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:

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

From this, we get:

\[ 2n^2 = m^2 \]

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{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:

\[ \frac{p}{q} = x \cdot \frac{r}{s} \]

Rearranging the terms:

\[ x = \frac{ps}{qr} \]

This implies that the right side is rational, which contradicts the fact that \( x \) is irrational.

Therefore, the product of a rational and an irrational is irrational.