Question

prove that√6 is an irrational

Answer 1

Assume that sqrt(6) is rational.

Then sqrt(6) = p/q where p and q are coprime integers.

sqrt(6)^2 = 6 = p²/q²

p² = 6q²

therefore p² is an even number since an even number multiplied by any other integer is also an even number. If p² is even then p must also be even since if p were odd, an odd number multiplied by an odd number would also be odd.

So we can replace p with 2k where k is an integer.

(2k)² = 6q²
4k² = 6q²
2k² = 3q²

Now we see that 3q² is even. For 3q² to be even, q² must be even since 3 is odd and an odd times an even number is even. And by the same argument above, if q² is even then q is even.

So both p and q are even which means both are divisible by 2. But that means they are not coprime, contradicting our assumption so sqrt(6) is not rational.

Answer 2

Hii!!

Here is ur answer :-

To prove :- √6 is irrational

Let us assume that √6 is rational

so, √6 = p/q , where p and q are prime, cooprime and q ≠0

=> √6 = p/q

=> p = √6q

=> (p) ^2 = (√6q) ^2

=> p^2 = 6q^2

Therefore, p^2 is divisible by 6

and p is divisible by 6

Let us take p = 6r

so,

=> 6r = √6q

=> (6r) ^2 = (√6q) ^2

=> 36r^2 = 6q^2

=> 36r^2/6 = q^2

=> q^2 = 6r^2

Therefore, q^2 is divisible by 6

and q is divisible by 6

So, p and q has a common factor 6

But this contradicts that p anf q has no common factor than 1

Therefore our assumption is wrong.

Hence, √6 is irrational. (Proved)

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Hope it helps!!