Question

whether √11 is a rational or irrational ?​

Answer 1

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

$\huge \bold{ \fbox { \underline{ \underline \pink{required \: answer}}}}$

√11 is an irrational number

$\bold{ \underline{ \underline \orange{proof}}}$

let √11 be a rational number and is of the form of p/q , where p and q are co-prime integers and q ≠ 0

$\begin{lgathered}\bold{so \: \sqrt{11} = \frac{p}{q} } \\ \\ \implies \: p = \sqrt{11} q\end{lgathered} \:$

on squaring both sides..

$</p><p>{p}^{2} = 11 {q}^{2} \: \: \: ............(i) </p><p> \:$

→ 11 divides p2

we know that..if a prime number is divides square of any positive integer..then prime number divides that positive integer..(theorem).....(ii)

→ 11 divides p

now again let, p = 11r

on squaring both side, p² = 121r²( p = 11q² from eq.i )

→ 11 divides q²

→ 11 divides q ( by eq.ii )

since , 11 divides p and q both ( they are prime numbers)

so our assumption is wrong..

→ √11 is an irrational number