Counterexamples | Year 11 | Ascent Library

01 / LEARN

The idea, step by step

What a counterexample must do

A universal claim says that every object in a domain has a property. To disprove it, find one object in that domain for which the property fails. An example outside the domain tells us nothing about the claim.

For an implication, your example must make the assumption true and the conclusion false. If the assumption is false, the example is irrelevant to whether that implication fails.

Search strategically

Try boundary values, zero, negative numbers, fractions and small non-square integers when they are allowed. These often reveal an unstated assumption. State the failed calculation explicitly rather than saying only “not always”.

After finding a counterexample, the universal claim is false; checking more cases cannot rescue it. If you cannot find one, that is not proof that none exists. You need a general argument to prove a true universal claim.

A common trap

A hundred successful trials establish a universal theorem.

Instead: Trials provide evidence, but one valid counterexample is enough to overturn a universal claim.

02 / WORKED EXAMPLES

See the thinking

Example 1

A fraction breaks a familiar pattern

Disprove: for every positive real $x$ , $x^{2} > x$ .

Choose $x = \frac{1}{2}$ , which is in the stated positive-real domain.
Its square is $\frac{1}{4}$ .
Since $\frac{1}{4} < \frac{1}{2}$ , the claimed inequality fails.

x = \frac{1}{2}

is a counterexample.

Example 2

An integer outside a tested range

A student tests $n = 1, 2, 3$ and claims $n^{2} + n + 11$ is prime for every positive integer $n$ . Disprove the claim.

Choose $n = 11$ .
Then $11^{2} + 11 + 11 = 143 = 11 \times 13$ .
This positive-integer input produces a composite value.

n = 11

disproves the claim.

Check your understanding

Disprove: every rectangle has four equal sides.

Show answer

A rectangle with length 3 cm and width 2 cm.

It has four right angles but unequal adjacent side lengths.

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Disprove: every even integer is divisible by 4.
2 marks
Disprove: for every real $x$ , $\sqrt{x^{2}} = x$ .
2 marks
Disprove: the sum of any two irrational numbers is irrational.
2 marks
Disprove: if $a b = 0$ , then both $a$ and $b$ are zero.
2 marks
A claim concerns positive integers. Why does $n = 0$ not disprove it?
2 marks
For integer $n$ , the first four values of $n^{2} - n + 5$ are prime. Find a positive-integer counterexample.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Disprove: every even integer is divisible by 4.
2 marks
6 is even, but $6 / 4$ is not an integer.
6.
Disprove: for every real $x$ , $\sqrt{x^{2}} = x$ .
2 marks
$\sqrt{{(- 3)}^{2}} = 3 \neq - 3$ . The correct general identity is $\sqrt{x^{2}} = ∣ x ∣$ .
$x = - 3$ .
Disprove: the sum of any two irrational numbers is irrational.
2 marks
Both numbers are irrational; their sum is 0, which is rational.
$\sqrt{2}$ and $- \sqrt{2}$ .
Disprove: if $a b = 0$ , then both $a$ and $b$ are zero.
2 marks
The product is zero although $b \neq 0$ . The valid conclusion is that at least one factor is zero.
$a = 0$ , $b = 5$ .
A claim concerns positive integers. Why does $n = 0$ not disprove it?
2 marks
A counterexample must satisfy all of the original assumptions, including positivity.
0 is outside the domain.
For integer $n$ , the first four values of $n^{2} - n + 5$ are prime. Find a positive-integer counterexample.
3 marks
$5^{2} - 5 + 5 = 25 = 5 \times 5$ , so the universal prime claim is false.
$n = 5$ , giving 25.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: geometric reasoning and real-number/proof material, regrouped into a school chapter; all exercises are original.

Official curriculum source ↗

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