Algebraic proof

Objective. Prove parity, divisibility and consecutive-integer results.

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Learn: the key idea

Algebraic proof uses symbols to represent general numbers so a statement is shown true for every case, not just examples. Even numbers are written 2n, odd numbers 2n + 1, and consecutive integers n, n+1, n+2, for any integer n. To prove a parity or divisibility result, expand the algebraic expressions and show the result factorises into the required form, e.g. proving the sum of two consecutive integers is always odd: n + (n+1) = 2n + 1, which matches the odd-number form 2k+1. Divisibility proofs similarly factor out the required number, such as showing 6n + 9 = 3(2n+3) is always a multiple of 3.

Worked example

Prove algebraically that the sum of three consecutive integers is always a multiple of 3.

1 $Let the integers be n, n+1, n+2$
2 $Sum them: n + (n+1) + (n+2) = 3n + 3$
3 $Factorise: 3n + 3 = 3(n+1)$
4 $Since 3(n+1) is 3 times an integer, the sum is always a multiple of 3$

Answer: 3(n+1), a multiple of 3

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Common mistake. Students often 'prove' a statement using only one numeric example, e.g. checking 3+4=7 is odd and stopping there; a genuine proof needs general algebraic expressions like 2n+1 covering every possible integer case.

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