Rational and irrational numbers
Master rational and irrational numbers in Year 11 VCE Specialist Mathematics. A rational number is a ratio of two integers \(\dfrac{a}{b}\); an irrational number, such as \(\sqrt{2}\), is not. It sits in the Algebra, number and structure area of study of the VCE Mathematics Study Design (VCAA), within the Proof and number topic of Unit 1.
You will learn to convert terminating and recurring decimals to fractions, prove a number is irrational by contradiction, and combine rationals and irrationals with confidence — key skills for the proof topic ahead.
Theory
Rational and irrational numbers split the real numbers in two for Year 11 Specialist Mathematics. A rational number is a ratio \(\dfrac{a}{b}\) of integers; an irrational number, such as \(\sqrt{2}\) or \(\pi\), is a real number that is not a ratio of integers. This page shows how to convert between fractions and decimals and how to prove a number is irrational.
A number is rational if it can be written as \(\dfrac{a}{b}\), where \(a\) and \(b\) are integers and \(b\neq 0\). The rationals are written \(\mathbb{Q}\); an irrational number is a real number that is not rational.
The decimal expansion is the quick test. A number is rational exactly when its decimal terminates (like \(0.75\)) or eventually recurs (like \(0.\overline{3}=\tfrac{1}{3}\)). An irrational decimal, such as \(\sqrt{2}=1.41421\ldots\), never ends and never repeats.
Roots follow one rule: \(\sqrt{n}\) is rational exactly when \(n\) is a perfect square. So \(\sqrt{9}=3\) is rational, but \(\sqrt{7}\) is irrational.
Combining the two kinds obeys closure facts: a rational \(\pm\) an irrational is always irrational, and a non-zero rational \(\times\) an irrational is always irrational — but two irrationals can combine to a rational, for example \(\sqrt{2}\times\sqrt{2}=2\).
The rationals are the ratios of integers, defined in set-builder notation as:
To convert a pure recurring decimal with a \(k\)-digit repeating block, multiply by \(10^{k}\) and subtract:
How to classify or convert a number
- Simplify first: evaluate any root or reduce any fraction, e.g. \(\sqrt{16}=4\) or \(\dfrac{6}{9}=\dfrac{2}{3}\).
- Read the decimal: if it terminates or recurs the number is rational; if it never ends and never repeats it is irrational.
- Convert a recurring decimal: set \(x\) equal to it, multiply by the right power of \(10\) so the tails line up, subtract, then solve for \(x\) and simplify.
- Prove irrationality by contradiction: assume the number equals \(\dfrac{a}{b}\) in lowest terms, deduce a common factor, and reach a contradiction.
Set \(x\) equal to the decimal, multiply by \(10\) to shift one digit, then subtract to clear the recurring tail:
| \(x\) | \(=\) | \(0.888\ldots\) |
| \(10x\) | \(=\) | \(8.888\ldots\) |
| \(10x-x\) | \(=\) | \(8.888\ldots-0.888\ldots\) |
| \(9x\) | \(=\) | \(8\) |
| \(x\) | \(=\) | \(\dfrac{8}{9}\) |
\(0.\overline{8}=\dfrac{8}{9}\), a ratio of integers, so it is rational.
One digit sits before the repeating block, so shift by \(10\) and by \(100\) to line the tails up, then subtract:
| \(x\) | \(=\) | \(0.5333\ldots\) |
| \(10x\) | \(=\) | \(5.333\ldots\) |
| \(100x\) | \(=\) | \(53.333\ldots\) |
| \(100x-10x\) | \(=\) | \(53.333\ldots-5.333\ldots\) |
| \(90x\) | \(=\) | \(48\) |
| \(x\) | \(=\) | \(\dfrac{48}{90}\) |
| \(=\) | \(\dfrac{8}{15}\) |
\(0.5\overline{3}=\dfrac{8}{15}\).
A root \(\sqrt{n}\) is rational exactly when \(n\) is a perfect square; test each in turn:
| \(\sqrt{20}\) | \(=\) | \(4.472\ldots\notin\mathbb{Q}\) |
| \(\sqrt{49}\) | \(=\) | \(7\in\mathbb{Q}\) |
| \(\sqrt{72}\) | \(=\) | \(8.485\ldots\notin\mathbb{Q}\) |
| \(\sqrt{100}\) | \(=\) | \(10\in\mathbb{Q}\) |
| \(\text{count}\) | \(=\) | \(2\) |
\(2\) of the values (\(\sqrt{49}\) and \(\sqrt{100}\)) are rational.
Apply the closure rules one at a time, computing each combination exactly:
| \(4+\sqrt{3}\) | \(=\) | \(\text{rational}+\text{irrational}\) |
| \(\Rightarrow\) | \(\text{irrational}\) | |
| \(\sqrt{2}\times\sqrt{2}\) | \(=\) | \(2\in\mathbb{Q}\) |
| \(\sqrt{2}+(-\sqrt{2})\) | \(=\) | \(0\in\mathbb{Q}\) |
\(4+\sqrt{3}\) is irrational, yet two irrationals can combine to a rational (here \(2\) and \(0\)).
Common pitfalls
Frequently asked questions
What is the difference between a rational and an irrational number?
A rational number can be written as a fraction \(\dfrac{a}{b}\) of integers, so its decimal terminates or recurs. An irrational number cannot — its decimal never ends and never repeats, like \(\sqrt{2}\) or \(\pi\).
How do you turn a recurring decimal into a fraction?
Set \(x\) equal to the decimal, multiply by the power of \(10\) that lines the repeating blocks up, subtract to remove the tail, then solve for \(x\). For example \(0.\overline{45}=\dfrac{45}{99}=\dfrac{5}{11}\).
How do you prove that \(\sqrt{2}\) is irrational?
Assume \(\sqrt{2}=\dfrac{a}{b}\) in lowest terms. Then \(a^2=2b^2\), so \(a\) is even; writing \(a=2c\) makes \(b\) even too. Now \(a\) and \(b\) share the factor \(2\), contradicting lowest terms, so \(\sqrt{2}\) is irrational.
Is the square root of a number always irrational?
No. \(\sqrt{n}\) is rational when \(n\) is a perfect square, so \(\sqrt{9}=3\) is rational. It is irrational only when \(n\) is not a perfect square, such as \(\sqrt{7}\).
Is a rational plus an irrational always irrational?
Yes. If a rational \(r\) plus an irrational \(s\) were rational, then \(s=(r+s)-r\) would be a difference of rationals, hence rational — a contradiction. So \(r+s\) is always irrational.
Can two irrational numbers add or multiply to a rational number?
Yes. \(\sqrt{2}\times\sqrt{2}=2\) and \(\sqrt{2}+(-\sqrt{2})=0\) are both rational, so the sum or product of two irrationals is not always irrational.