Verified Answer :

So let's assume that the square root of 6 is rational. By definition, that means there are two integers a and b with no common divisors where: a/b = square root of 6. But this last statement means the RHS (right hand side) is even, because it is a product of integers and one of those integers (at least) is even.

Hippalectryon

Let's assume that it's rational: \(\sqrt6=\frac{p}q,p\wedge q=1\)

Then \(p^2=6q^2\) hence \(p\) is even, which can be written \(p=2p'\)

Hence \(4p'^2=6q^2\) thus \(2p'^2=3q^2\)

Thus p,q are both even, which is absurd.

Hence sqrt(6) is irrational

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