Disproof by counterexample

A counterexample is one permitted case for which a claimed general statement is false. It disproves any claim that says something holds for every member of a set.

For example, the statement

Every prime number is odd.

is false because $2$ is prime and $2$ is not odd. The single value $2$ settles the question. There is no need to list any other primes.

Before you start

You should be able to:

• substitute numbers into algebraic expressions
• use the meanings of integer, even, odd, prime and factor
• distinguish an equation from an identity
• read inequalities and set restrictions accurately

Review proof and reasoning if you need practice with implications, definitions or algebraic proofs. This lesson develops counterexamples to A level standard.

Why one example can disprove, but cannot prove

Consider a universal statement of the form

$$\text{For every }x\in S,\quad P(x)\text{ is true}.$$

To disprove it, find one $c\in S$ for which $P(c)$ is false. In symbols,

$$c\in S\quad\text{and}\quad \neg P(c).$$

The two conditions matter equally:

1. $c$ must belong to the stated domain $S$.
2. The claimed conclusion must fail at $c$.

By contrast, checking several cases where $P(x)$ is true leaves all untested cases unresolved. The values $2$, $4$, $6$ and $8$ support the statement that every even positive integer is composite, but $2$ actually refutes it. A finite collection of successful trials is not a proof of an infinite claim.

The logical asymmetry

One failure disproves a universal claim. Any finite number of successes usually does not prove one.

The exam method

For a question asking you to disprove a statement by counterexample:

1. Identify the domain and every condition.
2. Choose the simplest value likely to expose the weakness.
3. Substitute it into both sides of the claim, or check the relevant properties.
4. State explicitly why the results contradict the claim.

A polished answer often fits into two lines:

$$\text{Take }x=c.\quad \text{Then [hypothesis holds], but [conclusion fails].}$$

Therefore $x=c$ is a counterexample, so the statement is false.

Worked example 1: test the whole claim

Disprove the statement:

For every integer $n$, if $n^2$ is divisible by $4$, then $n$ is divisible by $4$.

We need an integer whose square is divisible by $4$, but which is not itself divisible by $4$. Try the smallest even integer that is not a multiple of $4$:

$$n=2.$$

Then

$$n^2=2^2=4,$$

which is divisible by $4$, but $2$ is not divisible by $4$. Therefore $n=2$ is a counterexample and the statement is false.

Notice that $n=3$ would not work. Although $3$ is not divisible by $4$, its square is not divisible by $4$ either, so it does not satisfy the hypothesis.

Self check 1

Disprove: for every integer $n$, if $n^2$ is divisible by $9$, then $n$ is divisible by $9$.

Answer

Take $n=3$. Then $n^2=9$ is divisible by $9$, but $3$ is not divisible by $9$. Thus $n=3$ is a counterexample, so the statement is false.

Choosing counterexamples strategically

Random guessing is rarely necessary. The wording of a false statement often suggests useful test cases.

Feature of the claim	Values worth testing
Powers or products	$0$, $1$, $-1$
Inequality	negative numbers, $0$, equal values
Fractions	values making a denominator negative or zero
Square roots or absolute values	negative inputs, or values on either side of $0$
Integers and factors	small primes, small composites, boundary cases
A diagram or geometric claim	symmetric cases, extreme shapes, degenerate cases if allowed

These are prompts, not guarantees. Every chosen value must still satisfy the stated conditions.

Worked example 2: negative numbers reverse intuition

Disprove:

For all real numbers $a$ and $b$, if $a^2>b^2$, then $a>b$.

Squaring loses information about sign, so choose $a$ negative and $b$ positive. Let

$$a=-3,\qquad b=2.$$

Then

$$a^2=9>4=b^2,$$

so the hypothesis holds. However,

$$a=-3<2=b,$$

so the conclusion fails. Hence $(a,b)=(-3,2)$ is a counterexample.

The correct conclusion from $a^2>b^2$ is $|a|>|b|$, not necessarily $a>b$.

Worked example 3: an alleged algebraic identity

Disprove the claim

$$\sqrt{x^2+9}=x+3$$

for all $x\geq 0$.

Choose $x=1$. Both expressions are defined, but

$$\sqrt{1^2+9}=\sqrt{10}$$

while

$$1+3=4.$$

Since $\sqrt{10}\ne4$, the equality fails. Therefore $x=1$ is a counterexample.

The tempting but invalid step is

$$\sqrt{a+b}=\sqrt a+\sqrt b.$$

Square roots do not distribute over addition. Compare this with valid rules for products, where the required domain conditions must also be respected.

Self check 2

Give a counterexample to each statement.

1. For every real $x$, $x^2\geq x$.
2. For all positive real $a$ and $b$, $\dfrac{a+b}{2}\leq\sqrt{ab}$.
3. For every real $x$, $|x+2|=|x|+2$.

Answers

1. Take $x=\frac12$. Then $x^2=\frac14<\frac12=x$.
2. Take $a=1$ and $b=4$. Then $(a+b)/2=5/2>2=\sqrt{ab}$.
3. Take $x=-1$. Then $|x+2|=1$, while $|x|+2=3$.

Other answers are possible. In each case, the values must be in the stated domain and the numerical comparison must be shown.

Counterexamples to implications

A statement of the form

$$P\Longrightarrow Q$$

is false only when $P$ is true and $Q$ is false. This is the logical pattern behind many exam questions.

$P$	$Q$	Is this a counterexample to $P\Rightarrow Q$?
true	true	no
true	false	yes
false	true	no
false	false	no

If the hypothesis $P$ is false, the case cannot refute what the implication promises, because the promise applies only when $P$ holds.

Worked example 4: confusing a statement with its converse

The true statement

$$6\mid n\Longrightarrow 3\mid n$$

says that if $n$ is divisible by $6$, then it is divisible by $3$.

Its converse is

$$3\mid n\Longrightarrow 6\mid n.$$

To disprove the converse, choose a multiple of $3$ that is not a multiple of $6$. Take $n=9$. Then $3\mid9$, but $6\nmid9$. Thus $9$ is a counterexample to the converse.

This does not affect the truth of the original implication. An implication and its converse are separate statements.

Counterexamples involving several variables

When a statement quantifies over several variables, one valid tuple is enough. Keep the numbers small and make every restriction visible.

Worked example 5: division of inequalities

Disprove:

For all real numbers $a$, $b$ and $c$ with $c\ne0$, if $a<b$, then $a/c<b/c$.

The conclusion fails when division is by a negative number. Choose

$$a=1,\qquad b=2,\qquad c=-1.$$

The hypothesis $1<2$ holds and $c\ne0$. However,

$$\frac ac=-1 \quad\text{and}\quad \frac bc=-2,$$

so

$$\frac ac>\frac bc.$$

Therefore $(a,b,c)=(1,2,-1)$ is a counterexample. The claim would become true if the extra condition $c>0$ were imposed.

Worked example 6: a number theory claim

Disprove:

The sum of two prime numbers is always composite.

Take the primes $2$ and $3$. Their sum is

$$2+3=5,$$

which is prime, not composite. Therefore the pair $(2,3)$ is a counterexample.

Choosing two odd primes would not reveal the problem: their sum is an even integer greater than $2$, so it is composite. The exceptional prime $2$ is the key boundary case.

Domain errors and invalid counterexamples

Suppose the claim is

For every positive real number $x$, $x+1/x\geq2$.

The choice $x=-1$ gives $x+1/x=-2$, but it is not a counterexample because $-1$ is not positive. The choice $x=0$ is even less suitable: it lies outside the domain and makes $1/x$ undefined.

In fact, the claim is true. For $x>0$,

$$(x-1)^2\geq0$$

implies

$$x^2-2x+1\geq0.$$

Dividing by the positive number $x$ gives

$$x+\frac1x\geq2.$$

Failure to find a counterexample does not prove a statement, but it may suggest that a proof is needed instead.

Three invalid approaches

• Giving a value outside the domain.
• Giving a value for which the hypothesis is false.
• Saying that a statement “looks wrong” without showing a failing case.

Refuting a claim versus repairing it

A counterexample answers the question “Is the statement always true?” It can also reveal how to improve the statement.

Consider

$$x^2>y^2\Longrightarrow x>y.$$

The counterexample $x=-3$, $y=2$ shows that signs cause the failure. Possible true replacements include

$$x^2>y^2\Longrightarrow |x|>|y|$$

and, with extra hypotheses,

$$x\geq0,\ y\geq0,\ x^2>y^2\Longrightarrow x>y.$$

Finding the precise missing condition is a powerful way to understand the underlying mathematics, although an exam question asking only for a counterexample does not require a repaired theorem.

Mixed self check

For each statement, either give a counterexample or explain briefly why the suggested value is invalid.

1. Every integer of the form $n^2+n+41$ is prime.
2. If $x<y$, then $x^2<y^2$. Use $x=-2$, $y=1$.
3. If an integer $n$ is odd, then $n^2$ is odd. A student proposes $n=2$ as a counterexample.
4. For all non-zero real $x$, $x+1/x\geq2$.
5. The product of two irrational numbers is irrational.

Answers

1. Take $n=41$. Then $$n^2+n+41=41^2+41+41=41(41+2)=41\times43,$$ which is composite.
2. It is valid: $-2<1$, but $(-2)^2=4>1=1^2$. Thus $(x,y)=(-2,1)$ is a counterexample.
3. The value is invalid because $2$ is not odd, so it does not satisfy the hypothesis. The statement is true and can be proved by writing $n=2k+1$.
4. Take $x=-1$. It is non-zero, but $$x+\frac1x=-2<2.$$
5. Take $\sqrt2$ and $\sqrt2$. Both are irrational, but their product is $2$, which is rational.

Exam checklist

Before finishing a counterexample answer, ask:

• Does my example satisfy every condition in the question?
• Have I evaluated or checked the relevant property correctly?
• Does the claimed conclusion genuinely fail?
• Have I stated that this makes the original universal statement false?

One clean counterexample is stronger than a long list. Extra examples add opportunities for arithmetic errors without adding logical force.

Next steps

Counterexamples disprove universal claims. To establish that a statement is true, study proof by deduction and exhaustion and proof by contradiction. For presenting the reasoning clearly under exam conditions, continue to communicating mathematical reasoning.