# Square Root of Two is Irrational

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## 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.