Proof with consecutive integers

Objective. Prove parity and divisibility properties algebraically.

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Learn: the key idea

Proof with consecutive integers uses algebraic expressions like n, n+1, n+2 to represent integers in sequence, allowing general statements about sums, products or parity to be proven for all cases at once. Represent the integers algebraically, perform the required operation (sum, product, difference), then factorise or simplify to reveal the required property such as always being even or always a multiple of 3. For the sum of three consecutive integers n, n+1, n+2, the total simplifies to 3n+3 = 3(n+1), proving it's always a multiple of 3. This method proves a property for every possible integer, not just a few tested examples.

Worked example

Prove that the sum of two consecutive even integers is always a multiple of 4 only sometimes; instead prove it is always even.

1 $Let the integers be 2n and 2n + 2$
2 $Sum = 2n + 2n + 2 = 4n + 2$
3 $Factorise as 2(2n + 1)$
4 $This is 2 \times an integer, so the sum is always even$

Answer: The sum 4n + 2 = 2(2n + 1) is always even

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Common mistake. Students often 'prove' a statement by testing only a few numerical examples, like checking n = 1, 2, 3, rather than using general algebraic expressions, which cannot guarantee the result for all integers.

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