Question

Q.27. Prove that 5 is an irrational number.​

Answer 1

Answer:

Any rational number can be represented in the form of pq, where q≠0. So, represent √5 as pq and suppose p, q are co-primes. ... Let us prove that √5 is an irrational number, by using the contradiction method. So, say that √5 is a rational number can be expressed in the form of pq, where q ≠0.

Answer 2

Answer:

Let 5 be a rational number.

then it must be in form of

q

p where, q0 ( p and q are co-prime)

√5=q

p

√5×q=p

Suaring on both sides,

5q^2=p^2

--------------(1)

p^2 is divisible by 5.

So, p is divisible by 5.

p=5c

Suaring on both sides,

p^2=25c^2

--------------(2)

Put p^2 in eqn.(1)

5q^2 =25(c)^2

q^2 =5c^2

So, q is divisible by 5.

Thus p and q have a common factor of 5.

So, there is a contradiction as per our assumption.

We have assumed p and q are co-prime but here they a common factor of 5.

The above statement contradicts our assumption.

Therefore, 5 is an irrational number.