An engineer signing off a bridge, or a programmer promising their code can never crash, cannot just try it a few times and hope — they need to be sure it holds in every case, forever. That gap between "tested a handful of times" and "guaranteed always" is exactly what a proof is for.

Try this: is the sum of two odd numbers always even? You check a few — 3 + 5 = 8, 7 + 9 = 16, 11 + 1 = 12. All even. So it's true, right?

Not proved. You have checked three pairs out of infinitely many. Maybe somewhere out among the billions there is a sneaky pair that breaks the rule. Checking examples gives you evidence — a good hunch — but never certainty.

A proof is different. It is an airtight logical argument showing something is always true, with no exceptions, ever. And the oldest, most direct kind is proof by deduction: you build from known facts and definitions, one watertight step at a time, until the conclusion is forced to be true.

The one clever move: let algebra stand for "any"

You can't check infinitely many numbers one by one. So instead you write a single symbol that represents every number of a given kind at once, and argue about that:

Here n and m are whole numbers. The letter 2n isn't one even number — it is the shape of every even number there is. Prove something about 2n and you have proved it for all of them in a single stroke. That is the whole engine of deduction: reason about the general case, and every specific case comes free.

Worked example 1 — the sum of two even numbers is even

Claim: add any two even numbers and the answer is always even.

Let the two even numbers be 2m and 2n, where m and n are whole numbers. Add them and factor out the 2:

Since m + n is a whole number, the answer is 2 \times (\text{a whole number}) — which is exactly the definition of even. And because m and n stood for any whole numbers, the claim holds for every pair. Done — forever. ∎

Worked example 2 — the sum of two odd numbers is even

Claim: the sum of two odd numbers is always even.

Let the two odd numbers be 2m + 1 and 2n + 1, where m and n are whole numbers. Add them and gather the terms:

Now m + n + 1 is a whole number, so the sum is 2 \times (\text{a whole number}) — which is exactly what it means to be even. Since m and n stood for any whole numbers, this proves the claim for every pair of odd numbers.

Worked example 3 — a statement about consecutive integers

Claim: the sum of any two consecutive integers is always odd.

Two consecutive integers are n and n + 1. Add them:

And 2n + 1 is precisely the algebraic shape of an odd number. So the sum is odd, whatever n you started with — no example survives as a counterexample, because we never used a specific number in the first place.

Notice the pattern in all three proofs: define the objects with symbols, do honest algebra, then read the answer against a definition (even means "2 × whole number", odd means "2 × whole number + 1"). Master that rhythm and most A-level deduction proofs fall over.

Worked example 4 — proving an identity

Sometimes the claim is that two expressions are equal for all values. You prove it by starting on one side and transforming it into the other by valid algebra.

Claim: (n + 1)^2 - n^2 = 2n + 1 for every integer n.

The left side became the right side using nothing but expansion and cancellation — steps that are valid for any n. So the identity is proved for all integers at once. (As a bonus, it quietly reveals something lovely: the gap between consecutive square numbers is always the next odd number — 1, 4, 9, 16, \dots jump by 3, 5, 7, \dots.)