Algebraic proof and mathematical reasoning

A mathematical proof explains why a statement must be true in every case covered by the statement. Checking examples can suggest a pattern, but proof replaces “it seems to work” with a chain of facts that cannot fail.

This distinction becomes essential at A-level. You will be expected to justify identities, properties of integers, geometric results and conclusions drawn from models.

Prerequisites

You should be able to:

• expand and factorise expressions;
• collect like terms;
• use negative numbers accurately;
• distinguish an expression from an equation;
• use basic facts about factors and multiples.

Review algebraic manipulation and factorisation if rearranging expressions is not yet secure. Exact arithmetic will help with divisibility and parity.

Statement, evidence and proof

Consider the claim:

The sum of two odd integers is even.

Examples such as $3+7=10$ and $11+15=26$ are evidence. They do not prove the claim because infinitely many pairs remain unchecked.

A proof represents arbitrary odd integers. Every odd integer has the form $2k+1$, where $k\in\mathbb Z$. Let the two odd integers be $2a+1$ and $2b+1$, where $a,b\in\mathbb Z$. Then

$$(2a+1)+(2b+1)=2(a+b+1).$$

Since $a+b+1$ is an integer, the sum is $2$ times an integer. It is therefore even.

The argument works for every pair because $a$ and $b$ were arbitrary integers.

Translate words into algebra

Most elementary algebraic proofs begin by expressing a general number correctly.

Description	Algebraic form	Condition
an integer	$n$	$n\in\mathbb Z$
an even integer	$2n$	$n\in\mathbb Z$
an odd integer	$2n+1$	$n\in\mathbb Z$
a multiple of $k$	$kn$	$n\in\mathbb Z$
two consecutive integers	$n,\ n+1$	$n\in\mathbb Z$
three consecutive integers	$n,\ n+1,\ n+2$	$n\in\mathbb Z$
consecutive even integers	$2n,\ 2n+2$	$n\in\mathbb Z$
consecutive odd integers	$2n+1,\ 2n+3$	$n\in\mathbb Z$

The letters represent integers, not particular chosen values. Writing $2n+1$ without saying $n$ is an integer leaves the definition incomplete.

Why two arbitrary numbers need two letters

If two integers are independently chosen, use different variables. Writing two odd integers as $2n+1$ and $2n+1$ silently assumes they are equal.

Use

$2a+1\quad\text{and}\quad 2b+1$

unless the numbers have a stated relationship. For consecutive odd integers, one variable is correct:

$2n+1\quad\text{and}\quad 2n+3.$

A reliable proof structure

For a direct algebraic proof:

1. Define arbitrary objects satisfying the hypothesis.
2. Manipulate them using valid algebra.
3. Reach a form whose meaning follows from a definition.
4. State the requested conclusion.

The last step matters. An expression such as $2(2n^2+2n)$ is not a complete conclusion until you explain that the bracket is an integer, so the expression is even.

Worked example 1: the square of an odd integer

Prove that the square of any odd integer is odd.

Let the odd integer be $2n+1$, where $n\in\mathbb Z$. Then

$$\begin{aligned} (2n+1)^2 &=4n^2+4n+1\\ &=2(2n^2+2n)+1. \end{aligned}$$

Since $2n^2+2n$ is an integer, the result has the form $2k+1$. Therefore the square of every odd integer is odd.

Notice that the proof does not say “let $n$ be odd” and then calculate $n^2$. It opens up the definition of oddness so the algebra can use it.

Worked example 2: consecutive integers

Prove that the product of two consecutive integers is even.

Let the integers be $n$ and $n+1$, where $n\in\mathbb Z$.

Consecutive integers have opposite parity. If $n$ is even, then $n(n+1)$ has an even factor. If $n$ is odd, then $n+1$ is even, so again the product has an even factor.

In either case the product is divisible by $2$. Therefore $n(n+1)$ is even.

This proof uses cases because expanding to $n^2+n$ does not immediately display a factor of $2$. A good proof uses the structure that makes the conclusion visible.

Worked example 3: divisibility

Prove that the sum of three consecutive integers is divisible by $3$.

Let the integers be $n$, $n+1$ and $n+2$, where $n\in\mathbb Z$. Their sum is

$$\begin{aligned} n+(n+1)+(n+2) &=3n+3\\ &=3(n+1). \end{aligned}$$

Since $n+1$ is an integer, the sum is a multiple of $3$.

The phrase “divisible by $3$” tells you the target form: $3\times\text{integer}$.

Worked example 4: a difference of squares

Prove that the difference between the squares of two consecutive integers is odd.

Let the consecutive integers be $n$ and $n+1$, where $n\in\mathbb Z$. Taking the larger square minus the smaller gives

$$\begin{aligned} (n+1)^2-n^2 &=n^2+2n+1-n^2\\ &=2n+1. \end{aligned}$$

This has the form $2n+1$, so it is odd.

If the subtraction were reversed, the result would be $-(2n+1)=2(-n-1)+1$, which is also odd. Negative integers can be odd or even too.

Identities and equations are different

An identity is true for every permitted value of its variables. It is written with $\equiv$ when emphasis is useful:

$$(x+3)^2\equiv x^2+6x+9.$$

An equation is true only for particular values, or is a condition to solve:

$$(x+3)^2=25.$$

This equation holds when $x=2$ or $x=-8$, not for every $x$.

Worked example 5: prove an identity

Prove that

$$(a+b)^2-(a-b)^2\equiv4ab.$$

Start from one side and use established algebra:

$$\begin{aligned} (a+b)^2-(a-b)^2 &=(a^2+2ab+b^2)-(a^2-2ab+b^2)\\ &=a^2+2ab+b^2-a^2+2ab-b^2\\ &=4ab. \end{aligned}$$

Hence the identity is proved.

Expanding both sides independently until they match is also valid. Avoid beginning by writing the result you are supposed to prove and then performing non-reversible steps, since that can assume the conclusion.

Implication and equivalence

The logic symbols have precise meanings:

$$P\Rightarrow Q$$

means that whenever $P$ is true, $Q$ must be true. It does not say that $Q$ forces $P$.

For example,

$x=3\Rightarrow x^2=9,$

but $x^2=9$ also allows $x=-3$. Therefore the reverse implication is false.

The symbol

$P\Leftrightarrow Q$

means that each statement implies the other. For real $x$,

$2x+5=11\Leftrightarrow x=3.$

Each algebraic step here is reversible.

Necessary and sufficient conditions

If $P\Rightarrow Q$, then $P$ is sufficient for $Q$, and $Q$ is necessary for $P$.

Being divisible by $4$ is sufficient for an integer to be even. Being even is necessary for an integer to be divisible by $4$, but it is not sufficient because $6$ is even and is not divisible by $4$.

Disproving a universal statement

A statement of the form “for every…” is disproved by one counterexample: a permitted case in which the conclusion fails.

Worked example 6: find a counterexample

Disprove the claim:

If $n$ is an integer, then $n^2+n+1$ is prime.

Choose $n=4$:

$4^2+4+1=21=3\times7.$

Since $21$ is not prime, $n=4$ is a counterexample. Therefore the universal claim is false.

One failed case settles the question. A list of many failed cases is unnecessary.

A counterexample must satisfy the hypothesis

To disprove “every positive integer has property $P$”, using $-2$ is invalid because $-2$ is not a positive integer. Always check that your example belongs to the stated domain.

Showing that an existential statement is true

A claim beginning “there exists” needs only one valid example.

To prove that there exists an even prime number, give $2$ and verify both properties: $2$ is divisible by $2$, so it is even, and its only positive factors are $1$ and $2$, so it is prime.

The number of cases required depends on the quantifier:

Claim	To prove it	To disprove it
Every object has property $P$	prove all allowed cases	give one counterexample
Some object has property $P$	give one verified example	prove no allowed case works

Reasoning from assumptions

A proof must use only the given information, definitions and established results. A diagram, numerical pattern or calculator display can guide your thinking, but it cannot silently add an assumption.

Worked example 7: spotting an invalid converse

A student argues:

1. If an integer is divisible by $6$, it is divisible by $3$.
2. $15$ is divisible by $3$.
3. Therefore $15$ is divisible by $6$.

The first statement is true:

$6\mid n\Rightarrow3\mid n.$

The argument then reverses the implication without justification. The counterexample $15$ confirms the failure because $3\mid15$ but $6\nmid15$.

The original implication and its converse are different statements.

Proof by exhaustion

If there are finitely many possible cases, you may prove a claim by checking every case. The list must be complete.

Worked example 8: remainders modulo $3$

Prove that the square of an integer leaves remainder $0$ or $1$ when divided by $3$.

Every integer has exactly one of the forms

$3k,\qquad 3k+1,\qquad 3k+2,$

where $k\in\mathbb Z$.

Square each form:

$$\begin{aligned} (3k)^2&=3(3k^2),\\ (3k+1)^2&=3(3k^2+2k)+1,\\ (3k+2)^2&=9k^2+12k+4=3(3k^2+4k+1)+1. \end{aligned}$$

The first case has remainder $0$ and the other two have remainder $1$. These are all possible integer cases, so the claim is proved.

Trying only $0,1,2$ is suggestive, but the forms $3k$, $3k+1$ and $3k+2$ prove the result for all integers.

Common misconceptions

• Checking several values proves a rule. Examples provide evidence only. A universal claim needs an argument covering every allowed case.
• One letter can represent two arbitrary integers. This forces them to be equal. Use independent letters unless a relationship is given.
• The algebra is the whole proof. Explain why the final form has the required property.
• An identity can be proved by substitution. Substitution checks selected values, not all values.
• A counterexample needs to be unusual. The smallest convenient valid example is usually best.
• The converse of a true implication is automatically true. It must be proved separately.
• A diagram drawn to scale proves a geometric claim. A diagram may suggest a result, but proof must use stated facts and theorems.
• Dividing by a variable is always reversible. If that variable could be zero, division may discard a case.

Self-check

1. Represent the integers

Write algebraic forms for three consecutive odd integers.

Answer

One suitable answer is

$2n+1,\qquad2n+3,\qquad2n+5,$

where $n\in\mathbb Z$.

2. Prove a parity result

Prove that the product of an even integer and an odd integer is even.

Answer

Let the even integer be $2a$ and the odd integer be $2b+1$, where $a,b\in\mathbb Z$. Then

$$(2a)(2b+1)=2\bigl(a(2b+1)\bigr).$$

Since $a(2b+1)$ is an integer, the product is $2$ times an integer and is therefore even.

3. Prove divisibility

Prove that the difference between the squares of two odd integers is divisible by $8$.

Answer

Let the odd integers be $2a+1$ and $2b+1$, where $a,b\in\mathbb Z$. Then

$$\begin{aligned} (2a+1)^2-(2b+1)^2 &=4a^2+4a-4b^2-4b\\ &=4\bigl(a(a+1)-b(b+1)\bigr). \end{aligned}$$

Each product of consecutive integers, $a(a+1)$ and $b(b+1)$, is even. Their difference is therefore even, so for some integer $k$,

$a(a+1)-b(b+1)=2k.$

Hence the original difference is $4(2k)=8k$, which is divisible by $8$.

4. Disprove a claim

Disprove: “The sum of two prime numbers is always even.”

Answer

$2$ and $3$ are both prime, but

$2+3=5,$

which is odd. Thus the claim is false.

5. Diagnose the logic

A student writes: “Since $x^2=16$, $x=4$.” What is missing?

Answer

The equation has two real solutions:

$x=4\quad\text{or}\quad x=-4.$

Squaring loses sign information, so $x^2=16$ does not imply only $x=4$.

6. Prove an identity

Prove that

$$(x+1)^3-(x-1)^3\equiv6x^2+2.$$

Answer

$$\begin{aligned} (x+1)^3-(x-1)^3 &=(x^3+3x^2+3x+1)\\ &\quad-(x^3-3x^2+3x-1)\\ &=6x^2+2. \end{aligned}$$

Therefore the identity holds for every real $x$.

Where this leads

You are ready to move from foundation arguments to the main A-level proof methods:

• proof by deduction and exhaustion;
• disproof by counterexample;
• proof by contradiction;
• communicating mathematical reasoning.

Keep algebraic fluency close at hand. Many apparent proof difficulties are really difficulties in reaching the form that makes the conclusion visible.