Algebraic Proof

1. 1 Claim: 'The sum of two odd numbers is always even.' Test: 3+5=8 ✓, 7+11=18 ✓, 101+203=304 ✓. 'Have we proved it?' (No — we have only shown examples.) 'How many examples would be enough to prove it for ALL odd numbers?' (None — there are infinitely many.)
2. 2 Any even number = 2n. Any odd number = 2n+1 (where n is an integer). 'Why does 2n represent ANY even number?' Display representations: consecutive integers (n, n+1, n+2), consecutive even (2n, 2n+2), consecutive odd (2n+1, 2n+3).
3. 3 Prove: the sum of two odd numbers is always even. Let the two odd numbers be 2a+1 and 2b+1. Sum = 2a+1+2b+1 = 2a+2b+2 = 2(a+b+1). Since a+b+1 is an integer, 2(a+b+1) is even. ∎ 'This works for ALL odd numbers simultaneously.'
4. 4 Students annotate the proof: (1) Define variables — what does a and b represent? (2) Express in algebraic form. (3) Simplify. (4) Conclude — why does the final expression prove the claim? (5) QED/∎.
5. 5 Prove: (a) The sum of three consecutive integers is divisible by 3. (b) The square of an odd number is always odd. (c) The product of two even numbers is divisible by 4. Students work in pairs, then share proofs for peer critique.
6. 6 Flawed proof: 'Claim: 1+2 = 1. Let a = b. Then a² = ab. a²−b² = ab−b². (a−b)(a+b) = b(a−b). Divide both sides by (a−b): a+b = b. Since a=b: 2b=b. Divide by b: 2=1.' Where is the flaw? (Dividing by a−b, which equals zero.)
7. 7 Prove: the difference between the squares of two consecutive odd numbers is always divisible by 8. Students must express consecutive odd numbers (2n+1 and 2n+3) and expand.

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