Question

prove that √11 is an ir-rational number​

Answer 1

Answer:

√11 =p/q

where p and q are co-prime

and q not equal to 0

so, √11=p/q

on squaring both sides we get,

11=p2/q2

11q2=p2 .......... (1 )

q2=p2/11

p=11c

p2 is divisible by 11

p is divisible by 11

subtract, p=11c in equation ( 1 )

11q2=(11c) 2

11q2=121c2

q2=121c2/11

q2=11c

we observe that p and q have least common factor but this contract that p and q are co prime therefore our assumptions are wrong.

Answer 2

Required to prove :-

√11 is an ir-rational number

Method used :-

• Contradictory Method

Conditions used :-

• p , q are integers

• q ≠ 0

• p and q are co - primes

Solution :-

Let's assume on the contradictory that √11 is an ir-rational number .

Equal √11 with p/q

Where p , q are integers , q ≠ 0 , p and q are co - primes

So,

$\tt{ \sqrt{11} = \dfrac{p}{q}}$

By cross multiplication

√11q = p

Squaring on both sides

( √11 q )² = ( p )²

11q² = p²

Now recall the fundamental theorem of arithmetic

According to which ,

If a divides q²

a divides q ( also )

So,

11 divides p²

11 divides p ( also )

However ,

Let's take

p = 11k

where k is any positive integer

This implies

√11q = 11k

squaring on both sides

( √11 q )² = ( 11 k )²

11q² = 121k²

Transposing 11 to the right side

$\tt{ {q}^{2} = \dfrac{121\;{k}^{2}}{ 11}}$

q² = 11k²

Interchanging the terms on the both sides

11k² = q²

Here ,

11 divides q²

11 divides q ( also )

From the above we can conclude that ;

11 is the common factor of both p and q .

But according to the condition ;

p , q are co- primes which means they should have only 1 as the common factor .

Hence

This contradiction is due to wrong assumption that √11 is a rational number .

So, our assumption is wrong .

Hence ,

√11 is an ir-rational number