Proof: $ \sqrt{2} $ is an irrational number

Table of Contents

Let’s assume the definition of rational numbers is known. Real numbers that are not rational are called irrational numbers.

Here, we will prove that $\sqrt{2}$ is an irrational number.

Proof

We will prove by contradiction.

Assume that $\sqrt{2}$ is rational, then there exist coprime natural numbers $m, n$ such that it can be written as:

\sqrt{2} = \frac{m}{n}

$\sqrt{2} = \frac{m}{n}$

From this, we get:

2 n^{2} = m^{2}

$2n^2 = m^2$

The left side ( $2n^2$ ) is even, so is the right side ( $m^2$ ). In other words, we can express $m$ as $m = 2k$ for some natural number $k$ , and:

\begin{matrix} 2 n^{2} & = & (2 k)^{2} = & 4 k^{2} \Leftrightarrow & n^{2} & = & 2 k^{2} \end{matrix}

$\begin{array}{crcl} & 2 n ^2 & = & ( 2k )^2 \\ & & = & 4k^2 \\ \Leftrightarrow & n^2 & = & 2k^2 \end{array}$

From this, we can infer that $n$ is also even. However, this contradicts the assumption that $m, n$ are coprime. Therefore, $\sqrt{2}$ is irrational.

Similarly, it can be proven that $\sqrt{n}$ is irrational when $n$ is a prime number.

Generally, when $b$ is not a perfect square, $a \sqrt{b} \; (a, b \in \mathbb{N})$ becomes irrational. The product of a rational and an irrational is irrational, so it’s sufficient to prove that $\sqrt{b}$ is irrational. Also, if $b$ is the product of prime factors raised to the power of 2, then it’s sufficient to prove the case where $b$ is a prime number.

Proof

Assume $b$ is factorized into $k$ different prime numbers $b_1, b_2, \cdots , b_k$ as follows:

\[ b = b_1 b_2 \cdots b_k ]

Assume $\sqrt{b}$ is rational, then there exist coprime natural numbers $m, n$ such that it can be written as:

$\sqrt{b} = \frac{m}{n}$

From this, we get:

$b n^2 = m^2$

The left side $b n^2$ includes $b_1, b_2, \cdots, b_k$ as prime factors, so does the right side $m^2$ . In other words, $m$ can be expressed as:

$\begin{eqnarray*} m & = & c b_1 b_2 \cdots b_k \\ & = & cb \end{eqnarray*}$

Similarly to $m$ , we have:

$\begin{array}{crcl} & b n^2 & = & ( cb )^2 \\ & & = & c^2 b^2 \\ \Leftrightarrow & n^2 & = & c^2 b \end{array}$

From this, we can infer that $n$ is also a multiple of $b$ , but this contradicts the assumption that $m, n$ are coprime.

Thus, $\sqrt{n}$ becomes rational (an integer) only when $n$ is a perfect square.

In this case, we have proved the property for square roots, but the same proof can be applied to $\sqrt[m]{n} \; (m, n \in \mathbb{N})$ .

Additional Proposition

The product of a rational number and an irrational number is irrational.

Proof

We will prove by contradiction.

Let $p, q, r, s$ be integers, and $x$ be an irrational number. Assume that the product of a rational number $\frac{p}{q}$ and an irrational number $x$ is a rational number $\frac{r}{s}$ . Then, we have: