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Year 11 Specialist (Unit 1 & 2) Introduction to proof

Sets of numbers

20 practice questions 0 video lessons Theory + worked examples

Get to grips with the sets of numbers that underpin Year 11 Specialist Mathematics in Queensland (QCAA). These are the number systems — the naturals, integers, rationals and reals — that nest inside one another and give proof its precise language.

You will learn to use set notation, place a number in the smallest set that contains it, and tell rational numbers from irrational ones — the classifying skills the proof topic is built on.

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Theory

Sets of numbers are the building blocks of proof in Year 11 Specialist Mathematics (QCAA, Queensland). The number systems — the naturals \(\mathbb{N}\), integers \(\mathbb{Z}\), rationals \(\mathbb{Q}\) and reals \(\mathbb{R}\) — nest inside one another, and every real number is either rational or irrational. This page shows how to name and classify each number using set notation.

A set is a collection of numbers, and \(\in\) means “is an element of”. Four number systems are used constantly, each built from the one before it by adding new numbers.

The natural numbers \(\mathbb{N}=\{1,2,3,\dots\}\) are the counting numbers. Adding zero and the negatives gives the integers \(\mathbb{Z}=\{\dots,-2,-1,0,1,2,\dots\}\); the positive integers are written \(\mathbb{Z}^{+}\) and the negatives \(\mathbb{Z}^{-}\).

The rational numbers \(\mathbb{Q}\) are every number that can be written as a fraction \(\dfrac{a}{b}\) with \(a,b\in\mathbb{Z}\) and \(b\neq 0\). A number is rational exactly when its decimal terminates (like \(0.75\)) or recurs (like \(0.\overline{3}=\tfrac{1}{3}\)).

An irrational number is a real number that is not rational; its decimal never terminates and never repeats. Surds of non-square integers such as \(\sqrt{2}\), and \(\pi\), are irrational. Together the rationals and irrationals make up the real numbers \(\mathbb{R}\), and the sets nest as \(\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\).

Nested number sets Four boxes nested one inside the next: the natural numbers N sit inside the integers Z, inside the rationals Q, inside the reals R. Sample numbers 1, 2, 3 sit in N; 0 and -3 in Z; one half and -0.75 in Q; root 2 and pi in R. 1, 2, 3 0, −3 ½, −0.75 √2, π
The number sets nest: \(\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\), with sample numbers in each ring.
Sample numbers on a number line A number line from 1 to 2.5. Root 2 is about 1.41 and root 5 is about 2.24, both irrational and marked in gold-orange. Three halves equals 1.5 and 1.75 are rational, marked in navy. The surds fall between whole numbers. x 1 1.5 2 2.5 √2 3/2 1.75 √5 gold = irrational, navy = rational
Sample numbers on a line: the surds \(\sqrt{2},\sqrt{5}\) are irrational; \(\tfrac{3}{2},1.75\) are rational.

The rational numbers are defined by set-builder notation as the ratios of integers:

\[ \mathbb{Q}=\left\{\dfrac{a}{b} : a,b\in\mathbb{Z},\ b\neq 0\right\} \]
={ab:a,b,b0}

The reals are the rationals together with the irrationals, and the systems nest:

\[ \mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R} \]
Roots and decimals. \(\sqrt{n}\) is rational exactly when \(n\) is a perfect square (so \(\sqrt{9}=3\) is rational but \(\sqrt{7}\) is not). A recurring decimal is always rational; only a decimal that never ends and never repeats is irrational.

How to classify a number

  1. Simplify first: evaluate any root or convert a decimal, e.g. \(\sqrt{16}=4\) or \(0.75=\tfrac{3}{4}\).
  2. Whole number? If it is a positive counting number it is in \(\mathbb{N}\); if it is a whole number that may be zero or negative it is in \(\mathbb{Z}\).
  3. A fraction or terminating/recurring decimal? Then it is rational, in \(\mathbb{Q}\) (but not in \(\mathbb{Z}\) unless it is whole).
  4. Otherwise irrational: a non-terminating, non-recurring decimal such as a surd of a non-square or \(\pi\) is irrational, so its smallest set is \(\mathbb{R}\).
Example 1 — Classify each number
Which of \(0.8\), \(\sqrt{6}\) and \(-\dfrac{4}{9}\) are rational?
Solution

Rewrite each number in its simplest exact form, then test for a fraction:

\(0.8\)\(=\)\(\dfrac{8}{10}\)
\(=\)\(\dfrac{4}{5}\in\mathbb{Q}\)
\(-\dfrac{4}{9}\)\(\in\)\(\mathbb{Q}\quad(\text{already a fraction})\)
\(\sqrt{6}\)\(=\)\(2.449\ldots\)
\(=\)\(\text{never terminates or recurs}\)
\(\Rightarrow\ \sqrt{6}\)\(\notin\)\(\mathbb{Q}\)

\(0.8\) and \(-\dfrac{4}{9}\) are rational; \(\sqrt{6}\) is irrational.

Example 2 — Recurring decimal to a fraction
Show that \(x=0.\overline{45}=0.4545\ldots\) is rational.
Solution

Multiply by \(100\) so the repeating block lines up, then subtract to remove it:

\(x\)\(=\)\(0.4545\ldots\)
\(100x\)\(=\)\(45.4545\ldots\)
\(100x-x\)\(=\)\(45\)
\(99x\)\(=\)\(45\)
\(x\)\(=\)\(\dfrac{45}{99}\)
\(=\)\(\dfrac{5}{11}\)

\(x=\dfrac{5}{11}\), a ratio of integers, so \(x\in\mathbb{Q}\).

Example 3 — Smallest set containing a number
Name the smallest of \(\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R}\) that contains \(-\dfrac{3}{4}\), then \(\sqrt{50}\).
Solution

Test each number against the sets from smallest to largest:

\(-\dfrac{3}{4}\)\(\notin\)\(\mathbb{Z}\quad(\text{not a whole number})\)
\(-\dfrac{3}{4}\)\(\in\)\(\mathbb{Q}\)
\(\text{smallest}\)\(=\)\(\mathbb{Q}\)
\(\sqrt{50}\)\(=\)\(7.071\ldots\notin\mathbb{Q}\)
\(\sqrt{50}\)\(\in\)\(\mathbb{R}\)
\(\text{smallest}\)\(=\)\(\mathbb{R}\)

\(-\dfrac{3}{4}\) sits in \(\mathbb{Q}\); \(\sqrt{50}\) sits in \(\mathbb{R}\).

Example 4 — Which roots are rational
For how many of \(\sqrt{8}\), \(\sqrt{49}\), \(\sqrt{18}\) and \(\sqrt{64}\) is the value rational?
Solution

A root \(\sqrt{n}\) is rational exactly when \(n\) is a perfect square; test each:

\(\sqrt{8}\)\(=\)\(2.828\ldots\notin\mathbb{Q}\)
\(\sqrt{49}\)\(=\)\(7\in\mathbb{Q}\)
\(\sqrt{18}\)\(=\)\(4.242\ldots\notin\mathbb{Q}\)
\(\sqrt{64}\)\(=\)\(8\in\mathbb{Q}\)
\(\text{count}\)\(=\)\(2\)

\(2\) of the values (\(\sqrt{49}\) and \(\sqrt{64}\)) are rational.

Common pitfalls

Thinking every never-ending decimal is irrational. A recurring decimal such as \(0.\overline{6}=\tfrac{2}{3}\) goes on forever but repeats, so it is rational. Only a decimal that never ends and never repeats is irrational.
Assuming any root is irrational. A square root is only irrational when the number under it is not a perfect square. \(\sqrt{16}=4\) and \(\sqrt{25}=5\) are whole numbers, hence rational.
Forgetting the nesting is one-way. Every integer is rational, but not every rational is an integer (\(\tfrac{1}{2}\) is not). Likewise every real number is not rational — \(\sqrt{2}\) is real but irrational.

Frequently asked questions

What are the sets N, Z, Q and R?

\(\mathbb{N}\) is the natural (counting) numbers, \(\mathbb{Z}\) the integers, \(\mathbb{Q}\) the rationals (fractions), and \(\mathbb{R}\) the reals. They nest as \(\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R}\).

How do you tell if a number is rational or irrational?

A number is rational if it can be written as a fraction of integers, which happens exactly when its decimal terminates or recurs. If the decimal never ends and never repeats — like \(\sqrt{2}\) or \(\pi\) — it is irrational.

Is a recurring decimal rational?

Yes. Any recurring decimal equals a fraction of integers, for example \(0.\overline{3}=\tfrac{1}{3}\), so every recurring decimal is rational.

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 \(\sqrt{7}\).

What is the difference between Z+ and N?

\(\mathbb{Z}^{+}\) is the positive integers \(1,2,3,\dots\), which is the same as the natural numbers \(\mathbb{N}\) under the counting-number convention used in this course.

Is pi a real number?

Yes. \(\pi\) is a real number, but it is irrational: its decimal never terminates or repeats, so it cannot be written as a fraction.