Question

1. Prove that route 5 is irrational.

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.

Hence proved..!!

Answer 2

$\bf\huge\red{Hello(✷‿✷)}$

Prove that √5 is irrational.

Solution:-

Let us assume, to the contrary, that √5 is rational.

i.e., we can find integer a and b , where b is not equal to 0.

such that , √5 = a/b

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are co-prime.

So, b√5=a

Squaring on both sides, rearranging, we get 5b²=a².

Therefore, a² is divisible by 5 , and a is also divisible by 5.

So, we can write a=5c for some integer c.

Substituting for a , we get 5b² = 25c², i.e., b²= 5c².

This means that b² is divisible by 5, and so b is also divisible by 5.

Therefore, a and b have at least 5 as a common factor.

But this contradicts the fact that a and b are co-prime.

This contradiction has arisen because of our incorrect assumption that √5 is rational.

So, we conclude that √5 is irrational.