Question

prove that root 6 is an irrational number

jaldiiiii​

Answer 1

Answer:

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 co-prime, contradicting our assumption so sqrt(6) is not rational.