Question

Prove that √13 is irrational no.

Answer 1

Let us assume that √13 is a rational number.

That is, we can find integers a and b (≠ 0) such that √13 = (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 coprime.

√13b = a

⇒ 13b2=a2 (Squaring on both sides) → (1)

Therefore, a2 is divisible by 13

Hence ‘a’ is also divisible by 13.

So, we can write a = 13c for some integer c.

Equation (1) becomes,

13b2 =(13c)2

⇒ 3b2 = 169c2

∴ b2 = 13c2

This means that b2 is divisible by 13, and so b is also divisible by 13.

Therefore, a and b have at least 13 as a common factor.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that √13 is rational.
So, we conclude that √13 is irrational.

Answer 2

Hi Friend !!!!


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Let us assume that its rational number

√13 = p/q ( where p,q are co primes)

13 = (p/q)²

13 = p²/q²

13 q² =p² __________(1)

If 13 divides p² then 13 also divides p


so , p = 13a

put it in first equation

13q² = (13a)²

q² = 13²a²/13

q² = 13a²

If 13 divides q² then 13 also divides q

So, 13 will be the common factor

But it contradics the fact that they are co primes

So our assumption is wrong


So, √13 is irrational

Hence proved


Hope it helps u : )