Question

√15,Prove that given number is irrational.

Answer 1

Let us assume on the contrary that √15 is a rational number.Then there exist positive integers p and q such that

√15 = p/q

[Where p and q are coprime i.e their HCF is 1]

√15 = p/q

On squaring both sides

(√15)² = (p/q)²

15 = p²/q²

15q² = p²

15 | p² [15 | 15q²]

15 | p………………….(1)

[By theorem: If p divides a² then p divides a]

p = 15c for some positive integer c

On squaring both sides

p² = 225c²

15q² = 225c²

[15q² = p²]

q² =15c²

15 | q² [15 | 5c²]

15 | q ………………..(2)

[By theorem: If p divides a² then p divides a]

From eq (1) & (2) we find that p and q have at least 15 as a common factor. This contradicts the fact that p and q are coprime.

Hence, √15 is an irrational number.

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Hi ,

Let us assume , to the contrary , that

√15 is rational .

So , we can find coprime integers a and

b ( b ≠ 0 ) such that

√15 = a/b

√15 b = a

Squaring on both sides , we get

15 b² = a²

Therefore ,

15 divides a² ,

15 divides a .

So , we can write

a = 15c for some integer c .

Substituting for a , we get

15b² = ( 15 c )²

15b² = 225 c²

b² = 15 c²

This means that 15 divides b² , and so

15 divides b .

Therefore , a and b have at least 15 as

a common factor .

But this contradicts the fact that a and

b have no common factor other than 1.

This contradiction arose because of our

incorrect assumption that √15 is rational.

So , we conclude that √15 is irrational.

I hope this helps you.

: )