Rational and Irrational Numbers

The dividing line

Every real number is one or the other.

	Can be written as $\tfrac{a}{b}$?	Decimal
rational	yes	stops or repeats
irrational	no	runs on with no pattern

$$\tfrac{3}{4}, \; 7, \; -2.5, \; 0.333\ldots \quad \text{rational}$$ $$\sqrt{2}, \; \pi, \; \sqrt{10} \quad \text{irrational}$$

The name comes from ratio. A rational number is one expressible as a ratio of two integers; an irrational number is not.

Why a rational number’s decimal must stop or repeat

This is the part worth understanding rather than accepting, and the reason is short.

Divide $3$ by $7$ by long division. At each step you take a remainder and divide again. The possible remainders are $0, 1, 2, 3, 4, 5$ or $6$ — only seven of them, because a remainder is always less than the divisor.

So within at most seven steps, either:

• a remainder of $0$ appears and the decimal stops, or
• a remainder repeats, and from there the whole sequence of digits repeats with it.

$$\tfrac{3}{7} = 0.428571\,428571\ldots$$

There is no third possibility. A finite set of remainders cannot produce an endless pattern-free decimal, which is exactly why an irrational number can never be written as a fraction.

Repeating decimals are rational

A decimal that never ends can still be rational, provided it repeats:

$$0.333\ldots = \tfrac{1}{3} \qquad 0.142857\,142857\ldots = \tfrac{1}{7} \qquad 0.1666\ldots = \tfrac{1}{6}$$

The repetition is what makes the fraction findable, and the equality is exact rather than approximate. $0.333\ldots$ is not slightly less than $\tfrac{1}{3}$; it is $\tfrac{1}{3}$.

A bar over the repeating digits is the usual notation:

$$0.\overline{3} = \tfrac{1}{3} \qquad 0.\overline{142857} = \tfrac{1}{7}$$

Which square roots are irrational

The roots of perfect squares are whole numbers, so they are rational:

$$\sqrt{9} = 3 \qquad \sqrt{144} = 12$$

Every other square root is irrational. $\sqrt{2}$, $\sqrt{3}$, $\sqrt{5}$, $\sqrt{10}$ all run forever without repeating.

That $\sqrt{2}$ is irrational is provable, and the proof is one of the oldest in mathematics: assume $\sqrt{2} = \tfrac{a}{b}$ in lowest terms, and a short argument shows $a$ and $b$ must both be even — contradicting “lowest terms”. So no such fraction exists.

Approximating to compare and place

An irrational number’s decimal cannot be written out, so working with one means approximating it with a rational.

Where does $\sqrt{2}$ sit?

$$1.4^2 = 1.96 \qquad 1.5^2 = 2.25$$

So $\sqrt{2}$ is between $1.4$ and $1.5$. Narrow it:

$$1.41^2 = 1.9881 \qquad 1.42^2 = 2.0164$$ $$1.41 < \sqrt{2} < 1.42$$

Two decimal places is enough to plot it, order it against other numbers, or answer any practical question. Squaring the candidate is the check, and it needs no calculator.

Comparing irrationals

Approximate each one, then compare the approximations.

Which is larger, $\sqrt{7}$ or $2.5$?

$$2.5^2 = 6.25 \qquad \text{and} \qquad (\sqrt{7})^2 = 7$$

$7 > 6.25$, so $\sqrt{7} > 2.5$.

Squaring both is quicker than approximating, and it is exact. Comparing $\sqrt{7}$ with $\sqrt{8}$ is quicker still — the larger radicand gives the larger root.

Why the distinction matters

An exact answer keeps the radical. $\sqrt{2}$ is exact; $1.414$ is not, and neither is any decimal you could write.

That is why answers to the Pythagorean theorem and the quadratic formula are so often left as roots. A diagonal of $\sqrt{2}$ meters is a precise length; rounding it to $1.41$ throws information away, and errors from rounding early compound through later steps.

Irrational numbers also settle a question the number line raises. The rationals leave gaps — there is no fraction at the diagonal of a unit square — and the irrationals are what fill them. Between any two rationals lies an irrational, and between any two irrationals lies a rational.

Worked examples

Common mistakes

Sums and products

Add or multiply two rational numbers and the result is rational. Write them as fractions and look:

$$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd} \qquad \frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$$

The tops and bottoms are sums and products of whole numbers, so they are whole numbers too. The bottom is not zero, since neither $b$ nor $d$ was. That makes each answer a fraction of integers, which is what rational means.

Mix in an irrational number and the answer changes.

A rational plus an irrational is irrational. Suppose $r$ is rational and $x$ is irrational, and suppose $r + x$ came out rational — call it $q$. Then

$$x = q - r$$

and a difference of two rationals is rational. That says $x$ is rational, which it is not. So $r + x$ cannot be rational.

A nonzero rational times an irrational is irrational. The same move works. If $r \cdot x = q$ with $q$ rational, then $x = \tfrac{q}{r}$, a quotient of rationals, which is rational — impossible. The word nonzero matters, because $0 \cdot \sqrt{2} = 0$.

Combination	Result
rational $+$ rational	always rational
rational $\times$ rational	always rational
rational $+$ irrational	always irrational
nonzero rational $\times$ irrational	always irrational
irrational $+$ irrational	could be either
irrational $\times$ irrational	could be either

The last two rows really do go both ways. $\sqrt{2} + (-\sqrt{2}) = 0$ and $\sqrt{2} \cdot \sqrt{8} = \sqrt{16} = 4$ are rational. But $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$ is irrational, by the rule for a nonzero rational times an irrational.

Practice problems