Name: Surds: Introduction
Uploaded: 2019-07-19
Duration: 4 min 21 s
Description: We know that irrational numbers are written in the form of neither terminating nor recurring decimals. Let's learn to write another form of irrational number...

Question

Prove that there is no rational number whose square is

3

.

Accepted Answer

We are given to prove that there is no rational number whose square is $3$ .

Let us assume that there is a rational number $x$ such that $x^{2} = 3 \Rightarrow x = \sqrt{3}$ .

Hence, in order to prove that there is no rational number whose square is $3$ , it is sufficient to prove that $\sqrt{3}$ is an irrational number.

Let's assume $\sqrt{3}$ is a rational number.

As per the definition of rational numbers, we can express $\sqrt{3} = \frac{p}{q}$ , {where $p, q$ are co-primes and $q \neq 0$ }.

On squaring both sides, we get,

${(\sqrt{3})}^{2} = \frac{p^{2}}{q^{2}}$

$\Rightarrow 3 = \frac{p^{2}}{q^{2}}$

$\Rightarrow p^{2} = 3 q^{2} . . . . . . (i)$

Which means that, $p^{2}$ is a multiple of $3$ ,

$\Rightarrow 3$ divides $p$

Now, let $p = 3 m \Rightarrow p^{2} = 9 m^{2}$

Now putting this value of $p^{2}$ in equation $(i)$ ,

$9 m^{2} = 3 q^{2}$

$\Rightarrow 3 m^{2} = q^{2}$

$\Rightarrow 3$ divides $q^{2}$

$\Rightarrow 3$ divides $q$

This means that, $3$ is a common divisor of $p$ and $q$ , which contradicts the assumption done in the first step. { $p, q$ are co-primes}.

Hence, $\sqrt{3}$ is an irrational number.

Therefore, there is no rational number whose square is $3$ .

Prove that there is no rational number whose square is $3$ .

Important Questions on Number System

Learn and Prepare for any exam you want !

Prove that there is no rational number whose square is 3.

Important Questions on Number System

Learn and Prepare for any exam you want !

Prove that there is no rational number whose square is $3$ .