Question

Prove that √5 is irrational​

Answer 1

Let 5 be a rational number.

then it must be in form of  qp  where,  q=0     ( p and q are co-prime)

5=qp

5×q=p

Suaring on both sides,

5q2=p2           --------------(1)

p2 is divisible by 5.

So, p is divisible by 5.

p=5c

Suaring on both sides,

p2=25c2         --------------(2)

Put p2 in eqn.(1)

5q2=25(c)2

q2=5c2

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

Answer 2

Answer:

Proof is given below

Step-by-step explanation:

Let's assume √5 is irrational

√5 = p/q (where q ≠ 0 and p and q are co prime numbers) -> 1

On squaring both the sides

5 = (p/q)²

5 = p²/q²

5q² = p² -> 2

If p² is divisible by 5 then p is divisible by 5

Let p = 5m => p² = 25m -> 3

Substitute 3 in 2

25m² = 5q²

25m²/5 = q²

5m² = q²

q² = 5m²

If q² is divisible by 5 then q is also divisible by 5

This means that 5 is a factor of p and q.

This contradicts our assumption that √5 is a rational number

Hence √5 is an irrational number.

Hopefully

Brainliest