Question

prove that root 3 is irrational
number
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Answer 1

Step-by-step explanation:

Your answer is in the picture above...

Answer 2

$\huge\mathcal\purple{Answer:-}$

By using the method of contradiction,

assume that √3 is rational.

=> √3 = a/b [where b≠0 and a & b are coprimes]

=> √3b = a ...→(1)

Now, on squaring on both sides, we get

(√3b)² = a²

=> 3b² = a²

Therefore, 3 divides a.

Now,

Let a = 3k (for some integer k)

=> 3b² = 9k²

=> b² = 3k²

Therefore, 3 divides b.

Therefore, a and b have atleast 3 as a common factor.

But this contradicts that a and b are coprime.

This contradiction has arisen because of our wrong assumption that √3 is rational.

So, we conclude that √3 is irrational.

Hope this helps!❤