Question

prove that √3 is irrational please emergency please​

Answer 1

Let us assume on the contrary that √3 is a rational number

Then, there exist positive integers a and b such that

$\sqrt{3} = \frac{a}{b}$

where, a and b, are co prime that is their HCF is 1

Now,

$\sqrt{3} = \frac{a}{b}$

$3 = \frac{ {a}^{2} }{ {b}^{2} }$

$3 {b}^{2} = {a}^{2}$

3 divides a² [ Since 3 divides 3b² ]

3 divides a → (i)

a = 3c for some integer c

a² = 9c²

3b² = 9c² [ Since a² = 3 b² ]

b² = 3c²

3 divides b² [ Since 3 divides 3c² ]

3 divides b → (ii)

From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.

Hence,

√3 is irrational

I hope this helps you

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