Number Systems

Not all numbers are the same kind of number. Counting numbers, negative numbers, fractions and numbers like $\pi$ each belong to different sets, and those sets fit inside each other like nesting dolls. Knowing which set a number belongs to helps you predict how it behaves, and it explains some surprising facts, like why $0.999\ldots$ is exactly equal to $1$.

Key ideas

The number sets

Set	Symbol	What’s in it	Examples
natural numbers	$\mathbb{N}$	the counting numbers	$1, 2, 3, 4, \ldots$
whole numbers	$\mathbb{W}$	the natural numbers and zero	$0, 1, 2, 3, \ldots$
integers	$\mathbb{Z}$	the whole numbers and their opposites	$\ldots, -2, -1, 0, 1, 2, \ldots$
rational numbers	$\mathbb{Q}$	any number you can write as a fraction of two integers	$\dfrac{3}{4}$, $-2.5$, $0.\overline{3}$, $7$
irrational numbers	$\overline{\mathbb{Q}}$	numbers you can’t write as a fraction of integers	$\sqrt{2}$, $\pi$, $-\sqrt{7}$
real numbers	$\mathbb{R}$	all rational and irrational numbers together	every point on the number line

The three dots "$\ldots$" mean “and so on, forever”.

A rational number is any number that can be written as $\dfrac{a}{b}$, where $a$ and $b$ are integers and $b \ne 0$. The word comes from ratio. For example, $-2.5 = \dfrac{-5}{2}$ and $7 = \dfrac{7}{1}$, so both are rational.

How the sets fit together

Every natural number is also a whole number. Every whole number is also an integer. Every integer is also a rational number (write it over $1$). So the sets sit inside each other:

Each set sits inside the next. Rational and irrational numbers don’t overlap, and together they make the real numbers.

So a number can belong to several sets at once. The number $5$ is natural, whole, an integer, rational and real. But a number is either rational or irrational, never both.

Rational or irrational? Look at the decimal

Every rational number has a decimal that either stops or repeats:

• Terminating (stops): $\dfrac{3}{8} = 0.375$
• Repeating (a block of digits repeats forever): $\dfrac{3}{11} = 0.272727\ldots = 0.\overline{27}$

The bar over the digits shows which block repeats.

An irrational number has a decimal that goes on forever without a repeating block. For example, $\sqrt{2} = 1.41421356\ldots$ and $\pi = 3.14159265\ldots$ never settle into a pattern that repeats. Some irrational numbers have a pattern, just not a repeating one: $0.101001000100001\ldots$ (one more $0$ each time) never repeats the same block, so it’s irrational.

Square roots. The square root of a perfect square is rational: $\sqrt{49} = 7$ and $\sqrt{\dfrac{4}{9}} = \dfrac{2}{3}$. The square root of a whole number that is not a perfect square, like $\sqrt{2}$, $\sqrt{10}$ or $\sqrt{50}$, is irrational. (In Grade 11 you’ll learn to write square roots like $\sqrt{50}$ in a simpler form, in simplifying radicals.)

Pi. $\pi$ is irrational. The values $3.14$ and $\dfrac{22}{7}$ are only approximations: $\dfrac{22}{7} = 3.\overline{142857}$, which repeats, so it can’t be exactly $\pi$.

Density: there’s always another one in between

Between any two different rational numbers, there’s always another rational number. One easy way to find one is the average (add them and divide by $2$). For example, between $\dfrac{1}{3}$ and $\dfrac{1}{2}$:

$$\frac{1}{2}\left(\frac{1}{3} + \frac{1}{2}\right) = \frac{1}{2}\left(\frac{2}{6} + \frac{3}{6}\right) = \frac{1}{2} \times \frac{5}{6} = \frac{5}{12}$$

Now you can find a number between $\dfrac{1}{3}$ and $\dfrac{5}{12}$, and so on, forever. So between any two rational numbers there are infinitely many rational numbers. We say the rational numbers are dense. The real numbers are dense too.

The integers are not dense: there’s no integer between $3$ and $4$.

Infinity

Each of these sets is infinite: it never runs out. There’s no largest natural number, because whatever number $n$ you pick, $n + 1$ is bigger. Infinity ($\infty$) is not a number you can reach; it’s the idea of “going on without end”.

Limits: getting closer and closer

Add up $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \cdots$, one term at a time, and watch the running total:

Terms added	$1$	$2$	$3$	$4$	$5$
Total so far	$\dfrac{1}{2}$	$\dfrac{3}{4}$	$\dfrac{7}{8}$	$\dfrac{15}{16}$	$\dfrac{31}{32}$
Gap left to $1$	$\dfrac{1}{2}$	$\dfrac{1}{4}$	$\dfrac{1}{8}$	$\dfrac{1}{16}$	$\dfrac{1}{32}$

Each new term fills half of the gap that’s left. The total never goes past $1$, but it gets as close to $1$ as you like if you keep going. We say the limit of the total is $1$, and that the infinite sum equals $1$. You’ll meet sums like this again in Grade 11, in geometric series.

Each jump covers half of the distance that’s left, so the totals close in on $1$.

Why 0.999… equals 1

Look at $0.9,\ 0.99,\ 0.999,\ \ldots$ The gaps to $1$ are $0.1,\ 0.01,\ 0.001,\ \ldots$ Each one is $\dfrac{1}{10}$ of the one before. With infinitely many nines, the gap is smaller than any positive number you can name, so the gap must be $0$:

$$0.\overline{9} = 1$$

Another way to see it: $\dfrac{1}{3} = 0.\overline{3}$. Multiply both sides by $3$: $1 = 0.\overline{9}$.

Be careful: this is only true with nines that go on forever. The number $0.999$ (exactly three nines) is less than $1$.

Where this comes from

Irrational numbers in ancient Greece. The Pythagoreans, a group of Greek thinkers from around 500 BCE, believed at first that every length could be written as a ratio of whole numbers. Then, according to tradition, someone in their school showed that the diagonal of a square with side length $1$ (which is $\sqrt{2}$) can’t be written that way. This is traditionally regarded as the first discovery that some lengths can’t be written as a ratio of whole numbers, which is what we now call irrational numbers.

Zero. Many cultures used a symbol to mark an empty place in a number, including the Babylonians and the Maya. In India, the mathematician Brahmagupta (628 CE) went further and wrote rules for calculating with zero as a number in its own right, along with rules for negative numbers, which he described as debts. Our digits $0$ to $9$ come from this Indian system, passed on through Arabic mathematicians. Today, zero is everywhere: every computer stores information using only $0$s and $1$s.

Worked examples

Example 1: Classifying numbers

Put a check mark under every set each number belongs to: $-8$, $0$, $15$, $\dfrac{2}{3}$, $0.\overline{6}$, $\sqrt{36}$, $\sqrt{20}$, $\pi$, $-4.25$.

Solution. Simplify first where you can: $\sqrt{36} = 6$, and $0.\overline{6} = \dfrac{2}{3}$ (check: $2 \div 3 = 0.666\ldots$). Also $-4.25 = -\dfrac{17}{4}$, a fraction of integers.

Number	$\mathbb{N}$	$\mathbb{W}$	$\mathbb{Z}$	$\mathbb{Q}$	irrational	$\mathbb{R}$
$-8$			✓	✓		✓
$0$		✓	✓	✓		✓
$15$	✓	✓	✓	✓		✓
$\dfrac{2}{3}$				✓		✓
$0.\overline{6}$				✓		✓
$\sqrt{36} = 6$	✓	✓	✓	✓		✓
$\sqrt{20}$					✓	✓
$\pi$					✓	✓
$-4.25$				✓		✓

$\sqrt{20}$ is irrational because $20$ is not a perfect square ($16 \lt 20 \lt 25$). Notice that every number gets a check under $\mathbb{R}$.

Example 2: Rational or irrational?

Decide whether each number is rational or irrational. If it’s rational, write it as a fraction.

• (a) $0.125$
• (b) $0.454545\ldots$
• (c) $0.123456789101112\ldots$ (the counting numbers written one after another)
• (d) $\sqrt{0.25}$

Solution.

(a) The decimal stops, so it’s rational: $0.125 = \dfrac{125}{1000} = \dfrac{1}{8}$.

(b) The block $45$ repeats forever, so it’s rational. In fact $0.\overline{45} = \dfrac{5}{11}$. Check by dividing: $5 \div 11 = 0.4545\ldots$ ✓

(c) The digits follow a rule, but no block of digits repeats forever (the numbers keep getting longer). So it’s irrational.

(d) $0.5 \times 0.5 = 0.25$, so $\sqrt{0.25} = 0.5 = \dfrac{1}{2}$. It’s rational, even though it has a square root sign.

Example 3: Numbers in between

Find three rational numbers between $\dfrac{2}{5}$ and $\dfrac{1}{2}$.

Solution. Write both fractions with a common denominator. With denominator $20$, they’re $\dfrac{8}{20}$ and $\dfrac{10}{20}$. That leaves room for only one numerator in between ($9$), so use a bigger denominator. With $40$:

$$\frac{2}{5} = \frac{16}{40} \qquad \text{and} \qquad \frac{1}{2} = \frac{20}{40}$$

Three numbers in between are $\dfrac{17}{40}$, $\dfrac{18}{40} = \dfrac{9}{20}$ and $\dfrac{19}{40}$.

Check with decimals: $\dfrac{2}{5} = 0.4$ and $\dfrac{1}{2} = 0.5$, while $\dfrac{17}{40} = 0.425$, $\dfrac{9}{20} = 0.45$ and $\dfrac{19}{40} = 0.475$. All three are between $0.4$ and $0.5$. ✓

You could keep going with denominators $80$, $160$, and so on, so there are infinitely many answers.

Example 4: Walking toward a wall

You stand $8$ m from a wall. Each step, you walk half of the distance that’s left. Make a table for the first five steps. Do you ever reach the wall? What is the limit of the distance you’ve walked?

Solution. Each step halves the distance left:

Step	$1$	$2$	$3$	$4$	$5$
Step length (m)	$4$	$2$	$1$	$0.5$	$0.25$
Total walked (m)	$4$	$6$	$7$	$7.5$	$7.75$
Distance left (m)	$4$	$2$	$1$	$0.5$	$0.25$

After any number of steps, there’s still some distance left (half of the last amount), so you never quite reach the wall. But the distance left gets as small as you like, so the total walked gets closer and closer to $8$ m. The limit is $8$ m.

Common mistakes

Thinking every decimal that goes on forever is irrational. $0.333\ldots$ goes on forever, but it repeats, so it’s rational ($\dfrac{1}{3}$). Irrational decimals go on forever and never repeat.

Thinking every square root is irrational. $\sqrt{49} = 7$ and $\sqrt{0.25} = 0.5$ are rational. Simplify first: only the square root of a non-perfect square (like $\sqrt{20}$) is irrational.

Saying $\pi$ equals $3.14$ or $\dfrac{22}{7}$. Those are handy approximations, but they’re rational and $\pi$ isn’t. Write $\pi \approx 3.14$ (with "$\approx$"), not $\pi = 3.14$.

Mixing up natural and whole numbers. Zero is a whole number but not a natural number. Negative numbers are neither: $-3$ is an integer, but not a whole number.

Putting a number in only one set. The number $7$ isn’t “just a natural number”. It’s also whole, an integer, rational and real. Check every set, starting from the smallest.

Thinking an endless sum must be endlessly large. $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots$ has infinitely many terms, but the total never goes past $1$. When the terms shrink fast enough, the total can have a limit.

Practice

1. (Warm-up) List every set each number belongs to: $\mathbb{N}$, $\mathbb{W}$, $\mathbb{Z}$, $\mathbb{Q}$, irrational, $\mathbb{R}$.

• (a) $-12$
• (b) $0$
• (c) $\dfrac{5}{8}$
• (d) $\sqrt{11}$

Solution

(a) $-12$: integers, rational, real.

(b) $0$: whole numbers, integers, rational, real. (Not natural.)

(c) $\dfrac{5}{8}$: rational, real.

(d) $\sqrt{11}$: $11$ is not a perfect square ($9 \lt 11 \lt 16$), so it’s irrational and real.

2. (Warm-up) Rational or irrational?

• (a) $0.777\ldots$
• (b) $\sqrt{81}$
• (c) $\pi + 1$
• (d) $1.010010001\ldots$ (one more $0$ each time)
• (e) $-\sqrt{16}$

Solution

(a) Rational: the $7$ repeats. ($0.\overline{7} = \dfrac{7}{9}$.)

(b) Rational: $\sqrt{81} = 9$.

(c) Irrational: $\pi + 1 = 4.14159\ldots$ has the same never-repeating decimal digits as $\pi$.

(d) Irrational: there’s a pattern, but no block of digits repeats.

(e) Rational: $-\sqrt{16} = -4$.

3. (Core) Show that each number is rational by writing it as a fraction in lowest terms.

• (a) $0.35$
• (b) $0.\overline{6}$
• (c) $-2.25$

Solution

(a) $0.35 = \dfrac{35}{100} = \dfrac{7}{20}$.

(b) $0.\overline{6} = \dfrac{2}{3}$. Check: $2 \div 3 = 0.666\ldots$ ✓

(c) $-2.25 = -\dfrac{225}{100} = -\dfrac{9}{4}$.

4. (Core) Find two rational numbers between $-\dfrac{1}{4}$ and $-\dfrac{1}{5}$.

Solution

Use the common denominator $60$:

$$-\frac{1}{4} = -\frac{15}{60} \qquad \text{and} \qquad -\frac{1}{5} = -\frac{12}{60}$$

Two numbers in between are $-\dfrac{14}{60} = -\dfrac{7}{30}$ and $-\dfrac{13}{60}$.

Check with decimals: $-\dfrac{1}{4} = -0.25$ and $-\dfrac{1}{5} = -0.2$, while $-\dfrac{7}{30} = -0.2\overline{3}$ and $-\dfrac{13}{60} = -0.21\overline{6}$. Both are between $-0.25$ and $-0.2$. ✓ (Other answers are possible.)

5. (Core) True or false? Explain each answer.

• (a) Every integer is a rational number.
• (b) Every rational number is an integer.
• (c) Zero is a natural number.
• (d) Some real numbers are not rational.

Solution

(a) True. Any integer $n$ can be written as $\dfrac{n}{1}$.

(b) False. $\dfrac{1}{2}$ is rational but not an integer.

(c) False. The natural numbers start at $1$. Zero is a whole number.

(d) True. Irrational numbers like $\sqrt{2}$ and $\pi$ are real but not rational.

6. (Core) How many integers are there between $3$ and $4$? How many rational numbers? Explain the difference.

Solution

There are no integers between $3$ and $4$: the integers jump straight from $3$ to $4$.

There are infinitely many rational numbers, such as $3.5$, $3.25$, $3.125$, … Each time, you can take the average of $3$ and the last number to get a new one. The rational numbers are dense; the integers are not.

7. (Core) A square patio has an area of $30 \text{ m}^2$.

• (a) Is its side length rational or irrational?
• (b) Between which two whole numbers is the side length? Estimate it to one decimal place.

Solution

(a) The side length is $\sqrt{30}$ m. Since $30$ is not a perfect square, $\sqrt{30}$ is irrational.

(b) $5^2 = 25$ and $6^2 = 36$, and $30$ is between them, so the side is between $5$ m and $6$ m. Try $5.5$: $5.5^2 = 30.25$, a little too big. Try $5.4$: $5.4^2 = 29.16$, too small. $30$ is closer to $30.25$, so the side is about $5.5$ m. (A calculator gives $\sqrt{30} \approx 5.477$.)

8. (Challenge) Look at the sum $\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{27} + \dfrac{1}{81} + \cdots$, where each term is $\dfrac{1}{3}$ of the one before.

• (a) Find the total after $1$, $2$, $3$ and $4$ terms.
• (b) Find the gap between each total and $\dfrac{1}{2}$. What pattern do you see?
• (c) What is the limit of the total?

Solution

(a) Use the denominators $3$, $9$, $27$, $81$:

$$\frac{1}{3}, \qquad \frac{3}{9} + \frac{1}{9} = \frac{4}{9}, \qquad \frac{12}{27} + \frac{1}{27} = \frac{13}{27}, \qquad \frac{39}{81} + \frac{1}{81} = \frac{40}{81}$$

(b) The gaps to $\dfrac{1}{2}$ are

$$\frac{1}{2} - \frac{1}{3} = \frac{1}{6}, \quad \frac{1}{2} - \frac{4}{9} = \frac{1}{18}, \quad \frac{1}{2} - \frac{13}{27} = \frac{1}{54}, \quad \frac{1}{2} - \frac{40}{81} = \frac{1}{162}$$

Each gap is $\dfrac{1}{3}$ of the one before.

(c) The gaps shrink toward $0$, and the totals never pass $\dfrac{1}{2}$. So the limit is $\dfrac{1}{2}$.

9. (Challenge) Use $\dfrac{1}{3} = 0.\overline{3}$ and $\dfrac{2}{3} = 0.\overline{6}$ to explain why $0.\overline{9} = 1$. Then explain why there’s no number between $0.\overline{9}$ and $1$.

Solution

Add the fractions and add the decimals:

$$\frac{1}{3} + \frac{2}{3} = 1 \qquad \text{and} \qquad 0.333\ldots + 0.666\ldots = 0.999\ldots$$

The two sums are the same number, so $0.\overline{9} = 1$.

If two different numbers had a gap between them, you could find a number in between (density). But $0.\overline{9}$ is closer to $1$ than $0.9$, $0.99$, $0.999$, and every other decimal with a finite number of nines. So the gap between $0.\overline{9}$ and $1$ is smaller than $0.1$, $0.01$, $0.001$, …, which means it’s $0$. There’s no room for a number in between, because they’re the same number.