Here we further pursue the idea that multiplication can be thought of as scaling and apply this idea explicitly to expressions like 2⅔ × 2⅘ and ⅔ of ⅘.
Task 15A: Here scaling provides a vivid representation of multiplication, and one that is probably quite accessible to many pupils. However, many pupils will find it very difficult to evaluate the result precisely when the multipliers (or scale factors) are fractional rather than whole numbers.
For Aysha's modelling of 3×2, it is fairly easy to see that doubling the width of the letter A (and the square that surrounds it) and then trebling the new width, results in a letter (and rectangle) that is 6 times as wide as the original.
However, it is not so easy to find the precise value 2⅔ × 2⅘ using Aysha's approach. Before pupils try to do so, it is worth asking them to estimate the value based on the numbers in the expression and based on Aysha's diagram. The value will be greater than 2×2 = 4 and less than 3×3 = 9, and probably closer to the latter. The diagram suggest that the value is slightly less than 7½.
If we consider the scale above and the scale below Aysha’s scaled shape, the 1×2⅘ (or 14/5) mark on the top scale must coincide with the 1 (or 3/3) mark on the bottom scale. So if we split each 5th on the top scale into 3 equal parts, and split each 1/3 interval on the bottom scale into 14 equal parts, then the scales will have the same-size subdivisions: there will be 42 such subdivisions on the top scale between 0 and 2⅘, and the same number on the bottom scale between 0 and 1. These subdivisions will be 15ths (because there are 15 of these parts between 0 and 1 on the top scale). As there are eight 1/3 intervals along the bottom scale, each of which holds 14 of these subdivisions, a total of 8×14 of the subdivisions will fit along the width of the scaled figure and so its width is 8×14×⅟₁₅ = 112/15 = 7⁷⁄₁₅. These same numbers come into play when we evaluate 2⅔ × 2⅘ formally, by writing the mixed numbers as improper fractions:
2⅔ × 2⅘ = ⁸⁄₃ × ¹⁴⁄₅ = (8×14)/(3×5) = 112/15 = 7⁷⁄₁₅.
For the product 2⅔ × 2⅘, Eric's final shape can be partitioned into 8×14 small, equal parts (below, right). Fifteen of these parts cover Eric's original shape, so they represent 15ths. 8×14 15ths = 112 15ths = 7⁷⁄₁₅.
A bonus version for my grandson on his birthday.....
Task 15C: This again involves scaling, but this time we use 'of' to indicate 'multiply'. [We could have done this in the previous tasks too, for example by replacing '2⅔ × 2⅘' by '2⅔ lots of 2⅘', or even by '2⅔ of 2⅘' though this sounds a bit odd. Also, 'lots of' hints at repeated addition, rather than a single element being stretched.]
We can 'equalise' the scales above and below Aysha's final shape by splitting each interval in the top scale into 3 equal parts and each interval in the bottom scale into 4 equal parts. Fifteen of these new equal parts fit into the interval from 0 to 1 on the top scale, so each represents one fifteenth. And 2×4 = 8 of these 15ths fit along the width of the final shape, so it is ⁸⁄₁₅ units wide.
Formally, we can write
⅔ of ⅘ = (2×4)/(3×5) = ⁸⁄₁₅.
Task 15D: This involves the same product as in the previous task, but the result is represented by an area (as in Task 15B).
Task 15E: This is similar to Task 15D, but this time one of the multipliers (scale factors) is greater than 1 and the result is also greater than 1, so the result is a mixed number or improper fraction.
so 2²⁄₇ of ⅘ = 64/35 units = 1²⁹⁄₃₅ units.
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