Question

√5,Prove that given number is irrational.

Answer 1

Let us assume on the contrary that √5 is a rational number.Then there exist positive integers p and q such that

√5 = p/q

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

√5 = p/q

On squaring both sides

(√5)² = (p/q)²

5 = p²/q²

5q² = p²

5 | p² [5 | 5q²]

5 | p………………….(1)

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

p = 5c for some positive integer c

On squaring both sides

p² = 25c²

5q² = 25c²

[5q² = p²]

q² =5c²

5 | q² [5 | 5c²]

5 | 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 5 as a common factor. This contradicts the fact that p and q are coprime.

Hence, √5 is an irrational number.

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Hi ,

Let us assume that √5 is rational number.

√5:= p/q

[ where ( p,q ) = 1 , p,q€ Z

and q ≠ 0 ]

Squaring on both sides ,

5 = p²/q²

=> 5q² = p² ----( 1 )

=> 5 divides p² then 5 divides p ----( 2 )

Let p = 5m => p² = 25m²

Putting the value of p² in ( 1 ) , we get

5q² = 25m²

=> q² = 5m²

=> 5 divides q² => 5 divides q ----( 3 )

Thus from ( 2 ) p is a multiple of 5 and

from ( 3 ) q is also multiple of 5

i.e ., 5 is a common factor of p and q .

This contradicts to our assumption.

So , there is no common factor of p and q.

Hence , √5 is an irrational number.

I hope this helps you.

: )