Monday, 28 May 2018

ALG 16

This week we look at the structure of arithmagons. 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.
MONDAY: 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.
Yes, the intro is rather subtle, but we don't want to over-excite our muddled friend Capt Scarlet...

We can solve the question 'is there enough flour' by looking at the structure of the problem, which  means thinking algebraically. We can take a narrative approach, in this kind of way:
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.
Or we can express this with symbolic algebra, in this kind of way:
a + b + a + c = 130 + 160 = 290,
so the total number of grams, a + b + c = 290 – a,
so if a ≤ 40, a + b + c ≥ 250.
TUESDAY: The arithmagon quietly makes an appearance, but we stick with the Peter pancakes story for now. [But it's not pancakes for ever.]
We now know the combined weight for each pair of bags and we look at an approach for hence finding the total weight of the bags - and the weight of any one bag.
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 a, b or c turns out to be negative.... as we shall see.
WEDNESDAY: We can relax. The arithmogadon is here.
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 a, b and c. On Tuesday, we found the value of a + b + c (which then allowed us to find the value of each individual unknown) by adding the three pair-sums and halving:
a+b + b+c + c+a = 2(a + b + c), so a + b + c = (130+160+150)÷2 = 220.
This time we use the fact that c must be 30 more than b, and that the sum of b and c is 150, to find c and hence to find a + b + c.
If we are working formally, we can express this information as c = b + 30 and b + c = 150, which we can then combine and transform in this kind of way: c – 30 + c = 150, so 2c = 180, so c = 90.
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. 
THURSDAY: Here we look at that first scenario again, but in a more systematic way and without the 'real life' context.
We bring out the fact which may well have been noticed in ALG 16A, that the smaller the value of a, the greater the value of the sum a + b + c. [Formally, this can be said to stem from the fact that a+b + a+c = 130+160 = 290 = (a+b+c) + a. And as a increases by 1, the value of a + b + c decreases by 1.]
We also look at the limiting values of a for which the three 'vertices' (a, b and c) are all positive:
when a = 0, b = 130 ans c = 160 and the sum S = a + b + c = 0 + 130+160 = 290;
when a = 130, b = 0 and c = 30, and S = 130 + 0 + 30 = 160;
note: when a > 130, then b is negative and c is too when a passes 160.
What we haven't done here explicitly is look at the 'permissible' values of the three sums a+b, b+c and c+a, ie the value for which a, b and c are all positive. However, from the previous paragraph we can show that b+c will range from 30 to 290, with, of course, a+b=130 and a+c=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.
FRIDAY: We switch from an additive to a multiplicative arithmagon. Does Armageddon loom?
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 u, v, w taken in pairs, though part ii. is likely to prompt one to use formal algebra.
We can solve b)ii using an approach analogous to Tuesday's or Wednesday's approach.
Tuesday's approach: uvw = abbcca = (abc)², so  a² = uvw÷v² = uw/v.
Wednesday's approach: u/v = ab/bc = a/c; wu/v = aca/c = a².

Monday, 21 May 2018

ALG 15

We continue with Cartesian graphs for another week - sorry, bad planning! (But worth it, hopefully.)
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.
We go on to relate the visual to the symbolic by expressing the graphs symbolically and adding or subtracting the relations.
MONDAY: Here we have two linear functions, f(x) and g(x), 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 y = f(x) – g(x) cuts the y axis at the same point as f(x) cuts the axis [Why?], and cuts the x axis directly below the point of intersection of the given graphs [Why?].
It is interesting to note that f(x) is not as steep as g(x). What does this tell us about the slope of y = f(x) – g(x)?
TUESDAY: Here we are given some feedback on Monday's task and have the opportunity to consolidate our earlier ideas.
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.
We can use the symbolic representations in various ways. For example, knowing that f(x) = x + 10 and that g(x) = 2x, we can state that y = f(x) – g(x) = x + 10 – 2x which simplifies to y = 10 – x. We can then use this symbolisation to check whether our original sketch (ie the purple line) is correct, or, we could derive the symbolisation from the sketch, given that we now can see that the purple line goes through points with coordinates (0, 10), (10, 0) and (20, -10)
In the case of the function y = f(x) + g(x), 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 x and 2x, whose sum is 3x.
WEDNESDAY: Here we start with a very familiar straight line graph (of the function y = x) and 'perturb' it by adding a second, 'wilder' function. Interestingly, though, the effect of this second function is quite localised....
Note: I came across this lovely idea in Abraham Arcavi's 1994 Symbol Sense paper in For the Learning of Mathematics, 14, 3.
THURSDAY: 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/(x – 4) has a very large effect on the value of y when x is very close to 4, but its effect rapidly diminishes as we move away from x = 4. We then get a chance to build on this by sketching the graph of a closely related function.
The green curve below shows the graph for the new function. Since 2x – 8 = 2(x – 4), the term 1/(2x – 8) has only half the effect of the term 1/(x – 4).
FRIDAY: Here we subtract a linear function from a quadratic function.
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)?

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.

Next week: The Arithmagons are coming: Captain Scarlet's bitter foes (or was that the Mysterons?).

Sunday, 13 May 2018

ALG 14

This week's set of tasks was inspired by a task (below) from Doug French, that featured in his series The creative use of odd moments that appeared in Mathematics in School. (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.)
 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.
It is of course possible to find each line's equation by using the formal procedure of finding the values of m and c in y = mx + c. However, the way the tasks have been designed should encourage students to think about the relationship between the coordinates x and y 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 y = mx + c, but also x + y = a, when it applies, and perhaps more generally ax + by = c).
MONDAY: We start with a nice little diamond that is fairly near, but not too near, the axes...
The lines have gradient 1 or -1, so we can find the equations by seeing where the various lines cut the y-axis (giving us the value of c in y = mx + c). However, we can also (or instead) work more 'locally', by looking for the relationship connecting the x and y coordinates - eg for points A (8, 8) and D (8–2, 8+2), we can see that x + 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 y = x), ie the y coordinate has been increased by +4, for a given x coordinate, so y = x + 4 instead of y = x.
TUESDAY: We see what happens to the equations when we translate the square one unit up or across.
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.
For each line, we can again look for the (modified) relations between x and y, and/or visualise how the lines have moved and consider, in particular, where they cut the axes.
Notes: 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 x + y = 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 x, the value of y has increased by 1, so the sum x + y has increased by 1, from 16 to 17.
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 x + y = 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 x + y = 16 to x + y = 18, rather than to x + y = 17.
WEDNESDAY: 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, e, is a parameter rather than a variable.] We can of course check our new equations by letting e take the value 2.
The last part of the task throws up an interesting, if rather arcane, issue. Is the square resulting from the value e = -4 on and above the line y = x, or on and below? 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 'below'. 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 y = x ?
THURSDAY: We consider a small square with a different orientation from the 'diamonds' we've considered so far - so no longer a little gem?
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.
FRIDAY: 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...).

Tuesday, 8 May 2018

ALG 13

This 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.
MONDAY: We start with a standard linear relationship and multiply the x- and/or the y-values by 4.
Some students might be able to approach this formally. For example, in the case of Table B, we can think of the 'new' x (denote it by x', say) as replacing 4 times the old x, ie x' = 4x, and so y = 8x + 1 = 2×4x + 1 becomes y = 2x' + 1.
Or, less formally, "the new x-values are 4 times the old x-values, so multiplying the old x-values by 8 is the same as multiplying the new x values by 2; so y = 8x + 1 becomes y = 2x + 1".
Or students might notice, by examining the numbers in Table B, that 17 = 2×8 + 1, 41 = 2×20 + 1, etc.
IMPORTANT NOTE: 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.  
TUESDAY: Here, we start with the same relationship as in ALG 13A, but instead of multiplying by 4, we add 4 to the x- and/or the y-values.
Again, students might adopt a formal, a semi-formal, or an empirical approach. In the case of Table E, the relation changes to y = 8(x – 4) + 1, or y = 8x – 31.
WEDNESDAY: Here we start with a new linear relationship that is perhaps easier to describe succinctly.... Again, we multiply the x- and/or the y-values by 4.
The relationship for Table J is easy to spot or derive, the one for Table I less so.
THURSDAY: We stick with Wednesday's relationship, and as we did earlier, we add 4 to the x- and/or the y-values. Finding the resulting relationships is not particularly challenging on this occasion but we include the task for completeness.
FRIDAY: 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.
Note: 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.