Question

prove that ✓5 is an irrational number​

Answer 1

Answer:

Given: √5

We need to prove that √5 is irrational

Proof:

Let us assume that √5 is a rational number.

So it can be expressed in the form p/q where p,q are co-prime integers and q≠0

⇒ √5 = p/q

On squaring both the sides we get,

⇒5 = p²/q²

⇒5q² = p² —————–(i)

p²/5 = q²

So 5 divides p

p is a multiple of 5

⇒ p = 5m

⇒ p² = 25m² ————-(ii)

From equations (i) and (ii), we get,

5q² = 25m²

⇒ q² = 5m²

⇒ q² is a multiple of 5

⇒ q is a multiple of 5

Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number

√5 is an irrational number.

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Answer 2

Answer:

let us assume that √5 is rational, then it will be of the form a/b where a and b are integers and b not equal to zero.

Again let a and b have no common factor other than 1

therefore , √5 = a/b

on squaring both sides we get,

(√5)^2 = (a/b)^2 ------ 5 = a^2/b^2

5b^2 = a^2 -----(1)

So, 5 divedes a^2 => 5 divides a (from theorem 1)

Now , a can be written as 5m , where m is an integer

on putting a =5m in eq.(1) we get,

5b^2 = (5m)^2 => 5b^2 = 25m^2 => b^2 = 5m^2

So, 5 divides b^2 => 5 divides b (from theorem 1)

Thus , 5 is the common factor of a and b

but this contradicts the fact that a and b have no common factor other than 1.

The contradiction arises by assuming that √5 is rational.

Hence √5 is irrational

[ hence proved]