



Tools  Math 'Convincing and Proving' Critiquing 'Proofs' Tasks, Set #4 (solutions)
Always, Sometimes or Never True: Set #1 (solutions)  Set #2 (solutions) Critiquing 'Proofs': Set #3 (solutions)  Set #4 (solutions)
Malcolm Swan
Jim Ridgway
1. The Pythagorean' Theorem Here are three attempts to prove the Pythagorean theorem. When you add add three consecutive numbers, your answer is always a multiple of three. Look carefully at each attempt. Which is the best 'proof'? Explain your reasoning as fully as possible.
The best proof is attempt number ..........
Attempt 1 is a sufficient demonstration where the sides of the triangle are in the ratio 3:4:5 Attempt 3 is sufficient when the sides are in the ratio 1:1: Attempt 2 provides the only dissection which works for any right angled triangle.
Here are two attempts to prove the following result about the arithmetic mean and geometric mean of two numbers: Look carefully at each attempt. Which is the best 'proof'? Explain your reasoning as fully as possible.
The best proof is attempt number ..........
Attempt 2 is correct. Attempt 3 is the basis for a very elegant proof, but there are some holes and unnecessary jumps in it at present. The first is that it is always dangeraous to argue from adiagram because the diagram does not show all possible cases. In this example, the shaded area is clearly greater than 0 , but what if a and b were equal? In this case, the large square would be made entirely of four small squares and the shaded area would be zero. The diagram currently assumes that a > b. The special case when a = b needs to be considered. The final result thus should become instead of . The diagram suggests that the general identity would explain why for all possible values of a and b.
Here are three attempts to prove the following result for natural numbers. Look carefully at each attempt. Which is the best 'proof'? Explain your reasoning as fully as possible.
The best proof is attempt number ..........
Attempt 2 assumes what we are asked to prove (see the first line). It could, however, be modified to form a proof in the following way:
Let y = n(n + 1) = n^{2} + n. n or n+1 must be even since they are consecutive numbers. So y must be the product of an odd and even number. It is therefore even. So n^{2} + n must be even. Now if n^{2} is odd, then n (= y  n^{2}) must also be odd. (Since evenodd = odd)
Critiquing 'Proofs': Set #3 (solutions)  Set #4 (solutions)

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