Question

Prove that √3 is an irrational number.

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Answer 1

Hey mate here is ur sol :-

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Let us assume that √3 is a rational number. That is, we can find integers aand b (≠ 0) such that √3 = (a/b)where a and b are the Co primes numbers

√3b = a ⇒ 3b2=a2 (Squaring on both sides) → (1) Therefore, a2 is divisible by 3 Hence ‘a’ is also divisible by 3. So, we can write a = 3c for some integer c. Equation (1) becomes, 3b2 =(3c)2 ⇒ 3b2 = 9c2 ∴ b2 = 3c2 This means that b2 is divisible by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 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 √3 is rational. So, we conclude that √3 is irrational. 

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hopes it's helps uh ❗ ❗

pls mark my ans as brainliest ☆ ♡

Answer 2

we have to prove √3 is a irrational number it is in below"

now you have to separate the 3 and √ ok now you have another method that √3 multiply with any rational number like 2 or 2/5 it can not multiply with √3