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Secondary Magazine - Issue 128: It Stands to Reason


Created on 26 November 2015 by ncetm_administrator
Updated on 16 December 2015 by ncetm_administrator

 

Secondary Magazine Issue 128'The Thinker' by Japanexpertna.se
 (adapted), some rights reserved
 

It Stands to Reason

In Issue 127 we explored certain visual images that can help pupils develop conceptual understanding about factors and multiples, and hence reason about them with confidence and deeper perception. The easily observed structure of images reveals the much less obvious numerical structure that pupils can then exploit. Our aim is to help our pupils acquire the depth of understanding that enables them to use key relationships flexibly, so that they are not restricted to following without thinking algorithmic routines.

So let us continue from where we left off! Look at these two diagrams: the 15 square tiles of side-length 4 that fit in the 12-by-20 rectangle on the left have been rearranged to form the 4-by-60 rectangle on the right …

Having reasoned (see last issue) from the 12-by-20 rectangle, that 4 (which is now represented by the height of the 4-by-60 rectangle) is the highest common factor of 12 and 20 [HCF(12, 20)], pupils should then be asking themselves: what does the width of the 4-by-60 rectangle represent?

Give pupils time to think and discuss their thoughts about the answer to this question. The aim is for them to connect the width of the rectangle (as well as its height) to the numbers 12 and 20. They might find it helpful to make sketches, and then colour them in significant ways, such as these …

The key is seeing that …

60 = (3 × 4) × 5 = 12 × 5
60 = 3 × (4 × 5) = 3 × 20

… from which it follows (3 and 5 having no common factors) that the width of the 4-by-60 rectangle is the least common multiple of 12 and 20 [LCM(12, 20)]. If pupils know a definition of the least common multiple of two numbers p and q

The least common multiple of p and q, denoted LCM(p, q), is the smallest positive number, m, for which there exist positive integers np and nq such that p × np = q × nq = m.
[Wolfram Mathworld]

… they should be able to see, and articulate reasoning to explain, exactly why the width of a rectangle with area p × q and height equal to HCF(p, q) MUST BE the least common multiple of p and q. This is how they might reason about this example:

  • The area of a 12-by-20 rectangle is (3 × 4) × (5 × 4).
  • The smallest possible number which has factors 3 × 4 and 5 × 4 is 3 × 4 × 5, and 3 and 5 have no common factors. So LCM(12, 20) is 3 × 4 × 5.
  • We also know that 4 is HCF(12, 20).
  • If we draw a rectangle with the same area as a 12-by-20 rectangle, that is with area (3 × 4) × (5 × 4) = (3 × 4 × 5) × 4, but with height 4 (which is HCF(12, 20)), the width must be 3 × 4 × 5, which is LCM(12, 20).

One powerful consequence of this representation is that pupils can use relatively simple facts that they know about shapes to deduce a more complicated fact, one they probably did not previously know, about numbers:

  • Since both rectangles are formed with fifteen 4-by-4 squares they have the same area.
  • Therefore the area of the 4-by-60 rectangle is 12 × 20.
  • The area of a rectangle is its height × its width.
  • The height of the 4-by-60 rectangle is HCF(12, 20).
  • The width of the 4-by-60 rectangle is LCM(12, 20).
  • Therefore HCF(12, 20) × LCM(12, 20) = 12 × 20.

Thus they have deduced an example of the general relationship that

HCF(p, q) × LCM(p, q) = p × q.

This is the general reasoning (which pupils may not yet be ready to develop unsupported, but should be ready to explore with guidance):

Suppose HCF(p, q) is h.

  • Then p = a × h and q = b × h, where a and b have no common factors (otherwise h would not be the highest common factor of p and q.)
  • The area of a p-by-q rectangle is (a × h) × (b × h).
  • The smallest possible number which has factors a × h and b × h is a × b × h (remember a and b have no common factors). So a × b × h is LCM(p, q).
  • If we draw a rectangle with the same area as a p-by-q rectangle, that is with area (a × h) × (b × h) = (a × b × h) × h but with height h which is HCF(p, q)., the width must be a × b × h, which is LCM(p, q).

Cuboids

Quite different spatial reasoning about a 3-by-4-by-5 cuboid leads to, and so further illuminates, the same conclusion.

  • The numbers 12 and 20 are represented by the areas of pairs of opposite faces of the cuboid.
  • The HCF of 12 and 20, which is 4, is represented by one edge-length of the cuboid.
  • The ‘other’ factors of 12 and 20 (which are 3 and 5 respectively) are represented by the other two edge-lengths of the cuboid.
  • The volume of the cuboid, which is 5 × the area of the 3-by-4 face or 3 × the area of the 5-by-4 face then represents the LCM of 12 and 20, which is 3 × 4 × 5 = 60.

As in the previous example using rectangles, pupils have an opportunity to construct a chain of reasoning that shows why this finding is a necessary consequence of the given assumptions:

  • they know that 12 = 3 × 4 and 20 = 5 × 4, and that 4 is HCF(12, 20). By multiplying 3 × 4 by 5, or by multiplying 5 × 4 by 3, they must make the least common multiple of 3 × 4 and 5 × 4 because 3 and 5 have no common factors. So LCM(3 × 4, 5 × 4) = 3 × 4 × 5
  • the volume of a 3-by-4-by-5 cuboid is 3 × 4 × 5 (and by-the-way the areas of two pairs of opposite faces are 3 × 4 and 5 × 4)
  • so the volume is LCM(3 × 4, 5 × 4) = LCM(12, 20).

In general:

  • they know that, if p = a × h and q = b × h where h is HCF(p, q), then LCM(p, q) is a × b × h
  • the volume of a cuboid with edge-lengths a, b and h is a × b × h. So it represents the LCM of p and q.

Pupils should know that in order to multiply the volume of a cuboid by n, they must multiply just one dimension of the cuboid by n. Therefore to represent LCM(12, 20) × HCF(12, 20) we must stretch the cuboid by a factor of 4 in one direction only …

 
   

Both of the images of the cuboid-after-it-has-been-stretched-in-one-direction-only show that 4 × the volume of the original cuboid = 12 × 20. So they both show that HCF(12, 20) × LCM(12, 20) = 12 × 20.

Polygons in circles

Polygons created by joining equally spaced points on the edge of a circle are a different set of visual images that can help pupils see, and reason about, highest common factors and lowest common multiples.

When pupils consider 12 equally spaced points marked on one circle, and 18 on another …

… they can look for the greatest possible number of sides of a regular polygon that can be drawn in both circles with all its corners on marked points. That number of sides must be the greatest factor of both 12 and 18 (which are the number of points on each circle); that is, it must be HCF(12, 18). So, by making sketches of regular polygons in both circles, they will see that HCF(12, 18) = 6:

The number of equal spaces on the circle’s edge between each joined dot represents in one circle the fact that 12 = 2 × 6, and in the other that 18 = 3 × 6. If on the circle with 2 spaces between joined dots we now split every one of these spaces into 3 equal parts, and on the circle with 3 spaces between joined dots we now split every one of these spaces into 2 equal parts, we create the same total number of spaces on both circles. What does that number of spaces represent?

If we now split every space on both circles into 6 (the HCF of 12 and 18) equal parts [as we have done in the following diagrams – although the individual marks are difficult to see!], the total number of spaces is the product of 12 and 18.

These images illustrate the fact that HCF(12, 18) × LCM(12, 18) = 12 × 18.

Having reached the same conclusion about how the HCF, the LCM and the product of two particular numbers are related by reasoning about features of three completely different geometric images, your pupils should be convinced that it is true for those particular numbers. In the first two image-contexts pupils used simple geometrical facts that they knew to support them in reasoning as follows:

  • 12 = 3 × 4 and 20 = 5 × 4
  • HCF(3 × 4, 5 × 4) [which is HCF(12, 20)] = 4
  • 3 and 5 have no common factors
  • LCM(3 × 4, 5 × 4) [which is LCM(12, 20)] = 3 × 5 × 4
  • So HCF(12, 20) × LCM(12, 20) = 4 × (3 × 5 × 4) = (4 × 3) × (5 × 4) = 12 × 20

The third set of images helped them to reason that:

  • 12 = 6 × 2 and 18 = 6 × 3
  • HCF(2 × 6, 3 × 2) [which is HCF(12, 18)] = 6
  • 2 and 3 have no common factors
  • LCM(2 × 6, 3 × 6) [which is LCM(12, 18)] = 2 × 3 × 6
  • So HCF(12, 18) × LCM(12, 18) = 6 × (2 × 3 × 6) = (6 × 2) × (3 × 6) = 12 × 18

By focusing on, and discussing, the structure that is common to the numerical reasoning stimulated by visual images such as we have discussed, pupils can be supported to understand, and possibly construct using algebraic notation, the general argument below. They will see that in order to justify the general conjecture …

HCF(p, q) × LCM(p, q) = p × q

… they have always reasoned as follows:

  • Suppose p = a × h and q = b × h, where h is HCF(p, q) and a and b have no common factors
  • Then LCM(p, q) = a × b × h
  • HCF(p, q) = h, so HCF(p, q) × LCM(p, q) = h × a × b × h
                                                              = (a × h) × (b × h)
                                                              = p × q.

Your pupils are more likely to suggest, and follow, and recall, this general abstract argument after they have had sufficient opportunities to reason about concrete and pictorial representations of the HCF, LCM and product of particular values of p and q in particular visual images. In using such images to reason about pairs of numbers such as 12 and 20, pupils have first specialised in order to develop a clearer sense of the general argument.

Not only is the general relationship between the highest common factor, least common multiple and product of two numbers a conclusion that pupils can be challenged to reach through reasoning – and thereby deepen their understanding – but, once established and therefore known by them, it is a fact that they can and should use: their conceptual understanding should lead them to procedural fluency. For example, suppose that they want the least common multiple of 27 and 60. If they remember that LCM = product ÷ HCF, they can simply multiply 27 × 60 = 1620 and divide by the HCF, which is easily seen to be 3 (recall that the HCF must be a factor of the non-zero difference between any two multiples of the original numbers, and 60 – 27 × 2 = 6). LCM(27, 60) = 540 then follows.

Image credit
Page header by Japanexpertna.se (adapted), some rights reserved

 
 

 

 
 
 
 
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