Prove that \sqrt{3} is an irrational number.

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Rational Numbers

Question

Prove that

\sqrt{3}

is an irrational number.

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Step 1: Use the definition of rational number and prove the required result

Consider

\sqrt{3}

as a rational number.

We know that, a rational number can be expressed in the form \frac{p}{q} where p, q are co-prime integers.

\Rightarrow \sqrt{3} = \frac{p}{q}

$Squaring on both side we get$

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

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

$Eqaution (i), can be written as$ $q^{2} = \frac{p^{2}}{3}$

$This shows, p is divisible by 3. It can be expressed as p = 3m$

$Squaring on both sides$

p^{2} = (3 m)^{2} \Rightarrow p^{2} = 9 m^{2}

\dots (i i)

Substitute equation (ii) in eqaution (i) we get

3 q^{2} = 9 m^{2} \Rightarrow q^{2} = 3 m^{2}

Clearly , 3 is a factor of q

Thus we see that both p and q have comman factor 3 which is a contradiction that H.C.F of (p,q)=1

Hence,

\sqrt{3}

is not a rational number therefore is an irrational number.