Rational/irrational numbers and simple number proofs | Year 11

01 / LEARN

Understand the idea.

Rational numbers are ratios of integers

A rational number is a real number expressible as $\frac{a}{b}$ with integers $a, b$ and $b \neq 0$ . Integers, terminating decimals and recurring decimals fit this definition. An irrational real number admits no such representation.

For rational sums, differences and products, combine the integer fractions and check that the denominator remains nonzero. These closure properties are useful in contradiction proofs: if an expression were rational, isolate the known irrational part.

Do not apply closure rules to irrational numbers

Irrational numbers do not share the rational closure rules. Two irrational numbers can sum or multiply to a rational number. A nonzero rational multiple of an irrational number is irrational, because dividing back would otherwise make the original number rational.

To prove a square root irrational, suppose it has a fraction representation in lowest terms. If the equation forces a common factor in both numerator and denominator, it contradicts that initial choice. A calculator display cannot replace this exact argument.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A rational shift of an irrational number

Prove that $3 + \sqrt{2}$ is irrational.

Worked solution

1
Assume the sum is rational.
$3 + \sqrt{2} = r, r \in ℚ$
A contradiction proof starts by assuming the negation of irrationality.
2
Isolate the square root.
$\sqrt{2} = r - 3$
The difference of the rational numbers $r$ and $3$ would be rational.
3
Reject the assumption.
This contradicts the established irrationality of $\sqrt{2}$ . Hence $3 + \sqrt{2}$ is irrational.

Answer

$3 + \sqrt{2}$ is irrational.

EXAMPLE 02

The square root of two

Prove that $\sqrt{2}$ is irrational.

Worked solution

1
Assume a fraction in lowest terms exists.
$\sqrt{2} = \frac{a}{b}, a, b \in ℤ, b \neq 0, gcd (a, b) = 1$
Every rational number has a representation whose numerator and denominator have no common factor greater than one.
2
Square and show the numerator is even.
$a^{2} = 2 b^{2}$
An odd value $a = 2 m + 1$ would give $a^{2} = 2 (2 m^{2} + 2 m) + 1$ , which is odd. Since this square is even, $a$ must be even.
3
Substitute its even form.
$\begin{aligned} a & = 2 k, k \in ℤ \\ 4 k^{2} & = 2 b^{2} \\ b^{2} & = 2 k^{2} \end{aligned}$
Substitution shows that $b^{2}$ is even, so $b$ is also even by the same parity argument.
4
Identify the common factor.
$2 ∣ a, 2 ∣ b$
Both integers are divisible by two, contradicting the fraction being in lowest terms.
5
Conclude irrationality.
$\sqrt{2} \notin ℚ$
The assumed rational representation is impossible.

Answer

$\sqrt{2}$ is irrational.

03 / YOUR TURN

Try it yourself.

Check your understanding

Is $\sqrt{2} \times \sqrt{2}$ irrational?

Show answer & reasoning

No; the product is 2.

$\sqrt{2} \times \sqrt{2} = 2$ , which is rational. Irrational inputs do not guarantee an irrational product.

Keep practising

6 worksheet questions · 17 marks

Open worksheet Worked solutions

Read the questions online

Prove that the sum of two rational numbers is rational.
2 marks
Prove that a non-zero rational multiple of an irrational number is irrational.
3 marks
Show that $0. \bar{27}$ is rational.
3 marks
Give irrational numbers whose sum is rational, and another pair whose sum is irrational.
2 marks
Prove that $\sqrt{5}$ is irrational.
4 marks
Is the reciprocal of a non-zero irrational number irrational? Prove your answer.
3 marks

Read the worked solutions

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

Prove that the sum of two rational numbers is rational.
2 marks
$a d + b c$ and $b d$ are integers and $b d \neq 0$ .
$\frac{a}{b} + \frac{c}{d} = \frac{a d + b c}{b d}$ .
Prove that a non-zero rational multiple of an irrational number is irrational.
3 marks
Let $r \neq 0$ be rational and $x$ irrational. If $r x$ were rational, then $x = \frac{r x}{r}$ would be rational, a contradiction.
The product is irrational.
Show that $0. \bar{27}$ is rational.
3 marks
Let $x = 0.2727 \dots$ . Then $100 x - x = 27$ , so $x = \frac{27}{99} = \frac{3}{11}$ .
$\frac{3}{11}$ .
Give irrational numbers whose sum is rational, and another pair whose sum is irrational.
2 marks
0 is rational. A non-zero rational multiple of $\sqrt{2}$ is irrational.
$\sqrt{2} + (- \sqrt{2}) = 0$ ; $\sqrt{2} + \sqrt{2} = 2 \sqrt{2}$ .
Prove that $\sqrt{5}$ is irrational.
4 marks
Assume $\sqrt{5} = \frac{a}{b}$ in lowest terms. Then $a^{2} = 5 b^{2}$ .
A square divisible by the prime 5 has its root divisible by 5. Set $a = 5 k$ .
Then $b^{2} = 5 k^{2}$ , so 5 also divides $b$ , contradicting lowest terms.
It is irrational.
Is the reciprocal of a non-zero irrational number irrational? Prove your answer.
3 marks
If $\frac{1}{x}$ were rational and non-zero, its reciprocal $x$ would be rational, a contradiction.
Yes.

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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 ↗

In this lesson

Understand the idea.

Rational numbers are ratios of integers

Do not apply closure rules to irrational numbers

See the method in action.

A rational shift of an irrational number

Assume the sum is rational.

Isolate the square root.

Reject the assumption.

The square root of two

Assume a fraction in lowest terms exists.

Square and show the numerator is even.

Substitute its even form.

Identify the common factor.

Conclude irrationality.

Try it yourself.

Check your understanding

Keep practising