Note: GERMAN versions of the Week 16 tasks are available HERE.
In this week's tasks we consider the squares of numbers, as we did in Week 2, but this time with a more sustained focus on the difference of two squares. Formally, this relation can be derived from the fact that when we expand the brackets in (a + b)(a – b), the two ab terms cancel out and we are left with a² – b². However, we take a more grounded approach in this set of tasks, in that we infer the rule from specific numerical cases, in particular cases where a – b = 1. We then suggest a way of verifying the rule (for a – b = 1) using an argument based on structural arithmetic.
Monday: The desired value here is 32.6 + 31.6 = 64.2, using the difference of two squares. However, we don't expect students to know the relation at this point, which makes the task much more challenging.
One powerful approach is to use structural arithmetic:
32.6 × 31.6 = 31.6² plus an extra 31.6;
32.6 × 32.6 = 32.6 × 31.6 plus an extra 32.6
An equivalent approach is to use the area model: we can think of a 32.6 units square as consisting of a 31.6 units square with a 31.6 by 1 strip (blue) along one edge, and then a 32.6 by 1 strip (red) along an edge perpendicular to the first edge.
Tuesday: This mirrors the Week 2 Monday task. Do students have a sense that the difference between the squares of consecutive (or equally spaced) numbers increases as the numbers increase?
One approach that students might use is to consider some familiar square numbers and generalise from that. For example, the squares of the (positive) numbers 1, 2, and 3 are 1, 4, and 9, and 4 is closer to 1 than to 9. This suggests that 30² is closer to 29² than to 31².
Of course, we can’t always make a valid generalisation from a single case, though intuition suggests it must work here: how could the squares of other (positive) consecutive numbers possiby behave differently?! We can prove the result by, for example, using the area-model diagram in task 02B, or algebraically, as here:
The squares of the (positive) consecutive numbers a, a+1 and a+2 are
a², a² + 2a + 1 and a² + 4a + 4,
so the differences are 2a + 1 and 2a + 3,
and 2a + 3 > 2a + 1 (and |2a + 3| > |2a + 1| when a is positive).
Note: In the task we deliberately refer to the expressions 292, 302 and 312 as numbers, to make the point that we can think of expressions as numbers even when we haven’t found their value. You might need to clarify this for students.
Wednesday: Jay's equations confirm that 30² is closer to 29² than to 31² (Tuesday's task). Jay's rule can be expressed as a² – b² = a + b (for the kinds of numbers we are dealing with here, ie where a – b = 1).
It is likely that students will be able to spot Jay's rule (though they might not express it in such a formal way). Explaining why the rule works is more challenging. One approach that students might use is to again turn to more familiar numbers and to confirm that the rule works there too. But, of course, confirming a rule is not the same as proving it. We examine a structural argument that could form the basis of a proof in Thursday's task.
Thursday: This is a highly structured task and it is possible that some students who manage to complete the steps will do so without much insight. You might want to probe students' understanding by asking them to execute the steps for another pair of consecutive square numbers and, more revealing perhaps, for a pair of non-consecutive square numbers (as in Friday's task).
Friday: Here we extend the work by considering the squares of numbers whose difference is 2 rather than 1. You might want to extend this further by considering the squares of any two (whole?) numbers.
One way to explain Elle’s rule is to use Thursday’s structural arithmetic approach. So, for 262 – 242, say, whose value can be expressed as 2(26 + 24),
we can write this:
24×26 is 24×2 or 2×24 more than 242.
26×26 is 2×26 more than 24×26.
So 262 is 2×24 + 2×26 = 2(26 + 24) more than 242.
