Question

Prove that
$\sqrt{6}$
is an irrational no with reason .

Answer 1

hope it helps you frnd

Answer 2

Heya!


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Let √6 be rational

√6 = a/b

Where , a and b are co-prime.


√6 = a/b

Squaring both sides,

(√6)^2 = (a/b)^2

6 = a^2/b^2

6b^2 = a^2

[Theorem - Let p be a prime number , if p divides a^2 , then p divides a , where a is a positive integer.]

According to this theorem, if a^2 divides by 6 then a also divides by 6.

Thus, a = 6k. where k is any integer.

Put 'a' in the equation,

6b^2 = (6k)^2
6b^2 = 36k^2
b^2 = 36/6k^2
b^2 = 6k^2


Now, both a and b has 6 as common factor, thus our supposition that √6 is rational is wrong.

Hence , √6 is irrational.




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Hope it helps...!!!