Building Bridges
New Year, New Year’s resolutions, time for changes (ch … ch … changes) … so let’s be contrary and consider things not changing: the fact that 3 + 5 – 5 = 3, and that 3 × 5 ÷ 5 = 3. Let’s call these examples of, respectively, additive and multiplicative invariance. Additive invariance is a concept that most secondary pupils have grasped in KS2: it is not a surprise to them that 3 + 5 – 5 = 3. It is, nonetheless, well worth ensuring that your pupils are absolutely confident with less straightforward examples of the form a + b – a and a – b + c + b, and also those of the form a – b + c – d where b + d = a (or c). To confirm that your pupils are fluent with additive invariance, ask them a calculation such as (673 + 356) – (675 + 354): do they reason that the answer must be 0, because a sum of two terms is invariant if one term increases by an amount and the second term decreases by the same amount?
Part of the reason for looking at calculation strings such as these is to challenge your pupils’ likely misunderstanding of the unhelpful cliché “BODMAS”: they may think (indeed, they may have been told) that “you work out addition before subtraction” and so they think that they have to evaluate 3 + 5 – 5 as 8 – 5 = 3, rather than 3 + 0 = 3. The difference in efficiency will be stark when you ask them to evaluate 2368 – 9674 + 1332 + 9674! A better mnemonic for the order of operations is to think of a pyramid or a podium for gold, silver and bronze medals (a mnemonic that will work especially well in 2016!):
They need to be confident when, and why, calculation strings can be reordered, for example
 14 + 6 – 9 = 14 – 9 + 6 = 6 – 9 + 14 etc.
 8 × 6 ÷ 4 = 8 ÷ 4 × 6 = 6 ÷ 4 × 8 etc.
 14 + 6 – 9 ≠ either 14 – 6 + 9 or 9 + 6 – 14
 8 × 6 ÷ 4 ≠ either 8 ÷ 6 × 4 or 6 × 4 ÷ 8
 30 ÷ 3 ÷ 2 = 30 ÷ 2 ÷ 3, but not 3 ÷ 30 ÷ 2.
The order of operations is reflected in the medal winners’ hierarchy: gold before silver before bronze: 8 + 6 ÷ 2 = 11 not 7, and 5 × 32 = 45 not 225. This podium layout also captures distributivity: pupils see that × is literally written over + and –, and so they recall that multiplication is distributive over addition and subtraction, hence 3 × (a – b + c) ≡ 3a – 3b + 3c. Similarly, they see that division is written over addition and subtraction, and this should help them recall that
12 + 153 = (12 ÷ 3) + (15 ÷ 3) = 4 + 5, and that x − x^{2}x ≡ (x ÷ x) − (x^{2} ÷ x) ≡ 1 − x.
In contrast, 3 × (4 × 5) ≠ (3 × 4) × (3 × 5) because multiplication is not distributive over multiplication: × is not written over ×.
It is likely that your pupils will be less confident with multiplicative invariance, other than in straightforward strings such as 6 × 4 ÷ 4 and 6 ÷ 4 × 4. Calculations such as 7 ÷ 3 ÷ 2 × 24 and 632 × 123 – 316 × 246 will generate lots of worthwhile discussion and reasoning. Confident mastery of multiplicative invariance will enable your pupils to reasons in contexts where its applicability is not immediately apparent: if the density of metal G is 25% bigger than the density of metal S, by what % will the volume of a block of metal G be smaller than a block of metal S if the two blocks have the same mass?
The area representation of multiplication should help develop pupils’ understanding of multiplicative invariance. If your pupils think of the picture of a (real) 12 by 4 rectangle when considering the (abstract) multiplication 12 × 4, then they can see (with either real or abstract scissors) that doubling the width and halving the length keeps the area the same, i.e. that 12 × 4 = 6 × 8.
More complicated examples can be tackled either numerically or algebraically:
This experiment shows that if one factor is increased by a half (and so multiplied by a scale factor of 3/2), and the second factor is decreased by a third (and so multiplied by a scale factor of 2/3), then the product of the two factors – the area of the rectangle – will be unchanged. Once pupils have seen this represented pictorially (and, ideally, experienced it in the concrete with paper, scissors and glue), then the identity
a × b ≡ 23 a × 32 b
is easy for pupils to understand and explain. They can then conjecture, and prove, the result needed in the % context cited earlier: increasing the density by 25% is the same as multiplying it by 5/4, and therefore multiplying the volume by 4/5, i.e. decreasing it by 20%, will keep the product, which is the mass, invariant.
In conclusion they can articulate the general result: the product of two factors is invariant whenever one factor is multiplied by a scale factor and the second factor is multiplied by the reciprocal of the same scale factor. The mathematical language is spot on; for song lyrics, though, I’ll stick with Bowie.
Image credit
Page header by PauliCarmody (adapted), some rights reserved
