# Square Root of Two is Irrational

## Introduction to \(\sqrt 2 \):

Geometrically the square root of 2 is the length of a diagonal across a square with sides of one unit of length; this follows from the Pythagoras Theorem. It was probably the first number known to be irrational. As a good rational approximation for the square root of two, with a reasonably small denominator, the fraction \(\frac{{99}}{{70}}\) \(\left( { \approx 1.4142857} \right)\) is sometimes used. The true decimal expansion of \(\sqrt 2 \) truncated to 45 decimal places:

\[1.414213562373095048801688724209698078569671875...\]

Let us see why \(\sqrt 2 \) is irrational.

## Method of Proving by Contradiction

The method we use to prove this assertion is called *proof by contradiction*. In this method, we start with an assumption that is contrary to what we are actually required to prove. Then, using a series of logical deductions from this assumption, we reach an inconsistency – a mathematical or logical error – which enables us to conclude that our original assumption was incorrect, proving its contrary assertion to be true.

## Why \(\sqrt 2 \) is irrational?

In this case, we start by supposing that \(\sqrt 2 \) is rational. Thus, there will exist integers \({p}\) and \({q}\) (where \({q}\) is non-zero) such that \(\frac{p}{q} = \sqrt 2 \). We also make the assumption that \({p}\) and \({q}\) *have no common factor*. If they had any common factors, we could cancel them out. Thus, there is no problem in assuming that the only common factor of \({p}\) and \({q}\) is 1.

Now, squaring both sides, we have

\[\frac{{{p^2}}}{{{q^2}}} = 2 \Rightarrow {p^2} = 2{q^2}....(1)\]

We note that the right-hand side of the last equation is a multiple of 2, which means that the left-hand side is also a multiple of 2, which further means that \({p}\) itself must be a multiple of 2. This is because if the square of an integer is a multiple of 2, then that integer itself must also be a multiple of 2. Thus, we can assume that,

\[\begin{align}

&p = 2m,m \in Z\qquad{\text{ }}[{\text{Set of Integers}}]\; \hfill \\

&\!\!\Rightarrow {\left( {2m} \right)^2} = 2{q^2} \qquad {\text{ [From (1)]}} \hfill \\

&\!\!\Rightarrow 4{m^2} = 2{q^2} \hfill \\

&\!\!\Rightarrow {q^2} = 2{m^2} \hfill \\

\end{align}\]

Now, the right-hand side is a multiple of 2 again, which means that the left-hand side is a multiple of 2, which further means that \({q}\) is a multiple of 2, i.e.,

\[q = 2n,n \in Z\]

We have thus shown that both \({p}\) and \({q}\) are multiples of 2. But is that possible? We started by assuming that \({p}\) and \({q}\) have no common factors, whereas our logical deductions have shown that \({p}\) and \({q}\) have a common factor 2. This can only mean one thing: our **original assumption** of assuming \(\sqrt 2 \) as \(\frac{p}{q}\) (where \({p}\) and \({q}\) are integers) is **wrong**:

\[\sqrt 2 \ne \frac{p}{q}\]

Thus, \(\sqrt 2 \) does not have a rational representation – **\(\sqrt 2 \) is irrational**.

## Solved Example:

**Example 1:** Prove that \(\sqrt 3 \) is irrational.

We follow an exactly analogous approach to the one we just described. We start by supposing that \(\sqrt 3 \) is rational. Thus, \(\sqrt 3 = \frac{p}{q}\), where \({p}\) and \({q}\) are integers. We also assume that \({p}\) and \({q}\) and have no common factor. If they had any common factors, we could cancel them out. Thus, there is no problem in assuming that the only common factor of \({p}\) and \({q}\) is 1.

Now, squaring both sides, we have

\[3 = \frac{{{p^2}}}{{{q^2}}} \Rightarrow {p^2} = 3{q^2}....(1)\]

We note that the right-hand side of the equation (1) is a multiple of 3 which means that the left-hand side is also a multiple of 3, which further means that \({p}\) itself must be a multiple of 3. This is because if the square of an integer is a multiple of 3, then that integer itself must also be a multiple of 3. Thus, we can assume that,

\[\begin{align}

&p = 3m,m \in Z\qquad{\text{ }}[{\text{Set of Integers}}]\; \hfill \\

&\!\!\Rightarrow {\left( {3m} \right)^2} = 3{q^2} \qquad {\text{ [From (1)]}} \hfill \\

&\!\!\Rightarrow 9{m^2} = 3{q^2} \hfill \\

&\!\!\Rightarrow {q^2} = 3{m^2} \hfill \\

\end{align}\]

Now, the right-hand side is a multiple of 3 again, which means that the left-hand side is a multiple of 3, which further means that \({q}\) is a multiple of 3, i.e.,

\[q = 3n,n \in Z\]

We have thus shown that both \({p}\) and \({q}\) are multiples of 3. But this is a contradiction. We started by assuming that \({p}\) and \({q}\) have no common factors, whereas our logical deductions have shown that \({p}\) and \({q}\) have a common factor of 3. This means that our original assumption of assuming \(\sqrt 3 \) as\(\frac{p}{q}\) (where \({p}\) and \({q}\) are integers) is wrong:

\[\sqrt 3 \ne \frac{p}{q}\]

Thus, **\(\sqrt 3 \) is irrational**.

**Challenge: **Can you prove that \(\sqrt 7 \)** **is irrational?

**⚡Tip:** Use a similar approach as in example 1.