Pure · Proof · Year 1

Proof by Deduction

The workhorse of A-Level proof: start from something known to be true, argue in careful algebraic steps, and arrive at the result — with no gaps.

8 min read 3 worked examples

What "proof by deduction" means

Deduction means reasoning forward from statements you already accept as true — definitions, given facts, or established results — through valid algebraic steps, until you reach the statement you were asked to prove. Every line must follow logically from the ones before it. If a marker can ask "why is that true?" at any step and you can't answer, the proof has a gap.

This is the most common form of proof at A-Level, and it's what's expected unless a question specifically asks for proof by exhaustion or disproof by counter-example.

The three things a deductive proof needs

1. A clear starting point — something known, or a sensible way to represent the objects (e.g. "let an even number be $2n$").
2. Connected steps — each line follows from the last by algebra or a stated fact.
3. A conclusion that matches the claim exactly — end by stating the thing you were asked to show, ideally with a closing line like "which is what we needed to prove."

Representing numbers algebraically

Most deductive proofs start by writing the objects in the claim in a general algebraic form. Get these right and half the work is done:

An even number is $2n$. An odd number is $2n+1$. Two consecutive integers are $n$ and $n+1$. A multiple of 3 is $3n$. The key idea: $n$ stands for any integer, so proving it for $2n$ proves it for every even number at once.

Worked example 1

Prove that the sum of two consecutive integers is odd

Show that for any integer $n$, the sum of $n$ and $n+1$ is always an odd number.

ProofSum of two consecutive integers

Let the two consecutive integers be $n$ and $n+1$, where $n$ is an integer.

This represents any pair of consecutive integers.

Their sum is $n + (n+1) = 2n + 1$.

Collecting like terms.

$2n$ is even, so $2n+1$ is one more than an even number — which is odd.

By the definition of an odd number.

Therefore the sum of two consecutive integers is always odd.

Matches the statement we were asked to prove.

▪ proven

Examiner's eye

The mark isn't just for reaching $2n+1$ — it's for the final sentence that connects it back to "odd." Students routinely lose the last mark by stopping at the algebra without concluding.

Worked example 2

A harder one · uses expansion and factorising

Prove that the difference between the squares of two consecutive integers is equal to the sum of those two integers.

ProofDifference of consecutive squares

Let the integers be $n$ and $n+1$.

General representation of consecutive integers.

The difference of their squares is $(n+1)^2 - n^2$.

Larger square minus smaller.

$(n+1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1$.

Expanding and simplifying.

The sum of the two integers is $n + (n+1) = 2n+1$.

Adding them directly.

Both expressions equal $2n+1$, so they are equal.

The difference of squares equals the sum — as required.

▪ proven

Worked example 3

Exam-style · algebraic manipulation with a rational expression

Prove that for all real values of $x$, the expression $x^2 - 6x + 11$ is always positive.

ProofA quadratic that is always positive

Start with $x^2 - 6x + 11$ and complete the square.

Completing the square reveals the minimum value.

$x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x-3)^2 + 2$.

Halve the coefficient of $x$, square it, and adjust.

$(x-3)^2 \geq 0$ for all real $x$, since any real number squared is non-negative.

A square is never negative.

So $(x-3)^2 + 2 \geq 2$, which is always greater than $0$.

Adding 2 to something $\geq 0$ gives something $\geq 2$.

Therefore $x^2 - 6x + 11 > 0$ for all real values of $x$.

The claim is proven for every real $x$.

▪ proven

Why this technique

Completing the square is the standard deductive route for "always positive / always negative" proofs. The key insight is step 3 — spotting that a squared term can't be negative — which turns an algebraic expression into a watertight argument.

Common mistake

Testing a few numbers (like $3^2 - 2^2 = 5 = 3+2$) is not a proof — it only shows it works for those cases. A deductive proof must use a general $n$ so it holds for every integer. Checking examples is a good way to see what's going on, but never counts as the proof itself.

Check yourself

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

Let the three consecutive integers be $n$, $n+1$ and $n+2$.

Their sum is $n + (n+1) + (n+2) = 3n + 3$.

$3n + 3 = 3(n+1)$, which is a multiple of 3.

Factor out 3.

Therefore the sum of any three consecutive integers is a multiple of 3.

▪ proven

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Dynamics and Inclined Planes

Proof by Exhaustion