Question

Use a method of contradiction to show that $\sqrt 3$ is an irrational number.

Answer 1

To prove by method contradiction, we first assume that the statement is false. Then apply the definition of rational numbers. Then make necessary operations to arrive at a contradiction. A rational number can be written in the form $\dfrac{p}{q}$ where $p$ and $q$ are co-primes (integers with no common factor other than one) and $q \ne 0$.

Complete step-by-step answer:
To prove $\sqrt 3$ is irrational, suppose not.
That is, assume that $\sqrt 3$ is rational.
A rational number can be written in the form $\dfrac{p}{q}$ where $p$ and $q$ are co-primes (integers with no common factor other than one) and $q \ne 0$.
So we can write $\sqrt 3 = \dfrac{p}{q}$, with $p$ and $q$ are co-primes and $q \ne 0$.
Consider $\sqrt 3 = \dfrac{p}{q}$
Squaring both sides we get,
${\sqrt 3 ^2} = {(\dfrac{p}{q})^2}$
$\Rightarrow 3 = \dfrac{{{p^2}}}{{{q^2}}}$
Cross multiplying we get,
$\Rightarrow {p^2} = 3{q^2} - - - (i)$
This gives ${p^2}$ is a multiple of $3$.
Therefore $p$ is a multiple of $3$.
So we can write $p = 3r$, for some integer $r$.
Squaring both sides we get,
${p^2} = {(3r)^2}$
$\Rightarrow {p^2} = 9{r^2} - - - (ii)$
From equations $(i)$ and $(ii)$ we can write,
$3{q^2} = 9{r^2}$
Dividing both sides by $3$ we get,
${q^2} = 3{r^2}$
This gives ${q^2}$ is a multiple of $3$.
Therefore $q$ is a multiple of $3$.
Before we have seen that $p$ is a multiple of $3$.
But by our assumption, $p$ and $q$ are co-primes.
So we arrive at a contradiction.
Therefore our assumption is wrong.
That is $\sqrt 3$ is not rational.
$\therefore$ $\sqrt 3$ is irrational.
Hence we had proved the statement.

Note: A statement can be proved in different ways. Method of contradiction is one method. We are mainly using direct implication. But here we are asked to use the method of contradiction. Here the key point is for a number to be rational, when written in fraction form, the numerator and denominator cannot contain common factors.