Rational and irrational numbers
Get on top of rational and irrational numbers for Year 11 Specialist Mathematics in Queensland (QCAA). A rational number is a ratio of two integers; an irrational number, such as the square root of two, is not.
You will 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.