Real Numbers — Class 9, Unit 1

How to use this chapter

Every section here is set up the same way: a quick definition, a few worked examples, then the exercise problems from the textbook — all solved step by step. The small serif notes beside each line explain why we took that step, not just what the step is. Math equations are hand-typeset with LaTeX; short video walkthroughs accompany the harder examples.

1.1 Rational Numbers

Definition

A rational number is any number that can be expressed in the form $\dfrac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$. The set of rational numbers is denoted by $\mathbb{Q}$.

So integers, terminating decimals, and repeating decimals are all rational. The only numbers that get excluded from $\mathbb{Q}$ are ones whose decimal expansion is non-terminating and non-repeating — and we'll meet those in the next section.

Problem Which of the following are rational numbers? $\quad \dfrac{3}{7}, \quad -5, \quad 0.75, \quad 0.\overline{3}, \quad \sqrt{2}$

Test each number against the definition — can it be written as $\dfrac{p}{q}$ with integer $p, q$ and $q \neq 0$?

$\dfrac{3}{7}$ is already in the form $\dfrac{p}{q}$. Rational.

$-5 = \dfrac{-5}{1}$. Rational.

Every integer is rational — just place it over 1.

$0.75 = \dfrac{75}{100} = \dfrac{3}{4}$. Rational.

A terminating decimal can always be written with a power of 10 in the denominator, then simplified.

$0.\overline{3}$ means $0.3333\ldots$ Let $x = 0.\overline{3}$. Then $10x = 3.\overline{3}$, and $10x - x = 3$, so $x = \dfrac{3}{9} = \dfrac{1}{3}$. Rational.

$\sqrt{2} = 1.41421356\ldots$ — non-terminating and non-repeating. Not rational.

We'll prove this properly in the next section.

Rational: $\dfrac{3}{7},\ -5,\ 0.75,\ 0.\overline{3}$

1.2 Irrational Numbers

Definition

An irrational number is a real number that cannot be written as $\dfrac{p}{q}$ where $p, q \in \mathbb{Z}$ and $q \neq 0$. Equivalently, its decimal expansion is non-terminating and non-repeating.

Common irrationals: $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\pi$, $e$. The set of irrational numbers is denoted $\mathbb{Q}'$ (read "Q-prime"). The real numbers $\mathbb{R}$ are the union $\mathbb{Q} \cup \mathbb{Q}'$.

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

Suppose, for contradiction, that $\sqrt{2}$ is rational. Then $\sqrt{2} = \dfrac{p}{q}$ where $p, q$ are integers with no common factor (i.e. the fraction is in lowest terms) and $q \neq 0$.

We always start contradiction proofs by assuming the opposite of what we want to show.

Squaring both sides: $2 = \dfrac{p^2}{q^2}$, so $p^2 = 2q^2$.

Since $p^2 = 2q^2$, $p^2$ is even. But if $p^2$ is even, $p$ itself must be even (because the square of an odd number is odd). So write $p = 2k$ for some integer $k$.

Substitute back: $(2k)^2 = 2q^2$, giving $4k^2 = 2q^2$, so $q^2 = 2k^2$. By the same reasoning, $q$ must also be even.

But now $p$ and $q$ are both even, so they share a factor of 2 — contradicting our assumption that $\dfrac{p}{q}$ was in lowest terms.

The contradiction shows our assumption was false.

Therefore $\sqrt{2}$ cannot be rational. It is irrational. $\blacksquare$

Video walkthrough

4-minute walkthrough of this proof, with every step narrated. Coming soon.

1.3 Decimal Representation

Every real number has a unique decimal representation. The key fact: rational numbers have terminating or repeating decimals; irrational numbers have non-terminating non-repeating decimals.

Problem Convert $0.\overline{27}$ (i.e. $0.272727\ldots$) to a fraction in lowest terms.

Let $x = 0.\overline{27}$.

The repeating block has 2 digits, so multiply by $10^2 = 100$: $100x = 27.\overline{27}$.

We multiply by $10^n$ where $n$ is the length of the repeating block, so the repeating parts line up.

Subtract: $100x - x = 27.\overline{27} - 0.\overline{27} = 27$, so $99x = 27$.

Solve: $x = \dfrac{27}{99} = \dfrac{3}{11}$ (dividing top and bottom by their HCF, 9).

$0.\overline{27} = \dfrac{3}{11}$

1.4 Radicals & Radicands

Definition

In the expression $\sqrt[n]{a}$, the symbol $\sqrt{\ }$ is the radical sign, $n$ is the index, and $a$ is the radicand. If $n = 2$ we write just $\sqrt{a}$ for the square root.

Problem Simplify: (a) $\sqrt{50}$, (b) $\sqrt{72}$, (c) $\sqrt{18} + \sqrt{8}$.

(a) Factor the radicand to pull out a perfect square: $50 = 25 \cdot 2$, so $\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}$.

(b) $72 = 36 \cdot 2$, so $\sqrt{72} = 6\sqrt{2}$.

Always pull out the largest perfect square. If we'd used $72 = 4 \cdot 18$ we'd have to simplify again.

(c) Simplify each term first: $\sqrt{18} = 3\sqrt{2}$ and $\sqrt{8} = 2\sqrt{2}$. Then $\sqrt{18} + \sqrt{8} = 3\sqrt{2} + 2\sqrt{2} = 5\sqrt{2}$.

You can only add surds if the part under the radical matches — just like combining like terms.

(a) $5\sqrt{2}$, (b) $6\sqrt{2}$, (c) $5\sqrt{2}$

1.5 Laws of Exponents

For any $a, b > 0$ and real numbers $m, n$, the following hold:

The five laws

a^m \cdot a^n = a^{m+n}

\dfrac{a^m}{a^n} = a^{m-n}

(a^m)^n = a^{mn}

$(ab)^n = a^n b^n$

\left(\dfrac{a}{b}\right)^n = \dfrac{a^n}{b^n}

Problem Simplify $\dfrac{x^5 \cdot x^{-2}}{x^3}$.

Combine the numerator using law 1: $x^5 \cdot x^{-2} = x^{5 + (-2)} = x^3$.

Now divide using law 2: $\dfrac{x^3}{x^3} = x^{3-3} = x^0 = 1$.

Any non-zero number to the power zero is 1, by convention and consistency with the laws.

Answer: $1$

Ex 1.1 Exercise 1.1 — selected problems

Below are the textbook exercise problems with complete worked solutions. (Full exercise set — all 23 problems — will ship with the v1.1 update; these are the starters.)

Problem Classify each as rational or irrational: $\dfrac{4}{9}, \sqrt{16}, \sqrt{17}, \pi, -\dfrac{7}{2}, 1.75$

$\dfrac{4}{9}$ — already in $\dfrac{p}{q}$ form. Rational.

$\sqrt{16} = 4 = \dfrac{4}{1}$. Rational.

The radical looks irrational at first glance, but it evaluates to an integer.

$\sqrt{17}$ — 17 is prime, not a perfect square. Irrational.

$\pi \approx 3.14159265\ldots$ — non-terminating, non-repeating. Irrational.

$-\dfrac{7}{2}$ — already in form. Rational.

$1.75 = \dfrac{175}{100} = \dfrac{7}{4}$. Rational.

Problem Convert $0.\overline{6}$ to a fraction.

Let $x = 0.\overline{6}$. Then $10x = 6.\overline{6}$, so $10x - x = 6$.

$9x = 6 \Rightarrow x = \dfrac{6}{9} = \dfrac{2}{3}$.

$0.\overline{6} = \dfrac{2}{3}$

Problem Simplify $\sqrt{45} + 2\sqrt{20} - \sqrt{80}$.

Simplify each radical: $\sqrt{45} = 3\sqrt{5}$, $\sqrt{20} = 2\sqrt{5}$, $\sqrt{80} = 4\sqrt{5}$.

Substitute: $3\sqrt{5} + 2(2\sqrt{5}) - 4\sqrt{5} = 3\sqrt{5} + 4\sqrt{5} - 4\sqrt{5} = 3\sqrt{5}$.

Answer: $3\sqrt{5}$