Algebraic Proof

Grade 7

Why a single example is never enough

In maths, showing that something works for one number is not a proof. You need to show it works for every number. That is where algebraic proof comes in — you work with general expressions (like 2n for any even number) rather than specific values.

Key building blocks for proofs

You need to know how to express integers algebraically:

Any integer: n. Even number: 2n. Odd number: 2n + 1. Consecutive integers: n, n + 1, n + 2. Consecutive even numbers: 2n, 2n + 2, 2n + 4.

If your proof ends with an expression of the form 2(something), you have proved the result is even. If it ends as 2(something) + 1, you have proved it is odd. Keep that structure clear.

Worked example

EXAMPLE

Prove that the product of two consecutive odd numbers is always odd.

Let the two odd numbers be 2n + 1 and 2n + 3. Their product: (2n + 1)(2n + 3) = 4n² + 6n + 2n + 3 = 4n² + 8n + 3 = 2(2n² + 4n + 1) + 1. This is of the form 2(integer) + 1, so it is odd. Proved.

Disproof by counter-example

To disprove a claim, you only need one counter-example — a single value that makes the statement false. Finding one is enough; you do not need a full algebraic argument.

Try this one

Prove that for any integer \( n \), the expression \( n^2 + n \) is always even.

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Nova's hint:

This is an algebraic proof using factorisation and properties of consecutive integers.

Ask yourself: 'Can I factorise \( n^2 + n \) and what do I know about the product of two consecutive integers?'

For example, \( 3 \times 4 = 12 \) and \( 4 \times 5 = 20 \) are both even — why is this always so?

Try to Solve it with Nova?