Question

Prove that √7 and √11 are irrational numbers?

Answer 1

Answer:

let us assume that √7 be rational.

then it must in the form of p / q.

As definition of rational number says.. P is whole number q is non zero whole number.. And p and q is simplest ratio which is expressed.. That means there exists no prime factor common in p and q.

√7 = p / q

√7 x q = p

squaring on both sides

7q² = p² ------1.

p is divisible by 7

p = 7c [c is a positive integer] [squaring on both sides ]

p²= 49c²

subsitute p² in eqn(1) we get

7q² = 49 c²

q² = 7c²

q is divisble by 7

thus q and p have a common factor 7.

there is a contradiction to our assumption

as our assumsion p & q are co prime but it has a common factor.

so that √7 is an irrational.

$\sqrt{11}$ is irrational

By the method of contradiction..

Let √11 be rational , then there should exist √11=p/q ,where p & q are coprime and q≠0(by the definition of rational number). So,

√11= p/q

On squaring both side, we get,

11= p²/q² or,

11q² = p². …………….eqñ (i)

Since , 11q² = p² so ,11 divides p² & 11 divides p

Let 11 divides p for some integer c ,

so ,

p= 11c

On putting this value in eqñ(i) we get,

11q²= 121p²

or, q²= 11p²

So, 11 divides q² for p²

Therefore 11 divides q.

So we get 11 as a common factor of p & q but we assumpt that p & q are coprime so it contradicts our statement. Our supposition is wrong and √11 is irrational.

Answer 2

here is ur answer in attachment .

hope it help u

follow me ..

by this process u can find the √11 also ..

sorry I only did the √7 ..