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Proof by Deduction

Q: How do I know what to factorise towards?

Read the target. “Even” means aim for 2( ); “multiple of 3” means aim for 3( ); “always positive” means aim for a completed square plus a positive constant. The destination dictates the manipulation.

Q: Why does completing the square prove an inequality when substituting numbers doesn’t?

A squared bracket is 0 for every real number simultaneously, so (x-a)^2 + b b holds for all x in one stroke. Substituting values only ever checks finitely many points and can never cover them all.

Q: Is it okay to split into cases in a deduction?

Yes — arguing “if n is even… if n is odd…” is a legitimate deductive proof (and blends into proof by exhaustion when the cases are finite). Just make sure your cases together cover every possibility.

Proof by deduction is the direct route: start from a general case, do honest algebra, and rearrange until the property you want is staring back at you. It is the technique behind most “Show that…” marks on Paper 1.

30 min Video by TLMaths Proof

The big picture

If the previous lesson gave you the vocabulary of proof, deduction is the first full sentence you build with it. Nearly every “prove” question that is not obviously a counter-example or a contradiction is a deduction — you take a general object, transform it with algebra you already trust, and reveal that it must have the required property. The skill is not the algebra itself; it is choosing the manipulation that makes the answer visible.

A-Level Maths A1-06: Introducing Proof by DeductionWatch the walkthrough, then work through the notes below.

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What you'll be able to do

Prove a general statement directly by starting from a suitable algebraic form
Reveal “even”, “odd” or “multiple of $k$ ” by factoring out the right number
Prove inequalities such as $x^{2} - 6 x + 10 > 0$ by completing the square
Write a deduction that a reader can follow line by line to a clear conclusion

The idea: transform, don’t test

A deductive proof works forwards. You represent the general case algebraically, apply valid steps — expanding, factorising, completing the square — and arrive at a form that visibly has the property you were asked to prove. No cases, no assumptions, no examples: just a clean chain from the general to the specific claim.

The whole art is choosing the target form. If you must prove something is $even$ , aim to write it as $2 \times (integer)$ . If it must be a $multiple of 3$ , aim for $3 \times (integer)$ . If it must be $positive$ , aim for a square plus a positive number. Knowing the destination tells you which manipulation to reach for.

Why this works

Factoring out a 2 is not a trick — it is the definition of even made visible. “ $2 \times$ something whole” and “even” are literally the same statement, so once you reach that form the proof is finished.

The playbook

Four moves handle almost every deduction:

$1. Represent$ the general case with algebra (e.g. an odd number as $2 n + 1$ ). $2. Expand$ or combine everything into a single expression. $3. Factorise$ towards the target — pull out a 2, a 3, or complete the square. $4. Conclude$ by naming the property and referring back to the claim.

1Represent: let the odd number be

2 n + 1

n \in Z

2Expand:

(2 n + 1)^{2} = 4 n^{2} + 4 n + 1

3Factorise towards “odd”

= 2 (\dots) + 1

4 n^{2} + 4 n + 1 = 2 (2 n^{2} + 2 n) + 1

4Conclude:

2 n^{2} + 2 n

is an integer, so this is one more than an even number — odd.

Answer

(2 n + 1)^{2} = 2 (2 n^{2} + 2 n) + 1

, which is odd.

■

Revealing “multiple of $k$”

Statements about multiples are proved the same way — do the algebra, then factor out the number you need. The trick is to keep everything in one variable and resist the urge to test values.

1Represent the integers as

n

and

n + 1

2Sum of squares:

n^{2} + (n + 1)^{2} = n^{2} + n^{2} + 2 n + 1 = 2 n^{2} + 2 n + 1

3Rewrite as

2 (n^{2} + n) + 1

4Since

n^{2} + n

is an integer, this is odd.

Answer

n^{2} + (n + 1)^{2} = 2 (n^{2} + n) + 1

, which is odd.

■

Tip — If the target is “multiple of 4”, don’t stop at a factor of 2 — keep factorising until a 4 (or the exact number asked) is showing.

Proving inequalities by completing the square

A whole family of AQA proofs asks you to show an expression is always positive, or always at least some value. Testing numbers is useless here — you need to show it for $every$ real $x$ . The tool is completing the square, because a squared bracket can never be negative.

Once you write an expression as $(x - a)^{2} + b$ , the term $(x - a)^{2} \geq 0$ for all real $x$ , so the whole expression is at least $b$ . That single fact proves the inequality in one line.

(x - a)^{2} \geq 0 for all real x \Rightarrow (x - a)^{2} + b \geq b

A square is never negative — the engine behind “prove it is always positive”.

1Complete the square:

x^{2} - 6 x + 10 = (x - 3)^{2} - 9 + 10 = (x - 3)^{2} + 1

2For any real

x

(x - 3)^{2} \geq 0

3Adding 1:

(x - 3)^{2} + 1 \geq 1

4Since it is at least 1, it is certainly greater than 0.

Answer

x^{2} - 6 x + 10 = (x - 3)^{2} + 1 \geq 1 > 0

for all real

x

■

Why this works

The value of $b$ after completing the square is the minimum of the expression. So this method does more than prove positivity — it tells you the smallest value the expression can ever take, and the $x$ at which it happens.

Think like an examiner

The phrase “ $for all real x$ ” is a signal to complete the square, not to try numbers — a square bracket is the only clean way to argue about every real value at once.
When proving “multiple of $k$ ”, the mark for the conclusion requires the factor of $k$ to be $visible$ . Write $6 (\dots)$ explicitly rather than leaving it as $3 \times 2 \times (\dots)$ .
State that the leftover bracket “is an integer” before you call the result even/odd — examiners look for that justification.
A concluding sentence tying back to the question (“…so the expression is always positive”) is frequently the final mark.

Common misconceptions

Verifying an inequality by substituting a few values of

x

Complete the square to prove it for every real

x

at once. Substitution only ever checks single cases.

Stopping at

2 (2 n^{2} + 2 n)

and not adding the “

+ 1

” back when proving oddness.

Keep the full expression:

(2 n + 1)^{2} = 2 (2 n^{2} + 2 n) + 1

— the

+ 1

is what makes it odd.

Writing “

(x - 3)^{2} + 1 > 0

because squares are positive”.

Squares are

\geq 0

, not

> 0

(a square can equal 0). It is the

+ 1

that pushes the total above 0.

Expanding but never factorising, so the required property never becomes visible.

Always steer the algebra towards the target form — a factor of

k

, or a square plus a constant.

Deduction toolkit

(2 n + 1)^{2} = 2 (2 n^{2} + 2 n) + 1

Odd squared is odd.

(n + 1)^{2} - n^{2} = 2 n + 1

Difference of consecutive squares.

a x^{2} + b x + c = a (x + \frac{b}{2 a})^{2} + c - \frac{b ^{2}}{4 a}

Completing the square.

(x - a)^{2} \geq 0

A square is never negative.

Stretch yourself

Prove that $n^{3} - n$ is a multiple of 6 for every integer $n$ . (Hint: factorise first, then think about what three consecutive integers must contain.)

Hint — Factorise $n^{3} - n = (n - 1) n (n + 1)$ . Among any three consecutive integers, how many are even, and how many are multiples of 3?

Questions students ask

When should I use proof by deduction rather than another method?

Use deduction whenever a statement can be reached directly by algebra from a general case — which covers most “Show that…” questions. Reach for exhaustion when the cases are genuinely finite, contradiction when a direct route is blocked, and a counter-example when you suspect the statement is false.

How do I know what to factorise towards?

Read the target. “Even” means aim for

2 (\dots)

; “multiple of 3” means aim for

3 (\dots)

; “always positive” means aim for a completed square plus a positive constant. The destination dictates the manipulation.

Why does completing the square prove an inequality when substituting numbers doesn’t?

A squared bracket is

\geq 0

for every real number simultaneously, so

(x - a)^{2} + b \geq b

holds for all

x

in one stroke. Substituting values only ever checks finitely many points and can never cover them all.

Is it okay to split into cases in a deduction?

Yes — arguing “if

n

is even… if

n

is odd…” is a legitimate deductive proof (and blends into proof by exhaustion when the cases are finite). Just make sure your cases together cover every possibility.

Key takeaways

Deduction is a direct, forward argument from a general case to the claim.
Choose the target form: $2 (\dots)$ for even, $k (\dots)$ for a multiple of $k$ , a square $+$ constant for positivity.
Represent, expand, factorise, conclude — and justify that the leftover is an integer.
Prove “for all real $x$ ” inequalities by completing the square, using $(x - a)^{2} \geq 0$ .
Finish with a sentence that names the property and refers back to the statement.

How this fits the course

Build on

The Structure of a Proof

Completing the Squaresoon
Proof by Exhaustion

Leads to

Proof by Exhaustion

Test yourself

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