, <i>b</i> and <i>c</i>. On Tuesday, we found the value of <i>a + b + c</i> (which then allowed us to find the value of each individual unknown) by adding the three pair-sums and halving:<br /><i>a+b + b+c + c+a</i> = 2(<i>a + b + c</i>), so <i>a + b + c</i> = (130+160+150)÷2 = 220.<br />This time we use the fact that <i>c</i> must be 30 more than <i>b</i>, and that the sum of <i>b</i> and <i>c</i> is 150, to find <i>c</i> and hence to find <i>a + b + c</i>.<br />If we are working formally, we can express this information as <i>c = b</i> + 30 and <i>b + c</i> = 150, which we can then combine and transform in this kind of way: <i>c</i> – 30 + <i>c</i> = 150, so 2<i>c</i> = 180, so <i>c</i> = 90.<br />It is quite possible that these different approaches would have emerged informally as students investigated the original task, so we have now had a chance to make them more explicit. <br />-<br /><b>THURSDAY</b>: Here we look at that first scenario again, but in a more systematic way and without the 'real life' context.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-9oCBu3SFj7E/Ww8bxCBPSLI/AAAAAAAAAi8/W8O57AbRlIEWuWPKqjbBpJs1fKIXOzMUwCLcBGAs/s1600/ALG-16D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-9oCBu3SFj7E/Ww8bxCBPSLI/AAAAAAAAAi8/W8O57AbRlIEWuWPKqjbBpJs1fKIXOzMUwCLcBGAs/s1600/ALG-16D.jpg" /></a></div>We bring out the fact which may well have been noticed in ALG 16A, that the smaller the value of <i>a</i>, the greater the value of the sum <i>a + b + c</i>. [Formally, this can be said to stem from the fact that <i>a+b + a+c</i> = 130+160 = 290 = (<i>a+b+c</i>) + <i>a</i>. And as <i>a</i> increases by 1, the value of <i>a + b + c</i> decreases by 1.]<br />We also look at the limiting values of <i>a</i> for which the three 'vertices' (<i>a</i>, <i>b</i> and <i>c</i>) are all positive:<br /><blockquote class="tr_bq">when <i>a</i> = 0, <i>b</i> = 130 ans <i>c</i> = 160 and the sum S = <i>a + b + c </i>= 0 + 130+160 = 290;<br />when <i>a</i> = 130, <i>b</i> = 0 and <i>c</i> = 30, and S = 130 + 0 + 30 = 160;<br />note: when <i>a</i> > 130, then <i>b</i> is negative and <i>c</i> is too when <i>a</i> passes 160.</blockquote>What we haven't done here explicitly is look at the 'permissible' values of the three sums <i>a+b</i>, <i>b+c</i> and c<i>+a</i>, ie the value for which <i>a</i>, <i>b</i> and <i>c</i> are all positive. However, from the previous paragraph we can show that <i>b+c</i> will range from 30 to 290, with, of course, <i>a+b</i>=130 and <i>a+c</i>=160. In effect, the sum of the two smaller sums must be less than or equal to the larger sum - the same condition that applies to the lengths of the sides of a triangle.<br />- <br /><b>FRIDAY</b>: We switch from an additive to a multiplicative arithmagon. Does Armageddon loom?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-7_HXUyyJ6Ls/WxBl-J9UXBI/AAAAAAAAAjI/2zhrrx3m1UIOi-FBhdn1lhkKpjvyOQ70ACLcBGAs/s1600/ALG-16E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-7_HXUyyJ6Ls/WxBl-J9UXBI/AAAAAAAAAjI/2zhrrx3m1UIOi-FBhdn1lhkKpjvyOQ70ACLcBGAs/s1600/ALG-16E.jpg" /></a></div>Part a) can help us to get a feel for the relationships between the various elements. However, it is quite challenging to pin them down. In b), it is possible to solve part i. by spotting the common factors of <i>u, v, w</i> taken in pairs, though part ii. is likely to prompt one to use formal algebra.<br />We can solve b)ii using an approach analogous to Tuesday's or Wednesday's approach.<br />Tuesday's approach: <i>uvw</i> = <i>abbcca</i> = (<i>abc</i>)², so <i>a</i>² = <i>uvw÷v</i>² = <i>uw/v</i>.<br />Wednesday's approach: <i>u/v </i>=<i> ab/bc </i>=<i> a/c</i>; <i>wu/v </i>=<i> aca/c </i>=<i> a</i>².ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-47166101456577987792018-05-21T11:07:00.001+01:002018-05-25T23:33:36.820+01:00ALG 15We continue with Cartesian graphs for another week - sorry, bad planning! (But worth it, hopefully.)<br />In this week's tasks we foster a feel for graphs by 'adding' or 'subtracting' them. We begin with a visual approach, by focussing on the 'vertical' distance between two graphs - especially when this is zero, ie where the graphs intersect.<br />We go on to relate the visual to the symbolic by expressing the graphs symbolically and adding or subtracting the relations.<br /><b>MONDAY</b>: Here we have two linear functions, <i>f</i>(<i>x</i>) and <i>g</i>(<i>x</i>), though we don't know precisely what they are since the Cartesian axes have not been numbered. However, we can, for example, say that the graph of the new function <i>y</i> = <i>f</i>(<i>x</i>) – <i>g</i>(<i>x</i>) cuts the <i>y</i> axis at the same point as <i>f</i>(<i>x</i>) cuts the axis [Why?], and cuts the <i>x</i> axis directly below the point of intersection of the given graphs [Why?].<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-C4bsDVlmPtY/WwKZzv7xRfI/AAAAAAAAAgI/eGKW6WcleNU0EETIAR1MwYwf7LmfJq_-gCLcBGAs/s1600/ALG-15A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-C4bsDVlmPtY/WwKZzv7xRfI/AAAAAAAAAgI/eGKW6WcleNU0EETIAR1MwYwf7LmfJq_-gCLcBGAs/s1600/ALG-15A.jpg" /></a></div>It is interesting to note that <i>f</i>(<i>x</i>) is not as steep as g(<i>x</i>). What does this tell us about the slope of <i>y</i> = <i>f</i>(<i>x</i>) – <i>g</i>(<i>x</i>)?<br />-<br /><b>TUESDAY</b>: Here we are given some feedback on Monday's task and have the opportunity to consolidate our earlier ideas. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-bitG766g__w/WwNK65_94PI/AAAAAAAAAgU/DAs1wCbU9esOnaqmlP7pP-IGh2XREaCEQCLcBGAs/s1600/ALG-15B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-bitG766g__w/WwNK65_94PI/AAAAAAAAAgU/DAs1wCbU9esOnaqmlP7pP-IGh2XREaCEQCLcBGAs/s1600/ALG-15B.jpg" /></a></div>We are are also given one of the functions in symbolic form which allows us to determine the scale of the axes and hence to represent all the other functions symbolically. We can thus link the symbolic with our earlier visual/numerical/analytic approach.<br />We can use the symbolic representations in various ways. For example, knowing that <i>f</i>(<i>x</i>) = <i>x</i> + 10 and that <i>g</i>(<i>x</i>) = 2<i>x</i>, we can state that <i>y</i> = <i>f</i>(<i>x</i>) – <i>g</i>(<i>x</i>) = <i>x</i> + 10 – 2<i>x</i> which simplifies to <i>y</i> = 10 – <i>x</i>. We can then use this symbolisation to check whether our original sketch (ie the purple line) is correct, or, we could derive the symbolisation <i>from</i> the sketch, given that we now can see that the purple line goes through points with coordinates (0, 10), (10, 0) and (20, -10)<br />In the case of the function <i>y</i> = <i>f</i>(<i>x</i>) + <i>g</i>(<i>x</i>), we can determine that its line will have a gradient of 3, on the basis that we are adding lines with gradients of 1 and 2, or on the basis that its equation will contain the terms <i>x</i> and 2<i>x</i>, whose sum is 3<i>x</i>. <br />- <br /><b>WEDNESDAY</b>: Here we start with a very familiar straight line graph (of the function <i>y</i> = <i>x</i>) and 'perturb' it by adding a second, 'wilder' function. Interestingly, though, the effect of this second function is quite localised....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-WuGQf3Yffek/WwaxbVJDHJI/AAAAAAAAAg4/TTk4Kulnp387kUrIimbB-0q1Akf-OFzngCLcBGAs/s1600/ALG-15C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-WuGQf3Yffek/WwaxbVJDHJI/AAAAAAAAAg4/TTk4Kulnp387kUrIimbB-0q1Akf-OFzngCLcBGAs/s1600/ALG-15C.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>Note: I came across this lovely idea in Abraham Arcavi's 1994 <i>Symbol Sense</i> paper in <i>For the Learning of Mathematics</i>, 14, 3.<br />- <br /><b>THURSDAY</b>: Here we can check whether our sketch for Wednesday's task was on the right lines by comparing it to the red curve: this expresses the fact that the term 1/(<i>x</i> – 4) has a very large effect on the value of <i>y</i> when <i>x</i> is very close to 4, but its effect rapidly diminishes as we move away from <i>x</i> = 4. We then get a chance to build on this by sketching the graph of a closely related function. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-v0g5rdzPhwo/WwaxkwLSBCI/AAAAAAAAAg8/1MJ3-uYIR4kApr_aUl2f7ro2gncBmDgiwCLcBGAs/s1600/ALG-15D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-v0g5rdzPhwo/WwaxkwLSBCI/AAAAAAAAAg8/1MJ3-uYIR4kApr_aUl2f7ro2gncBmDgiwCLcBGAs/s1600/ALG-15D.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>The green curve below shows the graph for the new function. Since 2<i>x</i> – 8 = 2(<i>x</i> – 4), the term 1/(2<i>x</i> – 8) has only half the effect of the term 1/(<i>x</i> – 4).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-GDvUTOMUWiQ/Wwc1m9Y5J3I/AAAAAAAAAhM/ixg5pxsoDtY0Pq2BsLzjgsfyiZhYDEvUACLcBGAs/s1600/ALG-15D-ans.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="566" data-original-width="449" height="320" src="https://2.bp.blogspot.com/-GDvUTOMUWiQ/Wwc1m9Y5J3I/AAAAAAAAAhM/ixg5pxsoDtY0Pq2BsLzjgsfyiZhYDEvUACLcBGAs/s320/ALG-15D-ans.png" width="253" /></a></div><b>FRIDAY</b>: Here we subtract a linear function from a quadratic function.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-pM0cu-djlQs/Wwc199C9MVI/AAAAAAAAAhU/uXjeNvNz2G8bazXCiohWxm7_AQVZfiBTACLcBGAs/s1600/ALG-15E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-pM0cu-djlQs/Wwc199C9MVI/AAAAAAAAAhU/uXjeNvNz2G8bazXCiohWxm7_AQVZfiBTACLcBGAs/s1600/ALG-15E.jpg" /></a></div>The effect turns out to be surprisingly simple: just a translation of the curve (see below). Can we use algebra to find the translation (and to show the shape hasn't changed)?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-8GqA32n4dO8/WwiOWconOfI/AAAAAAAAAhs/LD1gHVoHPkkd4hc-FvZRWNMASdbyqiI_gCLcBGAs/s1600/ALG-15E-ans.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="616" data-original-width="379" height="320" src="https://4.bp.blogspot.com/-8GqA32n4dO8/WwiOWconOfI/AAAAAAAAAhs/LD1gHVoHPkkd4hc-FvZRWNMASdbyqiI_gCLcBGAs/s320/ALG-15E-ans.png" width="196" /></a></div>Interestingly, if one subtracts a similar linear function from a cubic, as below (blue – orange), the resulting cubic curve (grey) is not congruent, or even similar, to the original cubic curve.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-q5VyZOJF9MU/Wwc4WLyqYnI/AAAAAAAAAhg/njSnII7e5H8QgHJUahvpEK4KGtR8fUBFgCLcBGAs/s1600/ALG-15-cubic.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="651" data-original-width="422" height="400" src="https://3.bp.blogspot.com/-q5VyZOJF9MU/Wwc4WLyqYnI/AAAAAAAAAhg/njSnII7e5H8QgHJUahvpEK4KGtR8fUBFgCLcBGAs/s400/ALG-15-cubic.png" width="258" /></a></div><br /><b>Next week</b>: The <span style="font-family: "courier new" , "courier" , monospace;">Arithmagons</span> are coming: Captain Scarlet's bitter foes (or was that the <span style="font-family: "courier new" , "courier" , monospace;">Mysterons</span>?).<br /><br /><br />ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-32499275376388679862018-05-13T10:31:00.002+01:002018-05-18T09:20:40.479+01:00ALG 14This week's set of tasks was inspired by a task (below) from Doug French, that featured in his series <i>The creative use of odd moments</i> that appeared in <i>Mathematics in School</i>. (The jottings occurred during an interview with four high attaining Year 8 students, where we extended the task to find the equations of lines that formed other squares.)<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-uSAO5JeLxkY/Wvf_OzcdJjI/AAAAAAAAAew/yu6AqHfSEfYhzJP79DpPVaJW3SFL60KlgCLcBGAs/s1600/doug-square.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="689" data-original-width="718" height="383" src="https://2.bp.blogspot.com/-uSAO5JeLxkY/Wvf_OzcdJjI/AAAAAAAAAew/yu6AqHfSEfYhzJP79DpPVaJW3SFL60KlgCLcBGAs/s400/doug-square.png" width="400" /></a></div> In our set of tasks we focus on the equations of the lines on which the sides of a square lie, and we look at what happens to the equations when the square changes in simple ways - in particular, its position, size or orientation.<br />It is of course possible to find each line's equation by using the formal procedure of finding the values of <i>m</i> and <i>c</i> in <i>y</i> = <i>mx</i> + <i>c</i>. However, the way the tasks have been designed should encourage students to think about the relationship between the coordinates <i>x</i> and <i>y</i> of points on a given line. In turn, the focus on the coordinates, especially of points on the axes, should enhance students' understanding of the formal procedure and of the common forms in which equations are expressed (ie <i>y</i> = <i>mx</i> + <i>c</i>, but also <i>x</i> + <i>y</i> = <i>a, </i>when it applies, and perhaps more generally <i>ax</i> + <i>by</i> = <i>c</i>).<br /><b>MONDAY</b>: We start with a nice little diamond that is fairly near, but not too near, the axes...<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-UuEdWm9YuZg/WvjDhw-C4ZI/AAAAAAAAAfA/ZL-5YhO8mZED-HVNL0xq2Pb-WTphS4_7wCLcBGAs/s1600/ALG-14A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-UuEdWm9YuZg/WvjDhw-C4ZI/AAAAAAAAAfA/ZL-5YhO8mZED-HVNL0xq2Pb-WTphS4_7wCLcBGAs/s1600/ALG-14A.jpg" /></a></div>The lines have gradient 1 or -1, so we can find the equations by seeing where the various lines cut the <i>y</i>-axis (giving us the value of <i>c</i> in <i>y</i> = <i>mx</i> + <i>c</i>). However, we can also (or instead) work more 'locally', by looking for the relationship connecting the <i>x</i> and <i>y</i> coordinates - eg for points A (8, 8) and D (8–2, 8+2), we can see that <i>x</i> + y = 16. We might also notice that the line through D and C is 4 units 'above' the line through A and B (whose equation is <i>y</i> = <i>x</i>), ie the <i>y</i> coordinate has been increased by +4, for a given <i>x</i> coordinate, so <i>y</i> = <i>x</i> + 4 instead of <i>y</i> = <i>x</i>.<br />- <br /><b>TUESDAY</b>: We see what happens to the equations when we translate the square one unit up or across.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-JM9vn6jqui0/WvoBIIjQTDI/AAAAAAAAAfQ/nQPikzGRQw0YIAXw62MybyEMGt2Ad_XEACLcBGAs/s1600/ALG-14B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-JM9vn6jqui0/WvoBIIjQTDI/AAAAAAAAAfQ/nQPikzGRQw0YIAXw62MybyEMGt2Ad_XEACLcBGAs/s1600/ALG-14B.jpg" /></a></div>Here we provide both a check on Monday's task, and a chance to consolidate the ideas that arose, by varying the task in ways that lead to equations that are closely related to the previous ones.<br />For each line, we can again look for the (modified) relations between <i>x</i> and <i>y</i>, and/or visualise how the lines have moved and consider, in particular, where they cut the axes.<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">Notes</span>: Regarding the first approach, we can do this in an empirical way, by examining actual pairs of coordinates, or we can adopt a more general argument. For example, consider what happens to the equation of the line through AD, ie <i>x</i> + <i>y</i> = 16, when the square is moved one unit up. We can examine new coordinate pairs like (8, 9) and (6, 11), which suggests their sum has increased from 16 to 17; or we can argue more generally that for any given value of <i>x</i>, the value of <i>y</i> has increased by 1, so the sum <i>x</i> + <i>y</i> has increased by 1, from 16 to 17.<br />Visualising the lines can be very helpful, but sometimes care needs to be taken in drawing conclusions from the way a line has changed. For example, when the line with equation <i>x</i> + <i>y</i> = 16 is moved up one unit, it intersects each axis one unit further up or across, which can easily lead to the conclusion that the equation changes from <i>x</i> + <i>y</i> = 16 to <i>x</i> + <i>y</i> = 18, rather than to <i>x</i> + <i>y</i> = 17.<br />-<br /><b>WEDNESDAY</b>: Here we change the size of our little diamond. This doesn't change the task substantially, except that we express the new square's coordinates and the resulting equations in a (more) general form. [Note that when it comes to the equations, our new unknown, <i>e</i>, is a <i>parameter</i> rather than a variable.] We can of course check our new equations by letting <i>e</i> take the value 2.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-1nXoJVOEYqI/WvtlxwqehqI/AAAAAAAAAfg/7T-bqiqUHcY55LLKltyXVV-FbCpX7uATQCLcBGAs/s1600/ALG-14C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-1nXoJVOEYqI/WvtlxwqehqI/AAAAAAAAAfg/7T-bqiqUHcY55LLKltyXVV-FbCpX7uATQCLcBGAs/s1600/ALG-14C.jpg" /></a></div>The last part of the task throws up an interesting, if rather arcane, issue. Is the square resulting from the value <i>e</i> = -4 on and <i>above</i> the line <i>y</i> = <i>x</i>, or on and <i>below</i>? If we derive the new square from the equations that arose in the first part of the task, then the answer turns out to be '<i>below</i>'. Should we be bound by this, ie by the dictates of algebra?! Is this position for the square somehow more 'consistent' than placing it on and above the line <i>y</i> = <i>x</i> ?<br />-<br /><b>THURSDAY</b>: We consider a small square with a different orientation from the 'diamonds' we've considered so far - so no longer a little gem?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-xRLYLI_cgn0/Wvys7xUlflI/AAAAAAAAAfw/2HkgHTvLG107relBSays3DM4wpKEJrzwQCLcBGAs/s1600/ALG-14D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-xRLYLI_cgn0/Wvys7xUlflI/AAAAAAAAAfw/2HkgHTvLG107relBSays3DM4wpKEJrzwQCLcBGAs/s1600/ALG-14D.jpg" /></a></div>We're no longer dealing merely with slopes of 1 or -1. So a bit more of a challenge, whether we focus on pairs of coordinates and the relationship between them, or whether we consider where the lines cut the axes.<br />-<br /><b>FRIDAY</b>: We translate the square to a position far from the origin, so we are close to asking for general rules for our equations (which of course are already general rules...). <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-5wrq5RaAj4U/Wv3vz0UqtEI/AAAAAAAAAf8/vS0hCwF4Sxk36W_ed4r_bNKqXqRfH3dGACLcBGAs/s1600/ALG-14E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-5wrq5RaAj4U/Wv3vz0UqtEI/AAAAAAAAAf8/vS0hCwF4Sxk36W_ed4r_bNKqXqRfH3dGACLcBGAs/s1600/ALG-14E.jpg" /></a></div>x ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-56100127491952372382018-05-08T15:54:00.002+01:002018-05-11T11:29:53.248+01:00ALG 13This week we look at the relationship between pairs of values given in a table, and consider how the relationship changes as a result of systematic changes to the values. The new relationship can be found in a variety of ways: formally, by seeing the change in value as a transformation of one or both variables in the algebraic relation; informally, by giving meaning to the relationship and to how a set of numbers has been changed; empirically by searching for a new rule to fit the new set of values.<br />- <br /><b>MONDAY</b>: We start with a standard linear relationship and multiply the <i>x</i>- and/or the <i>y</i>-values by 4.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-gdKGFCj-qIQ/WvG0HlijlSI/AAAAAAAAAdo/qT2tuGjB3OMjVgOLF0dRrp8LSsEkaZlkQCLcBGAs/s1600/ALG-13A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-gdKGFCj-qIQ/WvG0HlijlSI/AAAAAAAAAdo/qT2tuGjB3OMjVgOLF0dRrp8LSsEkaZlkQCLcBGAs/s1600/ALG-13A.jpg" /></a></div>Some students might be able to approach this formally. For example, in the case of Table B, we can think of the 'new' <i>x</i> (denote it by <i>x</i>', say) as replacing 4 times the old <i>x</i>, ie <i>x</i>' = 4<i>x</i>, and so <i>y</i> = 8<i>x</i> + 1 = 2<span style="font-family: "times new roman"; font-size: 11.0pt;">×</span><style><!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:Calibri; panose-1:2 15 5 2 2 2 4 3 2 4; mso-font-charset:0; mso-generic-font-family:auto; mso-font-pitch:variable; mso-font-signature:-536870145 1073786111 1 0 415 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-unhide:no; mso-style-qformat:yes; mso-style-parent:""; margin:0mm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} .MsoChpDefault {mso-style-type:export-only; mso-default-props:yes; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} @page WordSection1 {size:612.0pt 792.0pt; margin:72.0pt 72.0pt 72.0pt 72.0pt; mso-header-margin:36.0pt; mso-footer-margin:36.0pt; mso-paper-source:0;} div.WordSection1 {page:WordSection1;} --></style><style><!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:Calibri; panose-1:2 15 5 2 2 2 4 3 2 4; mso-font-charset:0; mso-generic-font-family:auto; mso-font-pitch:variable; mso-font-signature:-536870145 1073786111 1 0 415 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-unhide:no; mso-style-qformat:yes; mso-style-parent:""; margin:0mm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} .MsoChpDefault {mso-style-type:export-only; mso-default-props:yes; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} @page WordSection1 {size:612.0pt 792.0pt; margin:72.0pt 72.0pt 72.0pt 72.0pt; mso-header-margin:36.0pt; mso-footer-margin:36.0pt; mso-paper-source:0;} div.WordSection1 {page:WordSection1;} --></style>4<i>x</i> + 1 becomes <i>y</i> = 2<i>x</i>' + 1.<br />Or, less formally, "the new <i>x</i>-values are 4 times the old <i>x</i>-values, so multiplying the old <i>x</i>-values by 8 is the same as multiplying the new <i>x</i> values by 2; so <i>y</i> = 8<i>x</i> + 1 becomes <i>y</i> = 2<i>x</i> + 1".<br />Or students might notice, by examining the numbers in Table B, that 17 = 2<span style="font-family: "times new roman"; font-size: 11.0pt;">×</span>8 + 1, 41 = 2<span style="font-family: "times new roman"; font-size: 11.0pt;">×</span><span style="font-family: "times new roman"; font-size: 12.0pt;"></span>20 + 1, etc.<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">IMPORTANT NOTE</span>: For brevity we have presented the three tables, B, C and D, on one slide. In class, it might be better to present the tables one at a time so that students can devote plenty of time to each rather than feeling they have to rush from one table to the next. <br />- <br /><b>TUESDAY</b>: Here, we start with the same relationship as in ALG 13A, but instead of multiplying by 4, we add 4 to the <i>x</i>- and/or the <i>y</i>-values.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-x4z4YdlNStY/WvG0Pqcp2_I/AAAAAAAAAds/RUuF8i1vJfk9CLZuWj7Y_4It54YoIM5-gCLcBGAs/s1600/ALG-13B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-x4z4YdlNStY/WvG0Pqcp2_I/AAAAAAAAAds/RUuF8i1vJfk9CLZuWj7Y_4It54YoIM5-gCLcBGAs/s1600/ALG-13B.jpg" /></a></div>Again, students might adopt a formal, a semi-formal, or an empirical approach. In the case of Table E, the relation changes to <i>y</i> = 8(<i>x</i> – 4) + 1, or <i>y</i> = 8<i>x</i> – 31.<br />-<br /><b>WEDNESDAY</b>: Here we start with a new linear relationship that is perhaps easier to describe succinctly.... Again, we multiply the <i>x</i>- and/or the <i>y</i>-values by 4.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-bt6Zg_HWQ5k/WvIY2ALEahI/AAAAAAAAAeA/PGS7YErb0CkgChNwUsePQq1T0PSfCsalgCLcBGAs/s1600/ALG-13C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-bt6Zg_HWQ5k/WvIY2ALEahI/AAAAAAAAAeA/PGS7YErb0CkgChNwUsePQq1T0PSfCsalgCLcBGAs/s1600/ALG-13C.jpg" /></a></div>The relationship for Table J is easy to spot or derive, the one for Table I less so.<br />-<br /><b>THURSDAY</b>: We stick with Wednesday's relationship, and as we did earlier, we add 4 to the <i>x</i>- and/or the <i>y</i>-values. Finding the resulting relationships is not particularly challenging on this occasion but we include the task for completeness.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-J2dFvDJUJJc/WvNdHVR4ASI/AAAAAAAAAeQ/7f4sN-opU1EOnrr7a2iYQ8fK6sqY6tHhwCLcBGAs/s1600/ALG-13D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-J2dFvDJUJJc/WvNdHVR4ASI/AAAAAAAAAeQ/7f4sN-opU1EOnrr7a2iYQ8fK6sqY6tHhwCLcBGAs/s1600/ALG-13D.jpg" /></a></div><b>FRIDAY</b>: We end the week with a task involving a much more challenging function. We provide four transformations for completeness, but, as suggested previously, it might be better to show the tables one at a time to prevent students feeling they have to rush from one to the next.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-cgX-vjFUVR8/WvTInamrV6I/AAAAAAAAAeg/5WItdgySnAkOslpiZJ7cdTQv5Gc0fmW2ACLcBGAs/s1600/ALG-13E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-cgX-vjFUVR8/WvTInamrV6I/AAAAAAAAAeg/5WItdgySnAkOslpiZJ7cdTQv5Gc0fmW2ACLcBGAs/s1600/ALG-13E.jpg" /></a></div><b>Note</b>: In all of this week's task we have looked at relationships expressed symbolically and with tables of values. It is also worth drawing graphs of the relationships.ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-6362449758569130242018-04-29T22:07:00.000+01:002018-05-07T13:13:23.408+01:00ALG 12This week we look at Cartesian graphs of straight lines again. The aim is to focus on the visual aspects of the graphs, in particular the <i>slopes</i> of various lines, and to find points on the lines by using notions of proportion and similarity, rather than by immediately expressing the lines as equations (though that would, of course, be a perfectly legitimate thing to do).<br />In many of the tasks we have deliberately chosen points with very different sets of coordinates [such as (0, 0) and (100, 101) in ALG 12A] so as to encourage students to visualise or sketch what's going on, rather than attempt to use accurate drawings. <br />- <br /><b>MONDAY</b>: Many students should be able to find coordinates for <i>some</i> points on the red line, eg (50, 50.5) and (200, 202). Encourage students to seek out lots of points as some might provide insights that would help with parts i. and ii., eg (10, 10.1) and (1, 1.01).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-wizHilcp_yM/WuYtwRFKxxI/AAAAAAAAAbY/Pruv2MOZcAEIwQdmiEPuuIRJQY6DLYi8gCLcBGAs/s1600/ALG-12A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-wizHilcp_yM/WuYtwRFKxxI/AAAAAAAAAbY/Pruv2MOZcAEIwQdmiEPuuIRJQY6DLYi8gCLcBGAs/s1600/ALG-12A.jpg" /></a></div>It is likely that some students will suggest that the value of the <i>y</i>-coordinate in part i. is 102. How can we show (eg with the aid of a sketch) that a point with coordinates (101, 102) is <i>not</i> on the red line? How can we decide whether it is above or below?<br />-<br /><b>TUESDAY</b>: We search for some more widely-spaced points.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-_R1pGJVBMYg/WueK4vpQGfI/AAAAAAAAAbo/tGOmZyakbec56_mUDDf4z6aDBIToTtn7wCLcBGAs/s1600/ALG-12B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-_R1pGJVBMYg/WueK4vpQGfI/AAAAAAAAAbo/tGOmZyakbec56_mUDDf4z6aDBIToTtn7wCLcBGAs/s1600/ALG-12B.jpg" /></a></div>A sketch might help; a table of values might help, to record moves towards the <i>y</i>-axis, say, in well-chosen steps (eg 10 to the left, 1 up).<br />PS: Here's a nice solution, posted on Twitter. Quite condensed but worth unpicking!<br /><a href="https://1.bp.blogspot.com/-_HavgY_VmWY/WujsMBwuRmI/AAAAAAAAAcE/JOGQzhCBudUcM6CvJsNtgagXNJfe3vtdACLcBGAs/s1600/12B-sol.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="146" data-original-width="344" height="168" src="https://1.bp.blogspot.com/-_HavgY_VmWY/WujsMBwuRmI/AAAAAAAAAcE/JOGQzhCBudUcM6CvJsNtgagXNJfe3vtdACLcBGAs/s400/12B-sol.png" width="400" /></a><br />-<br /><b>WEDNESDAY</b>: More points on a line, near and far...<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-LYcUeK_XhZM/WujqXvllA-I/AAAAAAAAAb4/LG_QRpUA294Xc7ggPcQfd4176XxXvvzkgCLcBGAs/s1600/ALG-12C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-LYcUeK_XhZM/WujqXvllA-I/AAAAAAAAAb4/LG_QRpUA294Xc7ggPcQfd4176XxXvvzkgCLcBGAs/s1600/ALG-12C.jpg" /></a></div>Follow the light green straight line.... Not that different from the previous task, but no harm in consolidating earlier ideas. And you might like to think up further variations....<br />-<br /><b>THURSDAY</b>: Two lines, by way of a change.....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-dkxbZ2l9QPs/Wuo5nLW5tDI/AAAAAAAAAcU/svIirOCPkHEBA60Ui5DJdCB5lWS2hoHWACLcBGAs/s1600/ALG-12D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-dkxbZ2l9QPs/Wuo5nLW5tDI/AAAAAAAAAcU/svIirOCPkHEBA60Ui5DJdCB5lWS2hoHWACLcBGAs/s1600/ALG-12D.jpg" /></a></div>Here we take a somewhat different approach by offering two methods for students to interpret. The first is quite grounded, the second is more abstract. You might, of course, prefer to ask students to come up with their own methods first, though the task itself is quite demanding.<br />-<br /><b>FRIDAY</b>: By way of a further change, we home in on the microscopic - though you might want to zoom out again ....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-U7GTA-r5KZc/WuuTxP6sDQI/AAAAAAAAAck/LDjzShTj0y0GtMiF_QhZJpCzyHFwLNEWgCLcBGAs/s1600/ALG-12E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-U7GTA-r5KZc/WuuTxP6sDQI/AAAAAAAAAck/LDjzShTj0y0GtMiF_QhZJpCzyHFwLNEWgCLcBGAs/s1600/ALG-12E.jpg" /></a></div>Don't be put off by the fractions! The underlying algebraic relationships are not that difficult.... And there are also some nice geometric approaches that can be used to solve the task.<br /><b>Commentary</b>: A closer look at the coordinates reveals that for one of the line segments the coordinates add up to 1, for the other the <i>y</i>-coordinate is 3 times the <i>x</i>-coordinate. Which two numbers add up to 1, where one number is 3 times the other? We can get there fairly rapidly using trial and improvement. Or we could draw a 'bar' or 'rod' or number line (although this requires quite a difficult switch in thinking - the 'bar' is <i>not</i> a line segment in the Cartesian plane!):<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-aXNuHtL8TOs/WvA0GahEElI/AAAAAAAAAc0/_8XjAk9sWDgoe6iDFQsX2JLK19CIR_wfACLcBGAs/s1600/quart-3quart.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="256" data-original-width="810" height="126" src="https://3.bp.blogspot.com/-aXNuHtL8TOs/WvA0GahEElI/AAAAAAAAAc0/_8XjAk9sWDgoe6iDFQsX2JLK19CIR_wfACLcBGAs/s400/quart-3quart.png" width="400" /></a></div><br />We can also express the relations between the coordinates more formally, as the simultaneous equations <i>x</i> + <i>y</i> = 1 and <i>y</i> = 3<i>x</i>, leading to <i>x</i> + 3<i>x</i> = 1, so <i>x</i> = 1/4, <i>y</i> = 3/4.<br />- <br />Here's a nice approach (by Matt L) that is more geometric. It involves constructing a new point on the line through the steeper line segment that is vertically below the left hand end-point of the other line segment. It also uses the fact that this line segment slopes at 45˚, so the two segments labelled '<i>a</i>' are equal.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-XjMFIyzKZuk/WvA1bh_io_I/AAAAAAAAAdA/NLfCoLpJJ2gsUhyeK1_8LozWzg67fxDpQCLcBGAs/s1600/matt-alg12e-BIT.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="480" data-original-width="362" height="320" src="https://4.bp.blogspot.com/-XjMFIyzKZuk/WvA1bh_io_I/AAAAAAAAAdA/NLfCoLpJJ2gsUhyeK1_8LozWzg67fxDpQCLcBGAs/s320/matt-alg12e-BIT.png" width="241" /></a></div><br />We can extend the two given line segments still further, until they touch a unit square. [In fact, this is what I started with when I designed the task!]<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-rnNCDXE8llA/WvA4Qaup1CI/AAAAAAAAAdM/ZEn6cr1wk0UYAUaqPtJq44Eq-E7g0R7wwCLcBGAs/s1600/ALG-12E-square.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="391" data-original-width="513" height="304" src="https://1.bp.blogspot.com/-rnNCDXE8llA/WvA4Qaup1CI/AAAAAAAAAdM/ZEn6cr1wk0UYAUaqPtJq44Eq-E7g0R7wwCLcBGAs/s400/ALG-12E-square.jpg" width="400" /></a></div>A couple of years ago there was a flurry of tasks on Twitter involving <i>dissected squares</i> of this sort, usually with one region shaded, the aim being to find 'What fraction of the square is shaded?' My favoured approach was to draw equally spaced parallel lines to produce <i>equal intercepts</i>, as here:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-4d1MhgVsq0I/WvA6CLh28KI/AAAAAAAAAdY/FKukcHHPNpQJwbVx2Ugng0UfWXOo0TG-QCLcBGAs/s1600/ALG-12E-square2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="391" data-original-width="513" height="303" src="https://2.bp.blogspot.com/-4d1MhgVsq0I/WvA6CLh28KI/AAAAAAAAAdY/FKukcHHPNpQJwbVx2Ugng0UfWXOo0TG-QCLcBGAs/s400/ALG-12E-square2.jpg" width="400" /></a></div>One could then argue along these lines: the diagonal of the square has been cut into 4 equal segments, so starting at the bottom left corner, say, of the square, point P is 1/4 units across and 3/4 units up. [Or the height of the triangle seated on the base of the square is 3/4 so its area is 3/8 of the unit square.]<br />A related and perhaps more direct approach (whether to locate P or to find areas) is to make use of the fact that this triangle on the base of the square is similar to the small 'upside down' triangle above it, with a base that is 3 times as long.<br />I remember someone coming up with an algebraic solution to one of these area tasks on Twitter. This took me by surprise and I was impressed by the power of such an approach. But it also seemed disappointing to 'abandon' the geometry! Of course, one might say the reverse about the ALG 12E task. Here an algebraic approach would seem an obvious choice and it is perhaps a bit perverse to go to the lengths of constructing a geometric approach instead!ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-58377030041956115792018-04-22T23:42:00.000+01:002018-04-26T22:30:09.444+01:00ALG 11<b>THIS WEEK</b>: We explore the phenomenon of <i>Letter as Object</i>, something we met briefly in ALG 9C, ie we look at contexts where a letter represents a pure number (of objects) or a quantity (the price or length or mass or some other numerical <i>quality</i> of an object), but where there might be a temptation to use the letter as a shorthand for the object itself (as in <i>'a</i> stands for apple, <i>b</i> stands for banana' - the classic <i>fruit salad algebra</i>). We will see that sometimes one can slide harmlessly between letter as object and letter as quantity, but that sometimes it leads to a completely fake algebra.<br /><b>MONDAY</b>: In this pair of tasks <i>m</i> stands for a numerical quality (mass) in the first task and for a pure number in the second. Though the answer is 5<i>m</i> in both cases, we have found that the second kind of task is substantially more demanding than the first - does that hold for your students?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-RneQ12RKq5A/Wt0JjkbsTxI/AAAAAAAAAZE/ItV6FsCpI_sUrIyT4SN5aHenbmtnCnxSgCLcBGAs/s1600/ALG-11A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-RneQ12RKq5A/Wt0JjkbsTxI/AAAAAAAAAZE/ItV6FsCpI_sUrIyT4SN5aHenbmtnCnxSgCLcBGAs/s1600/ALG-11A.jpg" /></a></div>In the first task, it is easy to interpret <i>m</i> as simply standing for <i>mints </i>rather than the mass of a mint, but still come up with the correct expression, 5<i>m</i>, albeit misinterpreted as 5 <i>mints</i>. In the second task students have to cope with the idea that <i>m</i> is most definitely a number, but one whose value we don't know, so that it is not possible to arrive at a specific numerical answer.<br /><b>TUESDAY</b>: Careful! Needles and pins. <i>Because of all my pride, the tears I gotta hide...</i> This turns out to be quite a Searching task, though let's hope it doesn't quite lead to tears!<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-GcpejgK1fSc/Wt5a1KTzxTI/AAAAAAAAAZU/Q-pXotIOv10toHQJQ02xumDYfsAB1ORYACLcBGAs/s1600/ALG-11B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-GcpejgK1fSc/Wt5a1KTzxTI/AAAAAAAAAZU/Q-pXotIOv10toHQJQ02xumDYfsAB1ORYACLcBGAs/s1600/ALG-11B.jpg" /></a></div>The task involves an expression in which the letters stand for numbers but where the temptation to treat them as objects is very strong (for us as well as our students) and where this leads to a complete misinterpretation of the expression. <i> </i><br /><i>Why can't I stop<br />And tell myself I'm wrong<br />I'm wrong, so wrong<br />Why can't I stand up<br />And tell myself I'm strong</i><br /><i>....</i><br />The extract below is from my article <i>Object lessons in algebra?</i> that appeared in <i>Mathematics Teaching</i> 98 in 1982. Still worth reading!<br /><div style="text-align: center;"><a href="https://4.bp.blogspot.com/-nPgAo2_TBKY/Wt7z4tAlU2I/AAAAAAAAAZk/hauIiEWlqns-gU4pTFyqOz21WlEW-HUXgCLcBGAs/s1600/cabbages.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="435" data-original-width="624" height="278" src="https://4.bp.blogspot.com/-nPgAo2_TBKY/Wt7z4tAlU2I/AAAAAAAAAZk/hauIiEWlqns-gU4pTFyqOz21WlEW-HUXgCLcBGAs/s400/cabbages.png" width="400" /></a> </div>Here is another variant of the task, but where we are asked to write a relation rather than interpret an expression. It is again <span style="color: #0000ee;">quite</span> challenging.<b><br /></b><br /><div class="separator" style="clear: both; text-align: center;"><b><a href="https://2.bp.blogspot.com/-ONIyZDvhgco/WuD4TGd0atI/AAAAAAAAAa4/KvIbW3GJkJcSlEvlTCgVzEa80xeyRxqkgCLcBGAs/s1600/ALG-11B-alt.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://2.bp.blogspot.com/-ONIyZDvhgco/WuD4TGd0atI/AAAAAAAAAa4/KvIbW3GJkJcSlEvlTCgVzEa80xeyRxqkgCLcBGAs/s400/ALG-11B-alt.jpg" width="400" /></a></b></div><br /><b>WEDNESDAY</b>: In this task, the misinterpretation that results from treating the letters as objects is not as jarring... The given expression stands for the <i>cost</i> (in number of pence) of 5 bananas and 2 coconuts, so it is wrong to simply translate it as <i>5 bananas and 2 coconuts</i>. Nevertheless this translation does make some kind of sense - the story is about 5 bananas and 2 coconuts, whereas the corresponding interpretation of the expression 2<i>p</i> + 5<i>n</i> in ALG 11B, as <i>2 pins and 5 needles</i>, does not fit that story at all.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-GAEbA8Lz3_8/Wt-4pxnzGYI/AAAAAAAAAZ0/cnPTvSVec08ScsTvADxxc6veEU7fkrZqACLcBGAs/s1600/ALG-11C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-GAEbA8Lz3_8/Wt-4pxnzGYI/AAAAAAAAAZ0/cnPTvSVec08ScsTvADxxc6veEU7fkrZqACLcBGAs/s1600/ALG-11C.jpg" /></a></div>It is interesting to consider the alternative version of task 11C shown below.<i><br /></i><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-CfRLUOVhaXE/Wt--jwbOCKI/AAAAAAAAAaE/x4QK8dTHJiUNyGCyJlb856ADaYItZxVkQCLcBGAs/s1600/ALG-11C-alt.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://1.bp.blogspot.com/-CfRLUOVhaXE/Wt--jwbOCKI/AAAAAAAAAaE/x4QK8dTHJiUNyGCyJlb856ADaYItZxVkQCLcBGAs/s400/ALG-11C-alt.jpg" width="400" /></a></div>Here students are quite likely to come up with the right expression<i>,</i> though not necessarily for the right reason - we can't be sure whether students who write 5<i>b</i> + 2<i>c</i> fully realise that this represents a number of pence and that it is not simply telling us about the number of bananas and coconuts bought. So this alternative version is not as useful as the original for revealing students' thinking. The task below, which appeared recently on Twitter, is even less effective as a diagnostic tool: here the correct option (B, I assume) can be chosen by simply treating it as an abbreviation of the verbal statements, ie by reading 5<i>t</i> + 2<i>c</i> = 3.70 as <i>5 teas and 2 coffees cost 3.70</i> (pounds), rather than appreciating that 5<i>t</i> + 2<i>c</i> actually represents the number of £s spent on the drinks.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-0OW56QZEYQU/Wt_B6HmNTFI/AAAAAAAAAaQ/Sxn25JCcoNUxcq0SFMn3ZED6DOibyXkqACLcBGAs/s1600/barton-coffee-tea-crop.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="387" data-original-width="523" height="295" src="https://2.bp.blogspot.com/-0OW56QZEYQU/Wt_B6HmNTFI/AAAAAAAAAaQ/Sxn25JCcoNUxcq0SFMn3ZED6DOibyXkqACLcBGAs/s400/barton-coffee-tea-crop.png" width="400" /></a></div><b>THURSDAY</b>: We present the first of two tasks involving sets of coloured rods, which we ask students to symbolise in different ways.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-JogOsW7zy60/WuDz5hopFwI/AAAAAAAAAas/f_C8o5WMdJ0QPELKwU5el5oRQWLvVmWsQCLcBGAs/s1600/ALG-11D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-JogOsW7zy60/WuDz5hopFwI/AAAAAAAAAas/f_C8o5WMdJ0QPELKwU5el5oRQWLvVmWsQCLcBGAs/s1600/ALG-11D.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>In a sense, this task is quite easy - we can get it right by interpreting the letters correctly, leading to 3<i>g</i> = <i>b</i> (on the basis that 3 times the length <i>g</i> equals the length <i>b</i>), but we can also get it right by writing 3<i>g</i> = <i>b</i> as shorthand for <i>3 green rods make a blue rod</i>, where <i>g</i> and <i>b</i> are perhaps being thought of merely as shortened <i>names</i> for the green and blue rods, and not necessarily as symbols for their lengths. Thus, in a task like this, students can appear to be thinking algebraically when perhaps they're not.<br />-<br /><b>FRIDAY</b>: Here we meet the same rows of green rods and blue rods, but this time we relate the number of green rods and blue rods, not their lengths.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-aY6L52_6B_g/WuJBGCyiSXI/AAAAAAAAAbI/aOkfmGfumzYMmgyyW8MtgLYsGr-SBoPNwCLcBGAs/s1600/ALG-11E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-aY6L52_6B_g/WuJBGCyiSXI/AAAAAAAAAbI/aOkfmGfumzYMmgyyW8MtgLYsGr-SBoPNwCLcBGAs/s1600/ALG-11E.jpg" /></a></div>In this task, we can feel a strong pull towards writing 3<i>g</i> = <i>b</i> again, on the basis that <i>3 green rods make a blue rod</i>, or <i>there are 3 green rods for every blue rod</i>. However, we need to fix firmly on the fact that in this version of the rods task, <i>g</i> and <i>b</i> are defined as <i>numbers of rods</i>, for example 3 and 1, or 6 and 2, or 9 and 3, etc. So <i>g</i> is 3 times <i>b</i>, ie <i>g</i> = 3<i>b</i>. This can feel counter-intuitive as it doesn't map onto our verbal descriptions of the situation.<br /><i></i>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-7322817712360039722018-04-15T19:55:00.000+01:002018-04-20T21:00:19.307+01:00ALG 10This week we describe various situations using symbolic algebra. We then 'play' with the algebra by choosing values that take us beyond the immediate situation: can we relate the algebra back to the situation? Does the situation still make sense?<br />The situations we've chosen are fairly straightforward, but the game we're playing is mathematically quite sophisticated. It's similar to starting with a familiar statement like 8 + <i>x</i> = 10, which works in natural numbers, and asking what happens in a case like 8 + <i>x</i> = 5: this can be made to work if we stretch our ideas about number, ie if invent new ones - the integers.<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">Note</span>: From an RME perspective, we could say that we are engaged in horizontal mathematisation (expressing a 'real' situation mathematically) and then in vertical mathematisation (developing the maths). <br />- <br /><b>MONDAY</b>: Here we play with an area formula, for a shape that can vary in size. We start with a straightforward application of the formula but then consider a case which only works if we allow (or invent) edges with negative lengths.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-UuaY1YDHkVQ/WtOdoOTrkFI/AAAAAAAAAWU/3Nj-3YhbZ7AAoEUs1zW5FhNYkzAqSMs-ACLcBGAs/s1600/Algrabya150-10A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-UuaY1YDHkVQ/WtOdoOTrkFI/AAAAAAAAAWU/3Nj-3YhbZ7AAoEUs1zW5FhNYkzAqSMs-ACLcBGAs/s1600/Algrabya150-10A.jpg" /></a></div>Some students might feel that negative lengths are simply <i>not allowed</i>. That is a perfectly defensible position, but it would restrict the mathematics that we are able to do. A simple response is to say we are going to enter (or invent) a new (mathematical) world where negative lengths <i>are allowed</i>. So there!<br />This is what the shapes in parts i and ii look like (if you allow negative lengths in part ii):<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-cBF4QcTiN5k/WtTDwma5IbI/AAAAAAAAAWk/0kjnv2rHXJMZM454JTTIWiFd9RPlrC-kgCLcBGAs/s1600/Algrabya150-10A-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://3.bp.blogspot.com/-cBF4QcTiN5k/WtTDwma5IbI/AAAAAAAAAWk/0kjnv2rHXJMZM454JTTIWiFd9RPlrC-kgCLcBGAs/s400/Algrabya150-10A-ans.jpg" width="400" /></a></div> -<br /><b>TUESDAY</b>: We look at another familiar area scenario, namely area of a trapezium (and, to keep it simple, a trapezium that is right-angled).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-L3MBR9svQKg/WtULi9UnleI/AAAAAAAAAW0/kCnXw58_3Z48kQeFuscvUKzsZe3v8rtbACLcBGAs/s1600/Algrabya150-10B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-L3MBR9svQKg/WtULi9UnleI/AAAAAAAAAW0/kCnXw58_3Z48kQeFuscvUKzsZe3v8rtbACLcBGAs/s1600/Algrabya150-10B.jpg" /></a></div>Here we need to accept the idea of a negative length again, but also the idea of a negative area (which we could sidestep or leave implicit in ALG 10A).<br />One way to find the area geometrically is to divide the trapezium into two triangles. For part i we can, for example, divide the trapezium into two triangles of area 30 and 15 square units (top-left diagram, below).<br />The top-right diagram shows what happens to the trapezium as point P moves until it is 5 units to the <i>left</i> of the formerly top-left vertex. The trapezium 'twists' over itself to form two triangles whose areas, we can argue, are 20 and -5 square units.<br />The two diagrams at the bottom of the slide, below, show an alternative interpretation for part ii. The yellow triangle corresponds to the 30 square units triangle in the top-left diagram. The green triangle corresponds to the 15 square units triangle in the same diagram, except its base has changed from 5 units to -5 units. If we 'cancel' the region where the yellow and green triangles overlap, we are left with the regions with area 20 and -5 square untis shown in the top-right diagram.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-V_84tkuMr6M/WtZOoIGnW0I/AAAAAAAAAXE/UYy8Xy9PSbgjBnHh3WQDWuuGyk_QsXotQCLcBGAs/s1600/Algrabya150-10B-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://1.bp.blogspot.com/-V_84tkuMr6M/WtZOoIGnW0I/AAAAAAAAAXE/UYy8Xy9PSbgjBnHh3WQDWuuGyk_QsXotQCLcBGAs/s400/Algrabya150-10B-ans.jpg" width="400" /> </a></div>- <br /><div class="separator" style="clear: both; text-align: left;"><b>WEDNESDAY</b>: Things get really interesting.... What's a 2-and-a-half sided regular polygon?<i>!</i> </div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-MRIOoO9utjQ/WtZaOOc0s6I/AAAAAAAAAXk/Mol0vqMo3GESTiehX7YZ2tqPOLXqWL0-wCLcBGAs/s1600/Algrabya150-10C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-MRIOoO9utjQ/WtZaOOc0s6I/AAAAAAAAAXk/Mol0vqMo3GESTiehX7YZ2tqPOLXqWL0-wCLcBGAs/s1600/Algrabya150-10C.jpg" /></a></div>I came across this beautiful idea, of replacing the whole number of sides, <i>n</i>, with a fraction, in one of David Fielker's articles in <i>Mathematics Teaching</i>, many years ago. For me, the idea is almost on a par with inventing negative or fractional indices. A simple but brilliant mathematical act!<br />You may recognise the form of the instructions if you are familiar with LOGO and Turtle Geometry. If you don't have a turtle to hand I hope you will have enacted the instructions yourself and traced the resulting paths on paper or in your head! This is what they turn out to look like:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-uSsWfBA9PdQ/Wteln6-zVvI/AAAAAAAAAX0/ojGcYSiU484cPJTAjJ7N-rNeES7COkktwCLcBGAs/s1600/Algrabya150-10C-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://3.bp.blogspot.com/-uSsWfBA9PdQ/Wteln6-zVvI/AAAAAAAAAX0/ojGcYSiU484cPJTAjJ7N-rNeES7COkktwCLcBGAs/s400/Algrabya150-10C-ans.jpg" width="400" /></a></div>- <br /><b>THURSDAY</b>: We again make the shift from whole numbers to fractions, this time on a familiar number grid.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-rpLEiiLO73U/WtemBJk96GI/AAAAAAAAAX4/lf49sJn5cQ8GF5Ix8jxsJJyA7_7z3e2XgCLcBGAs/s1600/Algrabya150-10D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-rpLEiiLO73U/WtemBJk96GI/AAAAAAAAAX4/lf49sJn5cQ8GF5Ix8jxsJJyA7_7z3e2XgCLcBGAs/s1600/Algrabya150-10D.jpg" /></a></div> It can be useful to consider what happens to the sum, <i>S</i>, when the T-shape moves across the grid (eg 1 square to the right, or 1 square up). We can think about this spatially (What happens to each of the numbers in the T-shape?) or algebraically (What happens to <i>S</i> when <i>n</i> in the expression 6<i>n</i>+120 is increased by 1, say, or by 10?). In the case of part i, <i>S</i> has increased by 240–210 = 30, which can be achieved by moving the T-shape 5 squares to the right... [Are there other ways?]<br />Here are positions for the T-shape for parts i and ii.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-17wTp3NvZI8/Wtj7t_RpbvI/AAAAAAAAAYM/JcuWpv2u9uYkyMl0U08gSAkK_uv7JdrogCLcBGAs/s1600/Algrabya150-10D-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://1.bp.blogspot.com/-17wTp3NvZI8/Wtj7t_RpbvI/AAAAAAAAAYM/JcuWpv2u9uYkyMl0U08gSAkK_uv7JdrogCLcBGAs/s400/Algrabya150-10D-ans.jpg" width="400" /></a></div><span style="font-family: "arial" , "helvetica" , sans-serif;">Note</span>: if we accept the principle behind the part ii answer, of allowing <i>fractions of a square</i>, we can find infinitely many positions for the T-shape, for parts i and ii, by moving the shape vertically (maybe just a tiny bit) as well as horizontally. What we're doing here, in effect, is to change the discrete 2-D grid into a continuous Cartesian plane.<br />- <br /><b>FRIDAY</b>: Here we consider square grids made of matchsticks - or parts of matchsticks.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-Nvy8eDwffYY/Wtj8Twma0RI/AAAAAAAAAYU/8cQ5DFD6ECQHtRShgMOR2CR52-ii7E9LQCLcBGAs/s1600/Algrabya150-10E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-Nvy8eDwffYY/Wtj8Twma0RI/AAAAAAAAAYU/8cQ5DFD6ECQHtRShgMOR2CR52-ii7E9LQCLcBGAs/s1600/Algrabya150-10E.jpg" /></a></div>I'm particularly fond of this pattern - it's one whose structure is fairly easy to discern <i>generically</i>, even though the relation between the dimension of the square and the number of matchsticks is <i>quadratic</i> rather than linear.<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">Note</span>: The slide below shows some interesting attempts to structure the grid by three Year 7 students (from a 'low attaining' set: set 3 of 4). I've written this up in chapter 3 of the Proof Materials Project <a href="http://www.mathsmedicine.co.uk/ioe-proof/PMPintro2.html">report</a>, <i>Looking for Structure</i>.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-wnp5rFUlfC8/Wtn-q1QP7II/AAAAAAAAAYk/gvBS-Q7IGWU5rUWdLNX1EE8l2KP3_735wCLcBGAs/s1600/y7sticks.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="408" data-original-width="849" height="305" src="https://4.bp.blogspot.com/-wnp5rFUlfC8/Wtn-q1QP7II/AAAAAAAAAYk/gvBS-Q7IGWU5rUWdLNX1EE8l2KP3_735wCLcBGAs/s640/y7sticks.png" width="640" /></a></div>Here's a solution to ALG 10E (below). The last part is, of course, the most interesting. Using an expression for the number of matchsticks for an <i>n</i> by <i>n</i> grid, we get 31.5 sticks for a 3.5 by 3.5 grid. We've constructed a drawing for the grid that fits that total by allowing fractional matchsticks - though it's up to you whether you are willing to accept this! The drawing consists of 24 whole sticks, 8 sticks split in half 'cross ways', another 6 sticks split in half length ways, and two quarter sticks (resulting from being split in half cross ways and length ways). This makes 24 + 4 + 3 + 0.5 sticks = 31.5 sticks. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-bLBa3Vhaa4c/WtpAXmWlkvI/AAAAAAAAAY0/MNFH_QDrkrYk-8Xo9Z-mS3NBHluD0wRqQCLcBGAs/s1600/Algrabya150-10E-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-bLBa3Vhaa4c/WtpAXmWlkvI/AAAAAAAAAY0/MNFH_QDrkrYk-8Xo9Z-mS3NBHluD0wRqQCLcBGAs/s1600/Algrabya150-10E-ans.jpg" /></a></div> -<br /><b>NEXT WEEK</b>: We revisit the phenomenon of <i>Letter as Object</i>, ie we look at contexts where a letter represents a pure number (of objects) or a quantity (the price or length or mass or some other numerical <i>quality</i> of an object), but where there might be a temptation to use the letter as a shorthand for the object itself (as in <i>'a</i> stands for apple, <i>b</i> stands for banana' - the classic <i>fruit salad algebra</i>). We will see that sometimes one can slide harmlessly between letter as object and letter as quantity, but that sometimes it leads to a completely fake algebra.ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-71792221130308100772018-04-08T23:47:00.000+01:002018-04-12T23:32:07.894+01:00ALG 9As it's <i>not quite</i> the summer term yet (some schools go back on Monday, some on Monday week), the theme this week is <i>not quite</i> algebra. We are going to look at some tasks from well-known (or well-publicised) textbooks that involve faux, fake, phony, manqué, mock, pretend, pseudo, quasi, cod algebra.<br /><b>MONDAY</b>: a peach from Singapore (or several peaches, if you can afford them).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-MFtfHjadYi4/WsqdY0-u5_I/AAAAAAAAAVE/02OUg8eEa9kZTjWBUbcq53T0W1r7EqnpACLcBGAs/s1600/Algrabya150-9A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-MFtfHjadYi4/WsqdY0-u5_I/AAAAAAAAAVE/02OUg8eEa9kZTjWBUbcq53T0W1r7EqnpACLcBGAs/s1600/Algrabya150-9A.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>What is going on here? Does the story make sense?<br />-<br /><b>TUESDAY</b>: We're still in Singapore, with a task where <i>x</i> is not so much an unknown as an utter mystery. What on earth might it stand for?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-Lyg8T4GLGoo/WsvkPN-hChI/AAAAAAAAAVU/Z0lFyCK8oA89NvQgzOjgIctdAnv88mNHwCLcBGAs/s1600/Algrabya150-9B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-Lyg8T4GLGoo/WsvkPN-hChI/AAAAAAAAAVU/Z0lFyCK8oA89NvQgzOjgIctdAnv88mNHwCLcBGAs/s1600/Algrabya150-9B.jpg" /></a></div>Mei Heng has unusual powers. The longer she works, the faster she works. However, if she works for less that 3 hour 40 minutes she seems to destroy rather than make T-shirts. Perhaps the Singapore government has created a mechanism that discourages part-time employment...<br />What of her pay, I wonder. Is it per hour or per T-shirt?<br /><span style="font-family: "arial" , "helvetica" , sans-serif;">NOTE</span>: I've written about this and a few other tasks from the same textbook in the journal <i>Mathematics in School</i> (November 2013, Volume 42, Issue 5, p25).<br />-<br /><b>WEDNESDAY</b>: The context here involves 'grids' made of rods joined by a variety of links. We are shown two specific grids and for each we are presented with a 'rule' stating how many rods and various links it consists of.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-q9zSAC9I2wg/Ws1Bqf8g6MI/AAAAAAAAAVk/8EGBCkhuB1sUS8s8fzMDDziW4mwNh4WsgCLcBGAs/s1600/Algrabya150-9C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-q9zSAC9I2wg/Ws1Bqf8g6MI/AAAAAAAAAVk/8EGBCkhuB1sUS8s8fzMDDziW4mwNh4WsgCLcBGAs/s1600/Algrabya150-9C.jpg" /></a></div>The 'rules' look 'algebraic' in that they contain letters. However, they are not general statements, nor do they involve any unknowns. And the letters that appear in the 'rules' don't stand for numbers. Instead, what we have here is the equivalent of <i>fruit salad algebra</i>, or what I have dubbed elsewhere as <i>letter as object</i>.<br />-<br /><b>THURSDAY</b>: Here we look at a task from an earlier edition of Wednesday's UK textbook series. The exercise is designed purely as a device for practising algebraic manipulation. But what does it convey to students about the purpose and utility of algebra - or, indeed, of geometry? <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-qGJOIU87d5o/Ws6FP9L8msI/AAAAAAAAAV0/nYSKUmbFYUcIMCClua4L1Ykl82EBAX_nwCLcBGAs/s1600/Algrabya150-9D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-qGJOIU87d5o/Ws6FP9L8msI/AAAAAAAAAV0/nYSKUmbFYUcIMCClua4L1Ykl82EBAX_nwCLcBGAs/s1600/Algrabya150-9D.jpg" /></a></div>It turns out that, treated with any kind of common sense, shape <b>e</b> collapses into nothing. Curiously, exercises of this sort, which reduce algebra to an exercise in manipulation, and which abuse geometry by treating it merely as a means to this end, are commonplace in UK textbooks.<br />-<br /><b>FRIDAY</b>: Finally, we consider a task from a recent UK adaptation of a Singapore textbook, which describes itself as <i>The Mastery Course for Key Stage 3</i>.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-yRxz-vzZJYA/Ws_Y6KuUHoI/AAAAAAAAAWE/S51s9sbwkRQuOCRLqyLXiSJ1FzhC9XHWgCLcBGAs/s1600/Algrabya150-9E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-yRxz-vzZJYA/Ws_Y6KuUHoI/AAAAAAAAAWE/S51s9sbwkRQuOCRLqyLXiSJ1FzhC9XHWgCLcBGAs/s1600/Algrabya150-9E.jpg" /></a></div>It helps if the algebra tasks we give students demonstrate the <i>purpose and utility</i> of algebra (to use a phrase coined by Janet Ainley and Dave Pratt). However, this is not always easy to bring about; in the case of this task, its absurd nature suggests that the exact opposite has been achieved!<br />-<br /><b>NEXT WEEK</b>: algebra takes control and we try to make sense of the consequences...ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-53850774196629855412018-03-25T23:46:00.002+01:002018-03-29T16:42:10.595+01:00ALG 8This week's set of tasks use a mapping diagram to represent liner functions. This model may not be familiar to some teachers and students but this can be advantage - it allows us to take a fresh look at the characteristics of linear functions and (though we don't do it here) can help us look at the Cartesian graph of linear functions with fresh eyes.<br /><b>MONDAY</b>: Here we are asked to read a mapping diagram and to determine the function that the given information represents. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-nKF8wWslT5Y/Wrgjznj0cVI/AAAAAAAAAS0/FkSQx-fpsVUuliBYjN7IQkdrN9CJrZuCwCLcBGAs/s1600/Algrabya150-8A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-nKF8wWslT5Y/Wrgjznj0cVI/AAAAAAAAAS0/FkSQx-fpsVUuliBYjN7IQkdrN9CJrZuCwCLcBGAs/s1600/Algrabya150-8A.jpg" /></a></div><b>TUESDAY</b>: Here we take a more detailed look at Monday's mapping diagram (which was for the function <i>y</i> = 3<i>x</i> – 2) and consider how its features can be used to identify the linear function. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-8LA__W1ZKLc/WroV27PI27I/AAAAAAAAATE/uP1XbhvT2yM0hatJZcTagqGYKY0iUDU9ACLcBGAs/s1600/Algrabya150-8B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-8LA__W1ZKLc/WroV27PI27I/AAAAAAAAATE/uP1XbhvT2yM0hatJZcTagqGYKY0iUDU9ACLcBGAs/s1600/Algrabya150-8B.jpg" /></a></div><b>WEDNESDAY</b>: I was amazed by this phenomenon when I first came across it (thanks to JH). Of course, as with many mathematical phenomena, the more you think about it the more obvious it becomes. But this one also becomes richer. All in all, an exciting and lovely result!<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-O2l57aFZraY/WrqM_oHXWmI/AAAAAAAAATU/M8Z0awoOKnQWBrYxNIol86xKzjiZ8C4LwCLcBGAs/s1600/Algrabya150-8C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-O2l57aFZraY/WrqM_oHXWmI/AAAAAAAAATU/M8Z0awoOKnQWBrYxNIol86xKzjiZ8C4LwCLcBGAs/s1600/Algrabya150-8C.jpg" /></a></div>We can visualise parts i and ii by translating point P <i>and</i> the arrows one unit to the right (part i) or one unit down (part ii). Alternatively, move both axes one unit to the left (part i) or one unit up (part ii). Which approach do you prefer?<br /><b>PS</b>: You will probably have noticed that P can be thought of as the <i>centre of enlargement </i>for the linear function. Image points (ie points on the <i>y</i> axis) are 3 times the distance from P as object points (ie the corresponding points on the <i>x</i> axis). So the scale factor of the 'enlargement' is ×3. Similarly, the distance between a pair of points on the <i>y</i> axis is 3 times the distance of corresponding points on the <i>x</i> axis.<br />In part i, the same distance-relationships hold, ie the scale factor is still ×3. However, in part ii the distance-relationships change, and therefore so does the scale factor. Has it got smaller or larger?<br /><b>THURSDAY</b> <b>1</b>: Here we extend the notion that the arrows on mapping diagrams of linear functions emanate from a 'centre of enlargement', by considering centres that lie between the two axes. What does that tell us about the 'scale factor', ie the value of <i>m</i> in <i>y = mx + c</i> ?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-CNPH_dh0E4g/WrwHlOz3EGI/AAAAAAAAATk/cxuQ8u9QMh4ncfDjuyWDur9rFb-l7DDGwCLcBGAs/s1600/Algrabya150-8D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-CNPH_dh0E4g/WrwHlOz3EGI/AAAAAAAAATk/cxuQ8u9QMh4ncfDjuyWDur9rFb-l7DDGwCLcBGAs/s1600/Algrabya150-8D.jpg" /></a></div> -<br /><b>THURSDAY 2</b> (instead of Good Friday): Here we consider the inverse of our original function.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-DElAe9QohUw/Wr0Doxp6fFI/AAAAAAAAAT0/TZBHB3V445MwyMTb9Jg-Vr6yHVqafFPUQCLcBGAs/s1600/Algrabya150-8E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-DElAe9QohUw/Wr0Doxp6fFI/AAAAAAAAAT0/TZBHB3V445MwyMTb9Jg-Vr6yHVqafFPUQCLcBGAs/s1600/Algrabya150-8E.jpg" /> </a></div><div class="separator" style="clear: both; text-align: left;">The inverse takes us back to where we started. One way of representing this there-and-back journey is by adding a second mapping diagram that is a reflection of the original, as here:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-aZXP7IunGq0/Wr0DtE3nf4I/AAAAAAAAAT4/RqZRZpHMo_UNfSa6vL3iGhP_8cWfMHnawCLcBGAs/s1600/Algrabya150-8E-flip.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="353" data-original-width="570" src="https://4.bp.blogspot.com/-aZXP7IunGq0/Wr0DtE3nf4I/AAAAAAAAAT4/RqZRZpHMo_UNfSa6vL3iGhP_8cWfMHnawCLcBGAs/s1600/Algrabya150-8E-flip.jpg" /></a></div><b>Note</b>: the inverse function itself is represented by just the second mapping diagram, but with the axes re-labeled (<i>x</i> axis at the top, <i>y</i> axis below), and with the arrows pointing downwards. As can be seen, the 'common point' (or 'centre of enlargement') is vertically below P, two units below the <i>y</i> axis. The scale factor here is 1/3, and if we look carefully, we can see that 0 is mapped onto 2/3, so the inverse function is <i>y</i> = 1/3 <i>x</i> + 2/3, or <i>y</i> = 1/3 (<i>x</i> + 2). We can verify this algebraically, by transforming the original function <i>y</i> = 3<i>x</i> – 2 into <i>x</i> = .... (and then transposing <i>x</i> and <i>y</i>).<br /><br />ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-30363788415916639932018-03-18T11:43:00.003+00:002018-03-30T12:08:29.520+01:00ALG 7This week we look at two tasks of classic form, both of which are rather contrived but which are also quite engaging if simply treated as a puzzles. The aim is to show the power of using a symbolic algebra approach, although in the case of the first task, below, we have deliberately devised a task which is also amenable to informal methods.<br />- <br /><div class="separator" style="clear: both; text-align: left;"><b>MONDAY</b>: This first task is of the classic 'How many cows and chickens' form, although we have tried to make it a bit more challenging by involving three rather than just two sets of 'elements', in this case, bikes, trikes and quadracycles.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-96Z6_z11ecI/Wq5GlRvB_1I/AAAAAAAAARk/YmmmD4dy0OUzocLZ_iDNhAIpcH0Ez785ACLcBGAs/s1600/Algrabya150-7A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-96Z6_z11ecI/Wq5GlRvB_1I/AAAAAAAAARk/YmmmD4dy0OUzocLZ_iDNhAIpcH0Ez785ACLcBGAs/s1600/Algrabya150-7A.jpg" /></a></div>As we spell-out in Tuesday's variant, this <i>Cycles</i> task can be solved algebraically by creating equations such as<br />T = Q + 5<br />31 = B + T + Q<br />76 = 2B + 3T + 4Q.<br />We can then substitute Q for T. Also, if we double the second equation, giving 64 = 2B + 2T + 2Q, we can eliminate B by subtracting this equation from the third equation, leaving us with 14 = T+ 2Q. <br />In Tuesday's variant, we also indicate that the task can be solved informally with this sort of argument:<br />Imagine that the cycles are all bikes;<br />we would then have 64 tyres (or wheels) which is 14 fewer than the actual number;<br />so we would need to replace some of the bikes with trikes and/or quads;<br />each trike would increase the number of tyres by 1, each quad would increase the number of tyres by 2;<br />so .... <br />There is a clear link between this narrative approach and the process that gave us the equation 14 = T + 2Q. Of course, students might not see this link immediately, though Freudenthal (in his <i>China Lectures</i> published in 1991) suggests that students can achieve this by solving a range of similar tasks in this informal way and generalising.<br />There is an interesting tension here. One of the strengths of a symbolic approach is that it 'frees' us from thinking about the context, once we have expressed the given information symbolically, and assuming we are fluent in manipulating the symbols.<br />On the other hand, the act of keeping hold of the context, or of referring back to it periodically, can help us (or more to the point, our students) to form the symbolisations and to check that these, and the subsequent manipulations, are valid.<br />- <br /><b>TUESDAY</b>: Here we present two methods for solving yesterday's <i>Cylces</i> task, one informal the other involving symbolic algebra, which we ask students to complete. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-FVz9oD2u9p8/WrBGcP758vI/AAAAAAAAAR0/oWBu2OFSHA8iXHqHObWiTz9GfO4QeOQXACLcBGAs/s1600/Algrabya150-7B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-FVz9oD2u9p8/WrBGcP758vI/AAAAAAAAAR0/oWBu2OFSHA8iXHqHObWiTz9GfO4QeOQXACLcBGAs/s1600/Algrabya150-7B.jpg" /></a></div>How students respond to these methods will depend on their knowledge and experience. Some students might gain a better appreciation of the power and efficiency of symbolic algebra, while others might gain some understanding of the algebraic approach by linking it to the more concrete, informal approach.<br />-<br /><b>WEDNESDAY</b>: Here we present two new tasks, both set in the context of 'sharing acorns' but where the given information is structured differently. We consider how amenable the tasks are to informal and formal approaches in the light of this difference in structure.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-DpffZOQ1fwY/WrGSJJxMWcI/AAAAAAAAASE/RtOsf5ut30UIxpzuc2dBJuJkY8oUDngQACLcBGAs/s1600/Algrabya150-7C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-DpffZOQ1fwY/WrGSJJxMWcI/AAAAAAAAASE/RtOsf5ut30UIxpzuc2dBJuJkY8oUDngQACLcBGAs/s1600/Algrabya150-7C.jpg" /></a></div>The first task can be solved in informal and formal ways that are very similar: the collection of 60 acorns can be partitioned into 1 plus 4 plus 5 equal shares, ie into 10 equal shares altogether; algebraically, we can construct something like this: <i>g</i> + 4<i>g</i> + 5<i>g</i> = 60, so 10<i>g</i> = 60.<br />The second task <i>can</i> be solved informally but it probably requires quite a lot of insight to do this successfully. On the other hand, it can be solved in a fairly routine way using symbolic algebra, assuming the solver has adequate technical skills.<br />Thus, we would argue that the second task is that elusive entity: a task (albeit a mere pointless puzzle!) where <i>the use of symbolic algebra comes into its own</i>. Of course, in the first task, we could also strengthen the need to record and symbolise, by increasing the number of recipients and by relating their shares to George's share in more varied ways.<br /><b>Note 1</b>: You may have noticed that the methods of sharing in the two tasks turn out to be equivalent. The second task has been made more complex by using what Dettori et al (2006) call 'circular references'. I came across this work in the excellent revue by Mason & Sutherland (2002), <a href="https://www.researchgate.net/profile/John_Mason3/publication/42788782_Key_aspects_of_teaching_algebra_in_schools/links/53f1f5750cf272810e4c7a23.pdf"><i>Key Aspects of Teaching Algebra in Schools</i></a>. You might like to devise similar tasks to challenge your students, by taking a straightforward sharing task and then using circular references to recast the description of some of the shares.<br /><b>Note 2</b>: The 'circular references' task used by Dettori (and reproduced in Mason & Sutherland) is this:<br /><a href="https://3.bp.blogspot.com/-G9eVEtEOOf0/Wr4WEXQd4RI/AAAAAAAAAUM/uERU8a7PMV0EAsqGAPKFkMc6VQSIIV70wCLcBGAs/s1600/dettori-fish.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="239" data-original-width="906" height="168" src="https://3.bp.blogspot.com/-G9eVEtEOOf0/Wr4WEXQd4RI/AAAAAAAAAUM/uERU8a7PMV0EAsqGAPKFkMc6VQSIIV70wCLcBGAs/s640/dettori-fish.png" width="640" /></a>Interestingly, an almost identical task occurs in an English translation (1822, p204) of a French translation of Leonhard Euler's <i>Vollständige Anleitung Zur Algebra</i> (1771). <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-pwgNMP0_dhw/Wr4XQFWPFKI/AAAAAAAAAUU/He5NMXygKB081dPItsKniRQVHRkT8NQpwCLcBGAs/s1600/euler-de-lalgrange-fish.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="112" data-original-width="665" height="107" src="https://4.bp.blogspot.com/-pwgNMP0_dhw/Wr4XQFWPFKI/AAAAAAAAAUU/He5NMXygKB081dPItsKniRQVHRkT8NQpwCLcBGAs/s640/euler-de-lalgrange-fish.png" width="640" /></a></div>The English book (below) contains 'Additions of M. De La Grange' and it is possible that the problem stems from him rather than Euler, as I can't find it in the original German edition.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-SXb9wpTL7Nc/Wr4ZqA8iMRI/AAAAAAAAAUo/LCqi91yJ6gk9mBrVdKhayc6mTlDb1_yugCLcBGAs/s1600/euler-elements-front.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="860" data-original-width="490" height="640" src="https://1.bp.blogspot.com/-SXb9wpTL7Nc/Wr4ZqA8iMRI/AAAAAAAAAUo/LCqi91yJ6gk9mBrVdKhayc6mTlDb1_yugCLcBGAs/s640/euler-elements-front.png" width="363" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><b>THURSDAY</b>: Here we introduce Auntie's way of sharing acorns. It is not equivalent to Grandpa's method, but it turns out to produce the same result for a particular number of acorns.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-DhE6gwoYFk0/WrLXMn7dR1I/AAAAAAAAASU/GXwCwIs1ui0_Ml0cB8w0mOaqkyocv8q7QCLcBGAs/s1600/Algrabya150-7D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-DhE6gwoYFk0/WrLXMn7dR1I/AAAAAAAAASU/GXwCwIs1ui0_Ml0cB8w0mOaqkyocv8q7QCLcBGAs/s1600/Algrabya150-7D.jpg" /></a></div>We can find the particular number of acorns using symbolic algebra, by solving the equation resulting from equating Peppa's shares (ie 4<i>g</i> and 2<i>g</i> + 20) or Chloe's shares (5<i>g</i> and 2<i>g</i> + 30) under the two ways of sharing. Both give the value <i>g</i> = 10, so that the total number of acorns is 100.<br />We have obviously carefully contrived the two sharing methods for them to produce the same outcome for all three recipients for a particular number (100) of acorns. This wouldn't happen by chance. However it raises an interesting (although rather subtle) question: given that the choice of Grandpa's or Auntie's sharing method makes no difference for 100 acorns, does it make a difference, for any individual recipient, for other numbers of acorns?<br /><b>FRIDAY</b>: Having established that Grandpa's and Auntie's method happen to produce identical shares when there are 100 acorns, today we consider what happens when there are more than 100 acorns: for each individual recipient, is one sharing method more favourable than the other?<br /><b>Note</b>: This task is quite complex and is probably more suitable as a challenge for teachers than as an activity in class. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-qk0IZlKHm0c/WrQ-d_o6V6I/AAAAAAAAASk/B1dh3vCEES42ov2tersvFiq30DLJiMfeQCLcBGAs/s1600/Algrabya150-7E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-qk0IZlKHm0c/WrQ-d_o6V6I/AAAAAAAAASk/B1dh3vCEES42ov2tersvFiq30DLJiMfeQCLcBGAs/s1600/Algrabya150-7E.jpg" /></a></div>In the case of George, he always gets one tenth of the acorns when Grandpa shares them out. This is not a large proportion, but still larger than what he gets from Auntie when there are less than 100 acorns. In the extreme case, when there are only 50 acorns, George gets none! George is better off with Auntie when there are more than 100 acorns.<br />The situation with Peppa turns out to be rather surprising. When there are 100 acorns, George's share is 10 acorns by either sharing method and she gets 40 acorns by either sharing method (ie 4×10 or 2×10 + 20). Now, when Georges share (<i>g</i>) is greater than 10, 4<i>g</i> will be greater than 2<i>g</i> + 20, which suggests that Grandpa's method will be more favourable than Auntie's when there are more than 100 acorns. However, it turns out that this is not the case! Why? Because Peppa's share depends on George's (as it's a function of <i>g</i>), but George's share differs depending on whether Grandpa or Auntie are sharing out the acorns. It turns out that Peppa's share comes to the same amount, whether it's Grandpa or Auntie doing the sharing, regardless of the number of acorns being shared out! For exmple, for 50 acorns she gets 20 under both methods, for 110 acorns she gets 44 under both methods. We leave it to the reader to verify/prove this.<br />For Chloe, it looks like Grandpa's method is more favourable than Auntie's when there are more than 100 acorns, as 5<i>g</i> > 2<i>g</i> + 30 when <i>g</i> > 10. But can we be sure? In the case of Grandpa's method, Chloe's share is 5/10 or 1/2 of the total number of acorns. In the case of Auntie's method, Chloe's share is 2/5 of the total plus 10 (you might like to verify this). When the total is more than 100, 1/2 of the total is greater than 2/5 of the total plus 10.<br /><br />ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-28657217223322757722018-03-11T22:58:00.003+00:002018-03-16T11:46:10.547+00:00ALG 6In this week's set of tasks we look at figurative patterns and express their structure in generic terms by using a <i>quasi variable</i> (in this case, 20). We look at equivalent ways of construing/expressing the structure.<br />Some of the week's patterns turn out to have a linear structure and some a quadratic structure - which raises the question, <i> </i><br /><i>How can we recognise this in the pattern itself, as well as in its symbolic representation</i>?<br />- <br /><b>MONDAY</b>: Here is the week's root task:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-257J4l3K0A4/WqWzh9BdcsI/AAAAAAAAAP8/mlMB4Io8i2UPFSXH44n4XPtoP5RcqMDsgCLcBGAs/s1600/Algrabya150-6A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-257J4l3K0A4/WqWzh9BdcsI/AAAAAAAAAP8/mlMB4Io8i2UPFSXH44n4XPtoP5RcqMDsgCLcBGAs/s1600/Algrabya150-6A.jpg" /></a></div><b>Note</b>: We have chosen 20 here (and in later tasks) on the basis that it is big enough to deter most students from simply counting each individual dot. Also, if students can find an efficient method for counting a 20-chain, it is likely that they can do so for any length of chain, perhaps even a chain of <i>n</i> Y-shapes. So, for many students, 20 is taking on the role of a variable.<br /><b>Note 2</b>: Another reason for going for such a 'far generalisation' early on, is to encourage students to look at the structure of the 20-chain as a whole. Of course, it is still possible to use an 'incremental' (or scalar or term-to-term) approach by noting that the 2-chain has 4 more dots than the 1-chain (and so the 20-chain has 19×4 more dots than the 1-chain). It would be interesting to see how the task would work without showing the 2-chain. We decided to put it in to avoid students thinking that there might be a hidden dot underneath the dot where the Ys join (in which case a 2-chain would simply have 2×5 dots and a 20-chain would simply have 20×5 dots).<br /><b>Note 3</b>: The presence of the 2-chain does open up other interesting possibilities: some students might argue that the 20-chain will have 10 times as many dots as the 2-chain; others might simply focus on the given numerical information, 1→5 and 2→9, and look for a mapping-rule that fits both, which can then be applied to 20.<br />-<br /><b>TUESDAY</b>: Here we are given two expressions for finding the number of dots in the 20-chain. The expressions are deliberately left open so as to show how the chain of Y-shapes can be structured. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-mTVDgwMsub4/Wqb71Mvq_vI/AAAAAAAAAQM/9M2KhdjnF-EB1xUu8x1p-BF1poO7DEMHQCLcBGAs/s1600/Algrabya150-6B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-mTVDgwMsub4/Wqb71Mvq_vI/AAAAAAAAAQM/9M2KhdjnF-EB1xUu8x1p-BF1poO7DEMHQCLcBGAs/s1600/Algrabya150-6B.jpg" /></a></div>It may be helpful to make an annotated sketch, as in the examples below, when describing each structure (ie when explaining Alma's and Bojan's methods).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-nN0PBVnDYKc/WqutYT4G8DI/AAAAAAAAARM/un9-W7efovsTlL8Py3JDjYUq8rqNYE10ACLcBGAs/s1600/Algrabya150-6B-structures.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://2.bp.blogspot.com/-nN0PBVnDYKc/WqutYT4G8DI/AAAAAAAAARM/un9-W7efovsTlL8Py3JDjYUq8rqNYE10ACLcBGAs/s400/Algrabya150-6B-structures.jpg" width="400" /></a></div>The given expressions are of course equivalent and you might want to challenge students to show how one expression can be transformed into the other. It is possible to structure the pattern in other (equivalent) ways, resulting in other (equivalent) expressions 'in 20'. How many can you find?<br />-<br /><b>WEDNESDAY</b>: Here we consider two further chains of Y-shapes, but this time the size of the Ys increases as the number of Ys in the chain increases. What does this do to the relationship between the number of Ys and the number of dots in a chain? Is it still linear? Can you tell from the pattern? Can you tell from the generic expression for the total number of dots?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-MNMkZhr1fDM/Wqgw-e8a9DI/AAAAAAAAAQc/qqt_d6k2o94fSx2qYdXfw7q0hTQWTDnYgCLcBGAs/s1600/Algrabya150-6C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-MNMkZhr1fDM/Wqgw-e8a9DI/AAAAAAAAAQc/qqt_d6k2o94fSx2qYdXfw7q0hTQWTDnYgCLcBGAs/s1600/Algrabya150-6C.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;"><b> Note</b>: Here we deliberately consider a <i>far generalisation</i> again, rather than ask for the number of dots in the 5-chain, say. The hope is that this will focus students' attention on the overall structure of the pattern rather than on the change from one chain to the next. Put another way, we are trying to see whether students will consider the pattern <i>generically</i>. Of course, it is perfectly legitimate to look at differences, ie to consider the pattern 'incrementally', but it's not particularly efficient here.</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;"><b>THURSDAY</b>: Here we look at some more dot patterns, but this time the Y-chains have morphed into single trees.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-QLUOTBuScm4/WqmiSAcEzJI/AAAAAAAAAQs/qBGkzPJKrYQEJmXXn9jxWXvnKtlx1U9ZgCLcBGAs/s1600/Algrabya150-6D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-QLUOTBuScm4/WqmiSAcEzJI/AAAAAAAAAQs/qBGkzPJKrYQEJmXXn9jxWXvnKtlx1U9ZgCLcBGAs/s1600/Algrabya150-6D.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">Yesterday, the relation between chain-length and the total number of dots turned out to be quadratic, in both cases. What about today? Can you tell just by looking at the pattern? Do the generic expressions for the 20-brown-dots trees confirm this?</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;"><b>FRIDAY</b>: Here we consider equivalent generic expressions for the number of dots in Fynn's 20-brown-dots tree. We try to relate them to the dot pattern - the first one is relatively straightforward, the second less so. We then try to transform one expression into the other.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-PD5Kiwkj7W8/WqsLY-7gA_I/AAAAAAAAAQ8/QbImIfe170g6ua06CZltdSpf-C50ix3ugCLcBGAs/s1600/Algrabya150-6E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-PD5Kiwkj7W8/WqsLY-7gA_I/AAAAAAAAAQ8/QbImIfe170g6ua06CZltdSpf-C50ix3ugCLcBGAs/s1600/Algrabya150-6E.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">We can show the structure contained in the expressions with annotated sketches like these:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-4WDJ_a0-hIY/Wqut0MirAEI/AAAAAAAAARQ/J9XHxcMzyeoHzyAxt6ynnH-fiS6znlH1QCLcBGAs/s1600/Algrabya150-6E-structures.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://4.bp.blogspot.com/-4WDJ_a0-hIY/Wqut0MirAEI/AAAAAAAAARQ/J9XHxcMzyeoHzyAxt6ynnH-fiS6znlH1QCLcBGAs/s400/Algrabya150-6E-structures.jpg" width="400" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">-</div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-80465657175364962412018-03-04T23:55:00.001+00:002018-03-09T18:23:40.662+00:00ALG 5In this week's set of ALG 5 tasks, we find specific unknown values by using the so-called bar model (curved bars in our case...) and by forming and equating algebraic expressions - and we look at alternative ways of forming the expressions.<br /><b>MONDAY</b>: The root ALG 5 task. A sketch might help or you might find another way to visualise or represent the situation - or you could use trial and improvement. You can view a dynamic version of the queue <a href="https://www.youtube.com/watch?v=6FwQhK6H85I">here</a>.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-a7_ueZwQC-M/Wp1ngM0fmeI/AAAAAAAAAN8/NozZUzh8jDcqtRev0MBk9gI2ia_ghquLACLcBGAs/s1600/Algrabya150-5A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-a7_ueZwQC-M/Wp1ngM0fmeI/AAAAAAAAAN8/NozZUzh8jDcqtRev0MBk9gI2ia_ghquLACLcBGAs/s1600/Algrabya150-5A.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div>Note: there's plenty of scope for varying the task.<br />Here's an easy variant:<br /><i>When has Deka queued for 11 times as long as Eric?</i><br /><i>- </i><br />Note 2: The bar model (or in this case, 'bent rod model') is a powerful device for visualising quantities, but what if you decide to make a sketch and the proportions in your drawing turn out to be not that good? That's probably OK, as long as you don't expect instant answers from the model and are prepared to slow down and modify the drawing (on paper or in your head).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-HggbXLV7rCg/Wp71M1rbGBI/AAAAAAAAAOc/Sg9BLBS1AWQG857c7M4ME-16QRsZcy-tQCLcBGAs/s1600/ALG5-baaaaa.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="297" data-original-width="1600" height="118" src="https://2.bp.blogspot.com/-HggbXLV7rCg/Wp71M1rbGBI/AAAAAAAAAOc/Sg9BLBS1AWQG857c7M4ME-16QRsZcy-tQCLcBGAs/s640/ALG5-baaaaa.png" width="640" /></a></div>-<br /><b>TUESDAY</b>: Here we ask for a symbolic algebra approach. Constructing the expressions can be quite challenging, though we can check to see whether they make sense by making use of the values found on Monday, ie by checking to see whether we get a value of 30 when <i>e</i> = 10. Then, having got the expressions, we can consider how they could themselves be used to derive the value <i>e</i> = 10, eg by forming an equation and solving it in some way ....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-NarKCpPE5TE/Wp3dLpboAKI/AAAAAAAAAOM/OSCGDKJjv80-MnM5Tcok4nCnPmSWp_v5ACLcBGAs/s1600/Algrabya150-5B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-NarKCpPE5TE/Wp3dLpboAKI/AAAAAAAAAOM/OSCGDKJjv80-MnM5Tcok4nCnPmSWp_v5ACLcBGAs/s1600/Algrabya150-5B.jpg" /></a></div>-<br /><b>WEDNESDAY</b>: In this variant, we solve the same problem by forming expressions in <i>d</i>, the time taken by Deka, instead of <i>e</i>, the time taken by Eric.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-r0dqEP8ojUM/Wp8nbJkTiuI/AAAAAAAAAOs/47vSwkdMTx8WOlX8bQaegk5qLqo1bE_kwCLcBGAs/s1600/Algrabya150-5C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-r0dqEP8ojUM/Wp8nbJkTiuI/AAAAAAAAAOs/47vSwkdMTx8WOlX8bQaegk5qLqo1bE_kwCLcBGAs/s1600/Algrabya150-5C.jpg" /></a></div>As well as solving the problem itself (whose solution we already know, of course), this gives us the opportunity to compare the two sets of expressions and to consider how an expression for <i>e</i> in terms of <i>d</i> is related to, and can be transformed into, the corresponding expression for <i>d</i> in terms of <i>e</i>.<br />-<br /><b>THURSDAY</b>: Here we use algebra to solve a slightly more complex problem which isn't quite as amenable to a bar-model approach as the previous task. So where previously we tried to give meaning to an algebraic approach, here we try to show its utility.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-H7HeuZ9RGZo/WqBcQxD-LjI/AAAAAAAAAO8/NqIfa_BbNIYplCItsVN0X4TxFl3UYBudwCLcBGAs/s1600/Algrabya150-5D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-H7HeuZ9RGZo/WqBcQxD-LjI/AAAAAAAAAO8/NqIfa_BbNIYplCItsVN0X4TxFl3UYBudwCLcBGAs/s1600/Algrabya150-5D.jpg" /></a></div> -<br /><b>FRIDAY</b>: Our final version of the ALG 5 task involves the same context but has a different structure. We again represent/solve it using the clock-diagram (or bar model) and in symbolic algebra - which approach do you find easier here?<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-aEpeS8Pzd3Y/WqJi5s-25QI/AAAAAAAAAPc/CPL5tMICKl87VLOnC7Urg3R3F82VOWMngCLcBGAs/s1600/Algrabya150-5E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-aEpeS8Pzd3Y/WqJi5s-25QI/AAAAAAAAAPc/CPL5tMICKl87VLOnC7Urg3R3F82VOWMngCLcBGAs/s1600/Algrabya150-5E.jpg" /></a></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: left;">We can also approach the task using a mix of trial and improvement and analysis: </div><div class="separator" style="clear: both; text-align: left;">At present (with <i>d</i> = 20 and <i>e</i> = 10) Deka and Eric have queued for a total of 30 minutes. If we wind forward 5 minutes, say, this will add 10 minutes to the total, so if we wind forwards a total of 15 minutes (making <i>d</i> = 20+15 and <i>e</i> = 10+15) they will have queued for a total of 60 minutes. This movie might help students see when to 'stop' the clock: <a href="https://www.youtube.com/watch?v=5J7dkwWxC7A">Long lunch with Deka and Eric</a>. </div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">Another approach would be to argue like this:</div><div class="separator" style="clear: both; text-align: left;">30+30 = 60</div><div class="separator" style="clear: both; text-align: left;">31+29 = 60</div><div class="separator" style="clear: both; text-align: left;">....</div><div class="separator" style="clear: both; text-align: left;">.... </div><div class="separator" style="clear: both; text-align: left;">35+25 = 60.</div><div class="separator" style="clear: both; text-align: left;">- </div><div class="separator" style="clear: both; text-align: left;">These approaches are not purely empirical - they make use of the task's structure so they can be classed as algebraic, I think. But they don't make use of symbolic algebra, even though the last method could be said to embody <i>d</i>+<i>e</i>=30 and <i>d</i>–<i>e</i> = 10. This highlights a real dilemma: <b>it is not that easy to devise accessible tasks where the need for symbolic algebra is compelling</b>. However, as we shall see in later weeks, one strategy is to make the task more complex, so that symbolising provides a way of keeping track of the information.</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">Notice: There's an interesting symmetry in the clock-diagram when <i>d</i> + <i>e</i> = 60:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-rb0rh1p6R40/WqLIZmmRqEI/AAAAAAAAAPs/1DJfXRGQaL42-DrFMRmjF9WW_b-t6QWSQCLcBGAs/s1600/de60.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="718" data-original-width="750" height="306" src="https://2.bp.blogspot.com/-rb0rh1p6R40/WqLIZmmRqEI/AAAAAAAAAPs/1DJfXRGQaL42-DrFMRmjF9WW_b-t6QWSQCLcBGAs/s320/de60.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-27536211761509517292018-02-25T22:30:00.001+00:002018-03-01T22:39:49.048+00:00ALG 4This set of ALG 4 tasks might require a somewhat greater understanding of algebra than the previous three ALG tasks, if students are to grasp some of the algebraic relations, especially in their more general form. However, even relative novices should be able to engage with the tasks and construe patterns even if this might be at a more empirical level.<br /><b>MONDAY</b>: This is the root ALG 4 task for this week. It can be approached in quite an exploratory way - later variants become more focused.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-LhEQ-VcTzN4/WpM13dri2dI/AAAAAAAAAMM/F6eHBGgNyfcobRVMgPMXWh8A81r4Ts5AwCLcBGAs/s1600/Algrabya150-4A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-LhEQ-VcTzN4/WpM13dri2dI/AAAAAAAAAMM/F6eHBGgNyfcobRVMgPMXWh8A81r4Ts5AwCLcBGAs/s1600/Algrabya150-4A.jpg" /></a></div>Here it is worth dwelling for a while on the form of the three equations, before sketching any lines. Students who know about the form of linear functions may notice that the greater the slope of the corresponding lines, the higher the point of intersection with the <i>y</i>-axis (ie as <i>m</i> increases, so does <i>c</i>, in <i>y</i> = <i>mx</i> + <i>c</i>). In turn this might suggest that the lines are 'hinged' about a common point, which turns out to be the case.<br />Students without this kind of knowledge can still engage meaningfully with the task, by sketching the lines given by the equations and examining the result.<br />- <br /><b>TUESDAY</b>: In this variant we show the three lines and ask students to explain the fact that they go through a common point, using both algebraic and geometric arguments.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-qZYUsLMfVvo/WpSQ1mWbN5I/AAAAAAAAAMc/G-BPxosmCQE5MxMsRfhpzVlSCYzb3d4RACLcBGAs/s1600/Algrabya150-4B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-qZYUsLMfVvo/WpSQ1mWbN5I/AAAAAAAAAMc/G-BPxosmCQE5MxMsRfhpzVlSCYzb3d4RACLcBGAs/s1600/Algrabya150-4B.jpg" /></a></div> As can be seen, the lines all go through the point (-3, 0). We can verify this by letting <i>y</i> = 0 in each of the equations and then solving for <i>x</i>, or by solving for <i>x</i> in the general equation <i>y</i> = <i>ax</i> + 3<i>a</i> or <i>y</i> = <i>a</i>(<i>x</i> + 3) after letting <i>y</i> = 0.<br />We can also explain this in terms of the slope of the lines and the triangle formed by the points with coordinates (-3, 0), (0, 0) and (0, <i>c</i>). We can think of the slope as being the change in <i>y</i> resulting from a change of +1 in the value of <i>x</i> (below). This means if we increase <i>x</i> by +3, in moving from (-3, 0) to the origin, we increase <i>y</i> by +3×<i>m</i>, and hence cut the <i>y</i> axis at (0, 3<i>m</i>).<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-xgubWiKxwIc/WpapEacBYOI/AAAAAAAAAM8/AumtSoH0lOwM8fDWm2-TwLDITHrGqsivwCLcBGAs/s1600/Algrabya150-4B-geo.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="358" data-original-width="556" height="257" src="https://4.bp.blogspot.com/-xgubWiKxwIc/WpapEacBYOI/AAAAAAAAAM8/AumtSoH0lOwM8fDWm2-TwLDITHrGqsivwCLcBGAs/s400/Algrabya150-4B-geo.jpg" width="400" /></a></div>-<br /><b>WEDNESDAY</b>: Here we find the common point for the lines given by another family of equations. But this time, rather than listing individual equations, the family is expressed in a general form. This might help students express the previous family in a similar form if they have not already done so (ie, as <i>y</i> = <i>ax</i> + 3<i>a</i>, as mentioned above). Also, as the two forms are very similar, it might help students see how, and perhaps why, the general form relates to the lines' common point.<br />[It might be argued that the form <i>y</i> = <i>a</i>(<i>x</i> – 2) is more salient than the given form <i>y</i> = <i>ax</i> – 2<i>a</i>, even though it diverges from the familiar <i>y</i> = <i>mx</i> + <i>c</i> form, as it immediately becomes apparent that <i>x</i> = 2 when <i>y</i> = 0, regardless of the value of <i>a</i>.] <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-SAqngCl_kCo/WpXr9UhWYuI/AAAAAAAAAMs/Fcj9UdzuehIfKc9Q5FXl2PLDrocBjF2VgCLcBGAs/s1600/Algrabya150-4C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-SAqngCl_kCo/WpXr9UhWYuI/AAAAAAAAAMs/Fcj9UdzuehIfKc9Q5FXl2PLDrocBjF2VgCLcBGAs/s1600/Algrabya150-4C.jpg" /></a></div>-<br /><b>THURSDAY</b>: Here we consider another family of equations whose general form, at least at first sight, seems rather different from the form of the two families we've considered so far. It turns out that the common point is not on the <i>x</i>-axis this time. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-lnhLmmzSUfE/Wpct-Hc72QI/AAAAAAAAANM/0XFTgP6KtCANSFdTZ-zdvdwpbZ2B1rwhQCLcBGAs/s1600/Algrabya150-4D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-lnhLmmzSUfE/Wpct-Hc72QI/AAAAAAAAANM/0XFTgP6KtCANSFdTZ-zdvdwpbZ2B1rwhQCLcBGAs/s1600/Algrabya150-4D.jpg" /></a></div>There are several ways of writing the general form of the family of equations, for example these:<br /><i>y</i> = <i>ax</i> + (3 – <i>a</i>), <i>y</i> = <i>a</i>(<i>x</i> – 1) + 3, <i>y</i> – 3 = <i>a</i>(<i>x</i> – 1).<br />What are the affordances of each?<br />-<br /><b>FRIDAY</b>: Here we reverse the process, by asking for the family of lines that goes through a specific point.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-QdcG-wfYmhs/Wph-YWw5fKI/AAAAAAAAANc/TkppB5_5rM8EKKhiACcT38IfchvX6yFdgCLcBGAs/s1600/Algrabya150-4E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-QdcG-wfYmhs/Wph-YWw5fKI/AAAAAAAAANc/TkppB5_5rM8EKKhiACcT38IfchvX6yFdgCLcBGAs/s1600/Algrabya150-4E.jpg" /></a></div> We can approach this in different ways. One way is to find the equations of some specific lines through (2, 5) and look for a pattern. Another would be to look back at Thursday's task and see what's general about it, ie how does the general equation (in whatever form) relate to the common point?<br />Another way of thinking about this is to consider lines through the origin (ie of the form <i>y</i> = <i>ax</i>) and to think of translating the lines 2 across and 3 up - how do these steps transform the equation?<br />An obvious extension of the task would be to consider a general point, with coordinates (<i>u</i>, <i>v</i>), say. If you've managed to solve the task for (2, 5), this should be fairly easy ....ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-32734979430119391802018-02-18T22:56:00.001+00:002018-02-23T16:31:44.696+00:00ALG 3In this week's set of tasks we try to get a better feel for a linear function by comparing it with a non-linear function. We see that as the independent variable changes at a steady rate, the values of the functions change at different rates, with one of them changing steadily.<br />The functions are set in a geometric context and we compare them geometrically (in two different ways, one of which is much more salient than the other), by expressing them symbolically, and by looking at numerical values in an ordered table.<br />-<br /><b>MONDAY</b>: This is the root task for the week. It can be quite tricky ....<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-yvrdB4LW8Aw/WooDnjab81I/AAAAAAAAAI4/EojKe0bMS_oxYdqNvPvVy1lgtqllXrbpQCLcBGAs/s1600/Algrabya150-3A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-yvrdB4LW8Aw/WooDnjab81I/AAAAAAAAAI4/EojKe0bMS_oxYdqNvPvVy1lgtqllXrbpQCLcBGAs/s1600/Algrabya150-3A.jpg" /></a></div>After students have drawn the shape for <i>u</i> = 3, you might want to show this:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-x9lIpavfSNI/WosSjE0ISnI/AAAAAAAAAJI/n2q0WxHiJEIyxcM6WXzmI9vbTJY_1BdEACLcBGAs/s1600/Algrabya150-3A-ans.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://3.bp.blogspot.com/-x9lIpavfSNI/WosSjE0ISnI/AAAAAAAAAJI/n2q0WxHiJEIyxcM6WXzmI9vbTJY_1BdEACLcBGAs/s400/Algrabya150-3A-ans.jpg" width="400" /></a></div>-<br /><b>TUESDAY</b>: Here we compare the area of our cross-shape with the area of another cross-shape:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-7VLPI6TcBWM/WotXyU-t-2I/AAAAAAAAAJY/lUhN5TAHLe0b9Wu4JmkN4Rmr50OOnob1gCLcBGAs/s1600/Algrabya150-3B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-7VLPI6TcBWM/WotXyU-t-2I/AAAAAAAAAJY/lUhN5TAHLe0b9Wu4JmkN4Rmr50OOnob1gCLcBGAs/s1600/Algrabya150-3B.jpg" /></a></div>You might want to present this on two separate slides, starting with the edited version below, so that students are not channeled too soon into answering the more structured second part. It is worth giving students plenty of time to explore the question, "Which shape has the larger area?".<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-r-x_uWoS3O0/WovVikpctxI/AAAAAAAAAJo/5XuxNW2ex0EWkCP0JJuzfll6corAzTVYgCLcBGAs/s1600/Algrabya150-3Bi.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="260" src="https://2.bp.blogspot.com/-r-x_uWoS3O0/WovVikpctxI/AAAAAAAAAJo/5XuxNW2ex0EWkCP0JJuzfll6corAzTVYgCLcBGAs/s400/Algrabya150-3Bi.jpg" width="400" /></a></div>After students have explored the task, they can check their hunches and confirm their insights with this <i>Dynamic crosses</i> <a href="https://www.youtube.com/watch?v=ucaukhX5Lgs">movie</a>. Some clips from the movie are shown below:<br /><a href="https://3.bp.blogspot.com/-Upz88ZKI4t4/WoxLUxTRF8I/AAAAAAAAAKA/hm1yhbHrQTkwcBPUlB_C9436y4Am3mu0gCLcBGAs/s1600/clips-u2468-noQ.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="1000" height="249" src="https://3.bp.blogspot.com/-Upz88ZKI4t4/WoxLUxTRF8I/AAAAAAAAAKA/hm1yhbHrQTkwcBPUlB_C9436y4Am3mu0gCLcBGAs/s640/clips-u2468-noQ.png" width="640" /></a>-<br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><b>WEDNESDAY</b>: Here we repeat Tuesday’s task but we treat the cross-shapes as the nets of two open boxes. </span><br /><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"></span></div><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-DhRER02XuHc/WoxvLSlcWdI/AAAAAAAAAKQ/DGMM_0ny2ZcwLmZxbaVIGWNaGSYUFJzXQCLcBGAs/s1600/Algrabya150-3C.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-DhRER02XuHc/WoxvLSlcWdI/AAAAAAAAAKQ/DGMM_0ny2ZcwLmZxbaVIGWNaGSYUFJzXQCLcBGAs/s1600/Algrabya150-3C.jpg" /></a></div><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">For one of the boxes, as <i>u</i> increases, the height of the box increases but the (square) base stays the same; for the other box, its base gets larger (in both dimensions) but the height stays the same. Thus one box changes in two dimensions simultaneously, while the other changes in only one. </span></div><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">These changes are far less obvious when one just looks at the 2D nets, unless one re-configures them. For example, the first cross-shape is changing in two directions as <i>u</i> changes. However, if one cuts and re-joins the net like this (below), the change can be seen to occur in just one dimension.</span></div><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"></span></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-g0Kn7yZNhYo/WoxvvatuCzI/AAAAAAAAAKY/gZ_AX8z9YOYJfUwATJ8Pqt2lPKjmrP6rgCLcBGAs/s1600/Algrabya150-3C-net-jig.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="143" data-original-width="467" src="https://2.bp.blogspot.com/-g0Kn7yZNhYo/WoxvvatuCzI/AAAAAAAAAKY/gZ_AX8z9YOYJfUwATJ8Pqt2lPKjmrP6rgCLcBGAs/s1600/Algrabya150-3C-net-jig.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">We can do the same for the net of the other box. As <i>u</i> increases, the net expands in two dimensions, because of the <i>u</i> by <i>u</i> square (tinted).</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-0klSPMxhfdo/Wo1uGWsfxaI/AAAAAAAAAKo/Tx2nobeYP2oVKB-7_AQKLZJHqeU2SSuowCLcBGAs/s1600/Algrabya150-3C-net-jig2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="164" data-original-width="620" src="https://4.bp.blogspot.com/-0klSPMxhfdo/Wo1uGWsfxaI/AAAAAAAAAKo/Tx2nobeYP2oVKB-7_AQKLZJHqeU2SSuowCLcBGAs/s1600/Algrabya150-3C-net-jig2.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">-</span><br /><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><b>THURSDAY</b>:This is Thursday's variant on the ALG 3 task. By now, students will, hopefully, have developed a good feel for how the areas of the two cross-shapes change with </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>: eg, that the first shape's area increases steadily for steady increases in </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>; and that the second shape's area changes more and more rapidly as </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span> increases, and so, though its area is smaller for values like <i>u</i> = 2, it overtakes the other area at some point, namely when <i>u</i> gets past 5.</span><br /><div class="separator" style="clear: both; text-align: center;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><a href="https://2.bp.blogspot.com/-2PPqv4PV-js/Wo3mHciAP0I/AAAAAAAAAK4/RdqvjZ3NvikWx42NY0cmlHFmHmehV-a8gCLcBGAs/s1600/Algrabya150-3D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-2PPqv4PV-js/Wo3mHciAP0I/AAAAAAAAAK4/RdqvjZ3NvikWx42NY0cmlHFmHmehV-a8gCLcBGAs/s1600/Algrabya150-3D.jpg" /></a></span></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">So in today's task, we represent the areas symbolically, with the aim of making these ideas about the changing areas more explicit and, conversely, giving meaning to the resulting algebraic expressions. This may also help students to get a better sense of the general form of what turn out to be linear and quadratic expressions.</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">- </span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><b>FRIDAY</b>: Here we try to consolidate some of the burgeoning ideas by revisiting the numerical values found in Tuesday's task, and judiciously extending these with the aid of an ordered table.</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-a3ZwmetcVVw/Wo9QMTtvXtI/AAAAAAAAALI/Yie6paLe4KoZzWHgdh6W9YVKzA1GtQltwCLcBGAs/s1600/Algrabya150-3E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-a3ZwmetcVVw/Wo9QMTtvXtI/AAAAAAAAALI/Yie6paLe4KoZzWHgdh6W9YVKzA1GtQltwCLcBGAs/s1600/Algrabya150-3E.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"> A possible next step would be to graph these sets of values - both as a way of illuminating the nature of the (linear and quadratic) relations, and as a way of giving meaning to graphs. And it would be worth relating this to the <i>Dynamic crosses</i> <a href="https://www.youtube.com/watch?v=ucaukhX5Lgs">movie</a> (with the mouse over the play/pause button).</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The completed table looks like this:</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-qyPDO2pSvkY/WpAJK1ibxJI/AAAAAAAAALY/0TOxE3ieTxcQtq1AJ1Kselg3971YCGLtQCLcBGAs/s1600/Algrabya150-3E-table.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="232" data-original-width="316" src="https://4.bp.blogspot.com/-qyPDO2pSvkY/WpAJK1ibxJI/AAAAAAAAALY/0TOxE3ieTxcQtq1AJ1Kselg3971YCGLtQCLcBGAs/s1600/Algrabya150-3E-table.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"> As we can see, as <i>u</i> increases by 1, 20</span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>+25 increases by the same amount, whether we start from </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>=5 or </span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>=10, whereas 20</span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>+</span><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><i>u</i></span>2 increases by an increasing amount ....</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">We can illustrate this with a graph, such as this off-the-peg version from Excel.</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-rBqPk-tW5lM/WpAKqBqaBJI/AAAAAAAAALk/3LR81Xs0pF87XMC83h0iqdbgzVTAAW4xgCLcBGAs/s1600/ALG-3E-graph.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="846" data-original-width="610" src="https://3.bp.blogspot.com/-rBqPk-tW5lM/WpAKqBqaBJI/AAAAAAAAALk/3LR81Xs0pF87XMC83h0iqdbgzVTAAW4xgCLcBGAs/s1600/ALG-3E-graph.png" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The graph raises some interesting questions:</span><br /><ul><li><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The blue line appears to be straight. Is this really the case and what would this mean (algebraically, and in the context of the cross-shape's area)? </span></li><li><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The red line seems to consist of ever-steeper line segments. Does this make sense? What if I had chosen different data-points - would I get different line-segments?? </span></li><li><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">What happens to the two graphs for negative values of <i>u</i> (algebraically, and in terms of the cross-shapes and their areas)?</span></li></ul><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">It can also be illuminating to see how equivalent forms of the algebraic expressions relate to (and can be derived from) the shapes of the crosses. This is for the first cross-shape:</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-jgOay4obLZk/WpAceBOUoCI/AAAAAAAAAL0/siUb3tP50Ekhk86rUvpw0A6f6FAper3rgCLcBGAs/s1600/Algrabya150-3C-net-jig-express.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="169" data-original-width="600" src="https://1.bp.blogspot.com/-jgOay4obLZk/WpAceBOUoCI/AAAAAAAAAL0/siUb3tP50Ekhk86rUvpw0A6f6FAper3rgCLcBGAs/s1600/Algrabya150-3C-net-jig-express.jpg" /> </a></div><div class="separator" style="clear: both; text-align: left;"> And this is for the second cross-shape: </div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-szqH1l_aaBE/WpAcvYApKhI/AAAAAAAAAL4/bNkEmZJZKQcpyF3v3YZIpHky6imC6mNPgCLcBGAs/s1600/Algrabya150-3C-net2-jig-express.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="299" data-original-width="525" src="https://3.bp.blogspot.com/-szqH1l_aaBE/WpAcvYApKhI/AAAAAAAAAL4/bNkEmZJZKQcpyF3v3YZIpHky6imC6mNPgCLcBGAs/s1600/Algrabya150-3C-net2-jig-express.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><br /></span></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-44054447094164285672018-02-15T12:56:00.002+00:002018-02-18T17:52:14.856+00:00ALG 2<div class="separator" style="clear: both; text-align: center;"></div>Here is an open version of our ALG 2 root task - a slightly more structured version will appear on Monday, with other variants later in the week.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-CWkxhu1Zf18/WoAjoTEpk6I/AAAAAAAAAGA/llLfnibVIagjGix-m4kQqnLqXjiSfLvxQCLcBGAs/s1600/Algrabya150-2Aopenopen.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://4.bp.blogspot.com/-CWkxhu1Zf18/WoAjoTEpk6I/AAAAAAAAAGA/llLfnibVIagjGix-m4kQqnLqXjiSfLvxQCLcBGAs/s1600/Algrabya150-2Aopenopen.jpg" /></a></div>The task starts with the idea that an unknown can vary, rather than the more common idea in school mathematics that it has a specific value that we want to find.<br />Also, rather than immediately asking students to perform various actions on the unknown (such as collecting like terms, manipulating or evaluating expressions, forming and solving equations), the task may help students realise that they can sometimes make sense of expressions from the outset - especially in this case, where the given expressions (<i>x</i> and 2<i>x</i>) are fairly simple, and thus might fairly readily lead to an interpretation like 'one angle in the triangle is twice the size of another angle'.<br />A movie of the triangle can be found <span style="color: #3d85c6;"><a href="https://www.youtube.com/watch?v=9Ua6wIwRwcA">here</a></span>. <br />-<br />MONDAY: Here is this week's root task. It is more structured than the version above but it should still help students discern the general relationship between two of the triangle's angles (ie between the expressions <i>x</i> and 2<i>x</i>). Part b) provides a shift towards the use of algebraic procedures, by asking students to find a particular (critical) value of <i>x</i>. We can solve this algebraically by forming the equation <i>x</i> + 2<i>x</i> = 180. However, it is important to acknowledge that we can also find <i>x</i> quite easily by other means, eg trial and improvement. <br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-11znlvWKxGc/WoFs_cMiGOI/AAAAAAAAAGQ/SqgkK7GN_NUVGR7ibhI9pi3a2-WejO-MwCLcBGAs/s1600/Algrabya150-2A.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-11znlvWKxGc/WoFs_cMiGOI/AAAAAAAAAGQ/SqgkK7GN_NUVGR7ibhI9pi3a2-WejO-MwCLcBGAs/s1600/Algrabya150-2A.jpg" /></a></div> We have made two movies showing the triangle as <i>x</i> varies. In <a href="https://www.youtube.com/watch?v=9Ua6wIwRwcA">one </a><a href="https://www.youtube.com/watch?v=9Ua6wIwRwcA">movie</a> the value of <i>x</i> goes up to the 'permissible' value. In the <a href="https://www.youtube.com/watch?v=PK2_6S1VTwY">other movie</a> the value of <i>x</i> goes beyond this value: what happens to the triangle when this occurs, and should such values be 'allowed'?!<br />-<br />TUESDAY: Like Monday's part b), we don't need formal algebra to solve this variant of ALG 2, but it gives us the opportunity to construct two more equations in <i>x</i>, so it becomes interesting to compare them and to discuss ways in which they can be solved.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-oNvjNibJWhk/WoIcqtLRv9I/AAAAAAAAAGg/JcugcRYfR9s_YPSMhxzDfHFli1N96SJdACLcBGAs/s1600/Algrabya150-2B.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-oNvjNibJWhk/WoIcqtLRv9I/AAAAAAAAAGg/JcugcRYfR9s_YPSMhxzDfHFli1N96SJdACLcBGAs/s1600/Algrabya150-2B.jpg" /></a></div> It is also worth engaging with the geometry of the situation. For example, why is the triangle always weighted to the right in the given diagrams? Also, we can see from the given triangles that the value of <i>x</i> will be close to 35 for one of the isosceles triangles, and somewhat less than 50 for the other - do our algebraic approaches confirm this?<br />-<br />WEDNESDAY: Having looked for isosceles triangles, this variant involves finding values of <i>x</i> for which the triangle is right-angled:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-2ikf-I8l2uk/WoMljrt48fI/AAAAAAAAAGw/81W3xPQdgtcK_UxvHkDx8tDJRFbNgskGwCLcBGAs/s1600/Algrabya150-2C.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-2ikf-I8l2uk/WoMljrt48fI/AAAAAAAAAGw/81W3xPQdgtcK_UxvHkDx8tDJRFbNgskGwCLcBGAs/s1600/Algrabya150-2C.jpg" /></a></div>We can find <i>x</i> by using trial and improvement, say, but we can also proceed algebraically by forming equations to represent the geometric relations resulting from angle B or angle C being a right angle. This gives 2<i>x</i> = 90 and <i>x</i> + 2<i>x</i> = 90 (or <i>x</i> + 2<i>x</i> + 90 = 180). Similarly, we can form equations for the earlier variants: <i>x</i> + 2<i>x</i> = 180 (ALG 2A), <i>x</i> + 2<i>x</i> + 2<i>x</i> = 180 and <i>x</i> + <i>x</i> + 2<i>x</i> = 180 (ALG 2B). These equations are all fairly easy to solve, but what is particularly nice about the more complex ones is that we may be able to persuade students of their <i>utility</i>: by forming and writing down equations we can ease the strain on our working memory.<br />It can also be useful to discuss and compare different ways of solving the equations, eg by inspection, by trial and improvement, or by transforming the equations.<br />-<br />THURSDAY: Here we change the relationship between the base angles of the triangle from <i>x</i>˚, 2<i>x</i>˚ to <i>x</i>˚, 4<i>x</i>˚. This gives students a chance to recap on earlier ideas and methods, as well as perhaps starting to think about how changes in the angle relationship might affect the size of the various angles that we are trying to find.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-R8KIm_S3kXU/WoS8oYzF2cI/AAAAAAAAAHE/c6-j1yaYfogXlExWDv4pVwmLuRjJOrqlQCLcBGAs/s1600/Algrabya150-2D.jpg" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-R8KIm_S3kXU/WoS8oYzF2cI/AAAAAAAAAHE/c6-j1yaYfogXlExWDv4pVwmLuRjJOrqlQCLcBGAs/s1600/Algrabya150-2D.jpg" /></a></div>Thursday enjoyed a couple of bonus tasks (below). The first can be varied in quite nice ways. The other [based on a seminal idea from David Fielker, writing some years ago in <i>Mathematics Teaching</i>] says something about the freedom that symbolic algebra can give us.<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-TkH5UxByp2k/WoXsF4VLyJI/AAAAAAAAAHg/vxaAOMpWMZUZ8rcOiygnJWqwplXdvfBnACLcBGAs/s1600/2A-bonus.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="295" data-original-width="888" height="212" src="https://1.bp.blogspot.com/-TkH5UxByp2k/WoXsF4VLyJI/AAAAAAAAAHg/vxaAOMpWMZUZ8rcOiygnJWqwplXdvfBnACLcBGAs/s640/2A-bonus.png" width="640" /></a></div>And here's another bonus (actually posted on Friday) which continues the regular polygon theme:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-3Fvs1r68BTE/WomqtsKAaFI/AAAAAAAAAIA/clE-6AZgn4ACbpduhbVeWjay4Q2N9Dt3wCLcBGAs/s1600/poly-exterior2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" height="208" src="https://3.bp.blogspot.com/-3Fvs1r68BTE/WomqtsKAaFI/AAAAAAAAAIA/clE-6AZgn4ACbpduhbVeWjay4Q2N9Dt3wCLcBGAs/s320/poly-exterior2.jpg" width="320" /></a></div>-<br />FRIDAY: This is our final (for now?) ALG 2 variant. It gives plenty of scope for consolidation and for exploring the relationship between the base angles, which is expressed here in a more general way (with the aid of parameters <i>a</i> and <i>b</i>). Let us know what you or your students find.<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: center;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-g3a5oRZSCmk/WoXuyDQYd1I/AAAAAAAAAHw/B40bLKnx_DsVdIOi8zDKfM6uHcZbIgk-QCLcBGAs/s1600/Algrabya150-2E.jpg" /></div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">SUNDAY Supplement: Here's a set of three tasks still involving the base angles of a triangle but where there's an inverse proportional relationship. Some of the resulting triangle-families can be quite surprising.</div><div class="separator" style="clear: both; text-align: left;">Here's the first task (below). Part <b>a)</b> is to get students to start thinking about the scenario. The given information means that <i>a</i>/20 = 50, so <i>a</i> = 1000. It might seem surprising that a relation involving '1000' works in the context of triangles, where the interior angle sum is a mere 180˚. Of course, the relation ∠A = <i>x</i>˚, ∠B = 1000˚/<i>x</i> doesn't work for all values of <i>x</i>. For example, if <i>x</i> = 1, then ∠B would be 1000˚. However it turns out the relation works for this range of values (approximately): </div><div class="separator" style="clear: both; text-align: left;">6 ≤ <i>x</i> ≤ 174.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-G_lAUGFu8CY/WomsoX7qHmI/AAAAAAAAAIM/Wszw82YKruYIRyZMFIs0yheOPg9HNtsPgCLcBGAs/s1600/Algrabya150-2Asunday01.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-G_lAUGFu8CY/WomsoX7qHmI/AAAAAAAAAIM/Wszw82YKruYIRyZMFIs0yheOPg9HNtsPgCLcBGAs/s1600/Algrabya150-2Asunday01.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">It is worth spending quite a lot of time on part <b>b)i</b>, to get a feel for how the base angles, and the overall shape of the triangle, change. The triangle turns out to be fairly 'flat'. When <i>x</i> is 'small', a change in <i>x</i> produces a rapid change in ∠B; when <i>x</i> is fairly 'large' ∠B seems to change hardly at all. A movie of the changing triangle can be found <a href="https://www.youtube.com/watch?v=ThoSbTK03jo">here</a>.</div><div class="separator" style="clear: both; text-align: left;">A good way to start part <b>b)ii</b> is to adopt a visual approach, eg by running the aforementioned movie and pausing it when the triangle looks to be isosceles. This happens in three places. A more precise estimate can then be found by using a numerical approach, which can be made more and more precise with the aid of a spreadsheet (with a column for <i>x</i>, for 1000/<i>x</i> and for 180 – <i>x</i> – 1000/<i>x</i>).</div><div class="separator" style="clear: both; text-align: left;">In the isosceles case where ∠A = ∠B, <i>x</i> = √1000. The value of <i>x</i> for the other two cases (∠A = ∠C and ∠C = ∠B) can be found algebraically by forming an equation in <i>x</i> and writing it in the usual quadratic form. However, it doesn't factorise....</div><div class="separator" style="clear: both; text-align: left;">- </div><div class="separator" style="clear: both; text-align: left;">Here's the second Sunday Supplement task. This is very exploratory. You might want to start by seeing what happens with some fairly 'random' values of <i>a</i>, and then try some more 'informed' values.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-2N4fuIglEwU/Wom3tSNntBI/AAAAAAAAAIc/1jXcL07jPWUXVeovyTdmT5IVSnQUmJsSACLcBGAs/s1600/Algrabya150-2Asunday02.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-2N4fuIglEwU/Wom3tSNntBI/AAAAAAAAAIc/1jXcL07jPWUXVeovyTdmT5IVSnQUmJsSACLcBGAs/s1600/Algrabya150-2Asunday02.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">After some explorations, you might want to look at two further movies, where <a href="https://www.youtube.com/watch?v=ouPjdKXrJPg"><i>a</i> = 100</a> and <a href="https://www.youtube.com/watch?v=Aonr9c4xcf0"><i>a</i> = 1</a>.</div><div class="separator" style="clear: both; text-align: left;">The choice of '100' is very deliberate and you might like to predict what effect this has on the shape of the triangle. The effect of choosing <i>a</i> = 1 is very strange.... - but the movie won't win any Oscars.</div><div class="separator" style="clear: both; text-align: left;">-</div><div class="separator" style="clear: both; text-align: left;">The third Sunday Supplement:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-5tN1a93-zxA/Wom5-stMPPI/AAAAAAAAAIo/Ey_g3w0wGbY2Pxbywh9AXqSAtus-yFGLgCLcBGAs/s1600/Algrabya150-2Asunday03.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-5tN1a93-zxA/Wom5-stMPPI/AAAAAAAAAIo/Ey_g3w0wGbY2Pxbywh9AXqSAtus-yFGLgCLcBGAs/s1600/Algrabya150-2Asunday03.jpg" /></a></div><div class="separator" style="clear: both; text-align: left;">In this task the numbers are nicer. You might be able to see the value of <i>x</i> for which ∠A = ∠B. The value of <i>x</i> for the other two isosceles cases (∠A = ∠C and ∠C = ∠B) can be found by forming a quadratic equation which this time does factorise. Or you could adopt a visual/numeric approach again, perhaps with the aide of <a href="https://www.youtube.com/watch?v=aJLsk3zv5rI">this</a> movie, or <a href="https://www.youtube.com/watch?v=VlSDZm09cg4">this</a> second version which provides a trace of vertex C.</div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-15759552069684742432018-02-11T13:46:00.001+00:002018-02-11T13:49:44.447+00:00SingalongAll together now ....<br />- <br /><span style="font-size: x-small;">[with thanks to Steve Miller and The Steve Miller Band]</span><br /><span style="font-size: x-small;">[and to Wren]</span><br /><span style="font-size: x-small;">- </span><br /><div class="separator" style="clear: both; text-align: left;"><iframe allowfullscreen='allowfullscreen' webkitallowfullscreen='webkitallowfullscreen' mozallowfullscreen='mozallowfullscreen' width='320' height='266' src='https://www.blogger.com/video.g?token=AD6v5dzg_NyIN2hevVfLXBrnLsPp0y3xTXBIDLsZUBWN7wkEHs4Yjz_FAmBkawzANcgg1ztvWXBlIWdrfHi97pEQow' class='b-hbp-video b-uploaded' frameborder='0' /></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0tag:blogger.com,1999:blog-9058930865720822726.post-7951505657704461822018-02-04T23:17:00.003+00:002018-02-08T23:47:21.071+00:00ALG 1<style><!-- /* Font Definitions */ @font-face {font-family:"Cambria Math"; panose-1:2 4 5 3 5 4 6 3 2 4; mso-font-charset:1; mso-generic-font-family:roman; mso-font-format:other; mso-font-pitch:variable; mso-font-signature:0 0 0 0 0 0;} @font-face {font-family:Calibri; panose-1:2 15 5 2 2 2 4 3 2 4; mso-font-charset:0; mso-generic-font-family:auto; mso-font-pitch:variable; mso-font-signature:-536870145 1073786111 1 0 415 0;} /* Style Definitions */ p.MsoNormal, li.MsoNormal, div.MsoNormal {mso-style-unhide:no; mso-style-qformat:yes; mso-style-parent:""; margin:0mm; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} .MsoChpDefault {mso-style-type:export-only; mso-default-props:yes; font-family:Calibri; mso-ascii-font-family:Calibri; mso-ascii-theme-font:minor-latin; mso-fareast-font-family:Calibri; mso-fareast-theme-font:minor-latin; mso-hansi-font-family:Calibri; mso-hansi-theme-font:minor-latin; mso-bidi-font-family:"Times New Roman"; mso-bidi-theme-font:minor-bidi; mso-ansi-language:EN-US; mso-fareast-language:EN-US;} @page WordSection1 {size:612.0pt 792.0pt; margin:72.0pt 72.0pt 72.0pt 72.0pt; mso-header-margin:36.0pt; mso-footer-margin:36.0pt; mso-paper-source:0;} div.WordSection1 {page:WordSection1;} --></style> <br /><div class="MsoNormal"><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;"><span style="color: blue;"><span style="color: #274e13;">MONDAY:</span></span> This is the ‘root’ task for the week. </span></div><div class="MsoNormal"><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-T6Y4F5DH5Lk/WneLOWnZKII/AAAAAAAAAEs/UKVWw5JHBMottXPO-yM2dJcrsH1j_LLQwCLcBGAs/s1600/Algrabya150-1A.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://3.bp.blogspot.com/-T6Y4F5DH5Lk/WneLOWnZKII/AAAAAAAAAEs/UKVWw5JHBMottXPO-yM2dJcrsH1j_LLQwCLcBGAs/s1600/Algrabya150-1A.jpg" /></a></div><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">The t<span style="font-size: small;">ask i</span>s designed to help students focus on the structure of the equation, rather than to reach for a familiar procedure for solving it (be it an informal strategy like trial and improvement or the use of formal transformation rules). We are not really interested in the solution per se - although it is fairly easy to find, if one uses the given information and observes the equation's structure.</span><br /><span style="font-family: "times new roman"; mso-ansi-language: EN-GB;">- </span></div><span style="font-family: "times new roman"; font-size: 12.0pt;"><span style="color: blue;"><span style="color: #274e13;">TUESDAY:</span></span> Tweaking a term in an equation can help us see what role the term plays in the equation and how it interacts with other terms. Here's another tweak of the original equation, one whose effect is relatively easy to see:</span><br /><div class="separator" style="clear: both; text-align: center;"><span style="font-family: "times new roman"; font-size: 12.0pt;"><a href="https://2.bp.blogspot.com/-LfwyQbUHagY/WnjizxIIrJI/AAAAAAAAAE4/JPaPulOSoLwhu54fgP3PMHruW7amWre2ACLcBGAs/s1600/Algrabya150-1B.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-LfwyQbUHagY/WnjizxIIrJI/AAAAAAAAAE4/JPaPulOSoLwhu54fgP3PMHruW7amWre2ACLcBGAs/s1600/Algrabya150-1B.jpg" /></a></span></div><span style="font-family: "times new roman"; font-size: 12.0pt;">You or students might like to think of some other tweaks - which ones seem easy to construe, and which are more opaque?</span><br /><span style="font-family: "times new roman"; font-size: 12.0pt;">-</span><br /><span style="font-family: "times new roman"; font-size: 12.0pt;">WEDNESDAY: The tweak lined up for the middle of the week is nice and simple. Actually, I've decided on an extra tweak. Both still quite simple, but they seem to work in subtly different ways. What's going on?!</span><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-Ha_JDljz6mY/WnpEzjXcB6I/AAAAAAAAAFQ/I48klr4_maEOaNR9Tbq1kRujhtzsghUEwCLcBGAs/s1600/ALG-1Cab.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="371" data-original-width="1112" height="212" src="https://2.bp.blogspot.com/-Ha_JDljz6mY/WnpEzjXcB6I/AAAAAAAAAFQ/I48klr4_maEOaNR9Tbq1kRujhtzsghUEwCLcBGAs/s640/ALG-1Cab.png" width="640" /></a></div>It might strike us as odd (at least momentarily), that doing the same thing <i>additively</i> to the terms 77 and 203 has no effect, while <i>multiplicatively</i> it does ....<br />- <br />THURSDAY: This is Thursday's variant. If it seems too simple, you could change 42 to 84, say, or wouldn't that make enough difference? I'd be interested to know.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-CnU8FJW01lw/Wntz5UHbiJI/AAAAAAAAAFg/Mi6OHT62ecwR9bk6RsAiyXERuFOYq2XRwCLcBGAs/s1600/Algrabya150-1D.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://1.bp.blogspot.com/-CnU8FJW01lw/Wntz5UHbiJI/AAAAAAAAAFg/Mi6OHT62ecwR9bk6RsAiyXERuFOYq2XRwCLcBGAs/s1600/Algrabya150-1D.jpg" /></a></div>-<br /><div class="separator" style="clear: both; text-align: center;"></div><span style="font-family: "times new roman";">FRIDAY: Here's the last ALG 1 variant (for now?). With some classes, you might decide to start with an open task such as this, rather than the earlier, more closed variants.</span><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-gRsFNEbRcc0/WnzeThh3lQI/AAAAAAAAAFw/_2D8RMyKVmg-0VvNgh3oMpUXMOdycCEUwCLcBGAs/s1600/Algrabya150-1E.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="392" data-original-width="602" src="https://2.bp.blogspot.com/-gRsFNEbRcc0/WnzeThh3lQI/AAAAAAAAAFw/_2D8RMyKVmg-0VvNgh3oMpUXMOdycCEUwCLcBGAs/s1600/Algrabya150-1E.jpg" /></a></div><span style="font-family: "times new roman";">And you might (at some stage) want something more open still, eg along the lines of "Change one of the numbers in the equation - what does this do to the solution?". You might also want to investigate equations with a different form from <i>ax</i> + <i>b</i> = <i>c</i>.</span> ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com9tag:blogger.com,1999:blog-9058930865720822726.post-34619046119754540812018-02-01T11:53:00.000+00:002018-02-01T11:59:58.653+00:00About this site<div class="entry-content"><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-hJuGPbpkClA/WnL_WkMrVQI/AAAAAAAAAEY/Y_hEw-0AvlATG-ka7uJ-jU3He6i8UQQWQCLcBGAs/s1600/alge2.png" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" data-original-height="239" data-original-width="547" height="137" src="https://3.bp.blogspot.com/-hJuGPbpkClA/WnL_WkMrVQI/AAAAAAAAAEY/Y_hEw-0AvlATG-ka7uJ-jU3He6i8UQQWQCLcBGAs/s320/alge2.png" width="320" /></a></div><span style="font-size: small;">The aim of this site is to publish tasks designed to help school students get a better feel for algebra.</span><br /><span style="font-size: small;">The plan is to post a new task at the beginning of a week, and then to post variations of the task during the week, together with comments and guidance. Let us know how the tasks work with your students.</span></div>ProfSmudgehttp://www.blogger.com/profile/00072867578929033601noreply@blogger.com0