Prove that \left(\right. 2 \sqrt{3} + \sqrt{5} \left.\right) is an irrational number. Also check whether \left(\right. 2 \sqrt{3} + \sqrt{5} \left.\right) \left(\right. 2 \sqrt{3} - \sqrt{5} \left.\right) is rational or irrational

Foundation

Mathematics Foundation

Properties of Real Number

Irrational Number

Algebraic Identities

Question

Prove that

(2 \sqrt{3} + \sqrt{5})

is an irrational number. Also check whether

(2 \sqrt{3} + \sqrt{5}) (2 \sqrt{3} - \sqrt{5})

is rational or irrational

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

a) Let us assume that

2 \sqrt{3} + \sqrt{5}

is rational number.

Let

P = 2 \sqrt{3} + \sqrt{5}

is rational

on squaring both sides we get

P^{2} = (2 \sqrt{3} + \sqrt{5})^{2} = (2 \sqrt{3})^{2} + (\sqrt{5})^{2} + 2 \times 2 \sqrt{3} \times \sqrt{5}

P^{2} = 12 + 5 + 4 \sqrt{15}

P^{2} = 17 + 4 \sqrt{15}

\frac{P^{2} - 17}{4} = \sqrt{15}

………..

(1)

Since

P

is rational no. therefore

P^{2}

is also rational &

\frac{P^{2} - 17}{4}

is also rational.

But

\sqrt{15}

is irrational & in equation

(1)

\frac{P^{2} - 17}{4} = \sqrt{15}

Rational

\neq

irrational

Hence our assumption is incorrect &

2 \sqrt{3} + \sqrt{5}

is irrational number.

P = (2 \sqrt{3} + \sqrt{5}) (2 \sqrt{3} - \sqrt{5})

P = 12 - 5 = 7

Hence P is rational as

\frac{p}{q} = \frac{7}{1}

& both

p

q

are coprime numbers.