7.2 Rational and irrational numbers

Why rational numbers are not enough

The integers are closed under addition, subtraction, and multiplication, but not under division. To solve equations such as

$$3x=2,$$

we enlarge the number system from $\mathbb Z$ to the rational numbers $\mathbb Q$. This enlargement is extremely useful: it allows fractions, it is stable under the four ordinary arithmetic operations, and it is still governed by integer divisibility through numerators and denominators.

However, even $\mathbb Q$ is not large enough for all familiar equations. The equation

$$x^2=2$$

has a real solution, namely the nonnegative square root $\sqrt2$, but that solution is not rational. The purpose of this section is to make this contrast precise. We first record the arithmetic stability of rational numbers, then use prime divisibility to prove that certain roots cannot be rational.

Rational and irrational numbers

The denominator condition $n\ne0$ is essential. Division by zero is not defined, so a rational representation must always have a nonzero denominator. Also, rational representations are not unique:

$$\frac12=\frac24=\frac{-3}{-6}.$$

When proving that a number is rational, it is enough to produce one valid fraction. When proving that a number is irrational, one must prove that no such fraction exists.

It is often convenient to reduce a rational number to lowest terms. If $x\in\mathbb Q$ and $x>0$, then we may write

$$x=\frac ab$$

with $a,b\in\mathbb Z^+$ and $\gcd(a,b)=1$. The condition $\gcd(a,b)=1$ means that all common factors have already been cancelled. Many irrationality proofs begin by assuming such a lowest-terms representation and then showing that a prime divides both a and b, which is impossible.

Closure of the rational numbers

This theorem says that rational numbers are closed under the four arithmetic operations, except that division by zero is still excluded.

To prove it, write

$$x=\frac mn,\qquad y=\frac pq,$$

where $m,n,p,q\in\mathbb Z$ and $n\ne0$, $q\ne0$. Then

$$x+y=\frac{mq+np}{nq}, \qquad x-y=\frac{mq-np}{nq}, \qquad xy=\frac{mp}{nq}.$$

The numerators and denominators displayed here are integers, and the denominator nq is nonzero. Hence all three numbers are rational.

For division, suppose $y\ne0$. Since

$$y=\frac pq,$$

the condition $y\ne0$ forces $p\ne0$. Therefore

$$\frac xy=\frac{m/n}{p/q}=\frac{mq}{np}.$$

Again the numerator and denominator are integers, and $np\ne0$, so $x/y$ is rational.

For example, $\sqrt2$ is irrational, but

$$\sqrt2+(-\sqrt2)=0$$

is rational. Also,

$$\sqrt2\cdot\sqrt2=2$$

is rational. The word "irrational" means "not in $\mathbb Q$"; it does not create a separate number system closed under the usual arithmetic operations.

Nonnegative nth roots

The word "nonnegative" is doing important work. For square roots, both 3 and $-3$ satisfy $x^2=9$, but only 3 is the nonnegative square root. Thus

$$\sqrt9=3,$$

not $\pm3$. The notation $\sqrt[n]{a}$ always refers to the unique nonnegative root when $a\ge0$.

For odd n, negative real roots can also be discussed, but this section only needs the nonnegative root of a nonnegative real number.

The existence of $\sqrt[n]{a}$ is recorded here as a fact about the real numbers. Its full proof belongs to a later analysis course. In this section, we use the notation to ask a different question: when can such a root be rational?

Irrationality of $\sqrt2$

The first major example is the classical proof that $\sqrt2$ is not rational. The proof uses Euclid's lemma from integer divisibility:

$$p\mid ab\quad\Longrightarrow\quad p\mid a\text{ or }p\mid b$$

when p is prime. In particular, if a prime p divides $a^2$, then p divides a.

Suppose, for contradiction, that $\sqrt2$ is rational. Since it is positive, we may write

$$\sqrt2=\frac ab$$

where $a,b\in\mathbb Z^+$ and $\gcd(a,b)=1$. Squaring both sides gives

$$2=\frac{a^2}{b^2}, \qquad\text{so}\qquad 2b^2=a^2.$$

Hence $2\mid a^2$. Since 2 is prime, Euclid's lemma implies $2\mid a$. Write $a=2c$ for some $c\in\mathbb Z^+$. Substituting this into $2b^2=a^2$ gives

$$2b^2=(2c)^2=4c^2.$$

Dividing by 2,

$$b^2=2c^2.$$

Thus $2\mid b^2$, and again Euclid's lemma gives $2\mid b$. We have shown that 2 divides both a and b, contradicting $\gcd(a,b)=1$.

Therefore the assumption that $\sqrt2$ is rational is false, and $\sqrt2$ is irrational.

The structure of the proof is more important than the particular number 2. A rational representation in lowest terms cannot have a prime factor forced into both numerator and denominator. Irrationality proofs often look for exactly that contradiction.

This example shows a useful principle: over the rational numbers, the two pieces 1 and $\sqrt3$ cannot secretly imitate each other. A rational part and an irrational $\sqrt3$ part must match separately.

Irrationality of prime nth roots

The same idea proves a stronger result for roots of prime numbers.

Suppose, for contradiction, that $\sqrt[n]{p}$ is rational. Since it is positive, write

$$\sqrt[n]{p}=\frac ab$$

where $a,b\in\mathbb Z^+$ and $\gcd(a,b)=1$. Raising both sides to the n-th power gives

$$p=\frac{a^n}{b^n}, \qquad\text{so}\qquad pb^n=a^n.$$

Hence $p\mid a^n$. By applying Euclid's lemma repeatedly, a prime dividing $a^n$ must divide a. Therefore $p\mid a$, so write

$$a=pc$$

for some $c\in\mathbb Z^+$. Substitute this into $pb^n=a^n$:

$$pb^n=(pc)^n=p^n c^n.$$

Cancel one factor of p:

$$b^n=p^{\,n-1}c^n.$$

Because $n>1$, the right-hand side is divisible by p. Hence $p\mid b^n$. Again, Euclid's lemma applied repeatedly gives $p\mid b$.

Thus p divides both a and b, contradicting $\gcd(a,b)=1$. Therefore $\sqrt[n]{p}$ is irrational.

When is $\sqrt n$ rational?

For square roots of positive integers, rationality has an exact answer: a positive integer has a rational square root precisely when it is already a perfect square.

First suppose n is a perfect square, say $n=m^2$ for some $m\in\mathbb Z$. Since $n>0$, we have $m\ne0$, and the nonnegative square root is

$$\sqrt n=|m|.$$

This is an integer, hence rational.

Conversely, suppose $\sqrt n$ is rational. Write it in lowest terms as

$$\sqrt n=\frac ab,$$

where $a,b\in\mathbb Z^+$ and $\gcd(a,b)=1$. Squaring gives

$$nb^2=a^2.$$

We claim that $b=1$. If $b>1$, then b has a prime divisor p. Since $p\mid b$, we have $p\mid b^2$, and the equation $nb^2=a^2$ implies $p\mid a^2$. By Euclid's lemma, $p\mid a$. Thus p divides both a and b, contradicting $\gcd(a,b)=1$. Therefore $b=1$.

So $\sqrt n=a$, and hence

$$n=a^2.$$

Thus n is a perfect square.

The theorem explains why $\sqrt4$, $\sqrt9$, and $\sqrt{49}$ are rational, while $\sqrt2$, $\sqrt3$, $\sqrt5$, $\sqrt6$, and $\sqrt{10}$ are not. The question is not whether the decimal expansion looks simple; it is whether the integer under the square root is a square of an integer.

Quick checks

Exercises

1. Prove directly from the definition that if $x,y\in\mathbb Q$, then $3x-5y\in\mathbb Q$.
2. Give examples showing that the sum and product of irrational numbers may be rational.
3. Prove that if $r\in\mathbb Q$, $s\notin\mathbb Q$, and $r\ne0$, then $rs\notin\mathbb Q$.
4. Let $a,b,c,d\in\mathbb Q$. Suppose $a+b\sqrt3=c+d\sqrt3$, and assume $\sqrt3$ is irrational. Prove that $a=c$ and $b=d$.
5. Prove that $\sqrt[3]{5}$ is irrational.
6. Let $n\in\mathbb Z^+$. Prove that $\sqrt n$ is rational if and only if n is a perfect square.
7. Determine whether each number is rational or irrational: $\sqrt{16}$, $\sqrt{18}$, $\sqrt[4]{7}$, and $2+\sqrt[3]{11}$.