Rational and Irrational Numbers

When you write $\dfrac{3}{4}$ or $-\dfrac{7}{2}$ you are using a rational number , a fraction of two integers. When you write $\sqrt{2}$ or $\pi$ , you are using something different. Both kinds of numbers live on the same line, and together they form the real numbers. This lesson sorts out what each kind is, how to spot one, and how to find many of them in any tiny stretch of the line.

Definitions

A rational number is a number that can be written in the form $r = \frac{p}{q}, \quad p, q \in \mathbb{Z}, \ q \ne 0.$ Examples: $\dfrac{3}{4}, -\dfrac{7}{2}, 5 = \dfrac{5}{1}, 0 = \dfrac{0}{1}, 0.25 = \dfrac{1}{4}$ .

A real number that is not rational is called irrational. So an irrational number is one that cannot be expressed as $\dfrac{p}{q}$ for any integers $p$ and $q$ . Examples include $\sqrt{2},\sqrt{3},\sqrt{5}, \pi$ , and $0.1010010001\ldots$ (each block of zeros one longer than the last).

The set of rationals is denoted $\mathbb{Q}$ , the set of irrationals has no standard one-letter name, and together they form the real numbers, $\mathbb{R}$ .

Concept and structure

Rationals are dense. Between any two distinct rationals there are infinitely many other rationals. The cheapest way to find one is the mean (or average): $\text{If } a < b, \quad \text{then } a < \frac{a+b}{2} < b.$ Now repeat with $\big(a, \tfrac{a+b}{2}\big)$ to get another rational between them. This trick produces as many rationals as you like in any open interval.

There are many irrationals too. A famous theorem of Cantor says that, even though both kinds are infinite, there are strictly more irrationals than rationals. Geometrically this means the rationals, dense as they are, still leave gaps that the irrationals fill.

Why $\sqrt{2}$ is irrational. Suppose $\sqrt{2} = \dfrac{p}{q}$ in lowest terms. Squaring gives $2q^2 = p^2$ , so $p^2$ is even, hence $p$ is even, say $p = 2k$ . Then $2q^2 = 4k^2$ , so $q^2 = 2k^2$ , hence $q$ is also even. But then $\dfrac{p}{q}$ wasn't in lowest terms , contradiction. So $\sqrt{2}$ is not rational. The same argument, with "even" replaced by "divisible by $p$ ", proves $\sqrt{p}$ is irrational for every prime $p$ . (We will revisit this proof in Chapter X.)

A practical way to spot irrationals. Numbers like $\sqrt{4} = 2$ or $\sqrt{49} = 7$ are rational because the square root is an integer. The number $\sqrt{n}$ is irrational precisely when $n$ is a positive integer that is not a perfect square. Thus $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}, \sqrt{7}, \sqrt{8}, \sqrt{10}, \ldots$ are all irrational.

A famous non-surd irrational: $\pi$ . The number $\pi$ is the ratio of a circle's circumference to its diameter. It is irrational (and, more strongly, transcendental), although the proof is much harder than the one for $\sqrt{2}$ . Be careful: $\dfrac{22}{7}$ is just an approximation to $\pi$ , not equal to it.

Constructing irrationals between two rationals. Suppose you want an irrational between $\dfrac{1}{2}$ and $\dfrac{3}{4}$ . Pick the rational mid-point $\dfrac{5}{8}$ and add a tiny known irrational, say $\dfrac{\sqrt{2}}{100}$ . Since (rational) $+$ (small irrational) is irrational and we kept the change small, the result still lies between $\dfrac{1}{2}$ and $\dfrac{3}{4}$ .

Worked examples

Example 1. Find three rationals between $\dfrac{1}{3}$ and $\dfrac{1}{2}$ .

Take averages repeatedly. Mid-point: $\dfrac{1}{2}\!\left(\dfrac{1}{3} + \dfrac{1}{2}\right) = \dfrac{5}{12}$ . Between $\dfrac{1}{3}$ and $\dfrac{5}{12}$ : $\dfrac{1}{2}\!\left(\dfrac{1}{3} + \dfrac{5}{12}\right) = \dfrac{3}{8}$ . Between $\dfrac{5}{12}$ and $\dfrac{1}{2}$ : $\dfrac{11}{24}$ . So $\dfrac{3}{8}, \dfrac{5}{12}, \dfrac{11}{24}$ work.

Example 2. Find a rational and an irrational between $0.1$ and $0.11$ .

Rational: $\dfrac{0.1 + 0.11}{2} = 0.105$ . Irrational: $0.10100100010001\ldots$ , non-terminating, non-repeating, and clearly between $0.1$ and $0.11$ .

Example 3. Is $\sqrt{45}$ rational or irrational? Why?

$\sqrt{45} = \sqrt{9 \cdot 5} = 3\sqrt{5}$ . Since $5$ is not a perfect square, $\sqrt{5}$ is irrational, and a non-zero rational times an irrational is irrational. So $\sqrt{45}$ is irrational.

Example 4. Show that $3 + 2\sqrt{2}$ is irrational, given $\sqrt{2}$ is irrational.

If $3 + 2\sqrt{2} = r$ were rational, then $\sqrt{2} = \dfrac{r-3}{2}$ would also be rational (difference and quotient of rationals are rational). That contradicts the irrationality of $\sqrt{2}$ .

Example 5. Find one rational and one irrational between $\sqrt{2}$ and $\sqrt{3}$ .

Use decimals: $\sqrt{2} \approx 1.414$ , $\sqrt{3} \approx 1.732$ . Rational $1.5 = \dfrac{3}{2}$ lies between them. Irrational: $1.5050050005\ldots$ also lies between them.

Try it yourself

Insert five rationals between $\dfrac{3}{5}$ and $\dfrac{4}{5}$ .
Are the following rational or irrational? $\sqrt{36}, \sqrt{40}, \sqrt[3]{27}, \pi - 2, 0.\overline{3}$ .
Give two examples of irrationals whose sum is rational.
Give two examples of irrationals whose product is rational.
Find an irrational between $\dfrac{2}{7}$ and $\dfrac{3}{7}$ .
Show that $5 - \sqrt{3}$ is irrational.
Is $\dfrac{\sqrt{12}}{\sqrt{3}}$ rational or irrational? Justify.
Insert three rationals and three irrationals between $0.\overline{1}$ and $0.\overline{2}$ .
State whether true or false: "The product of two irrationals is always irrational." Give a reason.
If $\sqrt{2} + \sqrt{3}$ were rational, derive a contradiction.

Pitfalls / Insight

"Rational" does not mean "with a nice decimal". $\dfrac{1}{3} = 0.\overline{3}$ never terminates but is rational.
Irrational $+$ irrational need not be irrational. $(\sqrt{2}) + (3 - \sqrt{2}) = 3$ .
Be precise about $\pi$ . $\pi \ne \dfrac{22}{7}$ ; the latter is a handy approximation only.

Insight. Think of rationals as "fractions you can write" and irrationals as "fractions you cannot , only approximate". The number line is so full that almost every point you randomly pick is irrational; the rationals are only the labelled points.