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.
Let's assume that it's rational: [tex]\sqrt6=\frac{p}q,p\wedge q=1[/tex]
Then [tex]p^2=6q^2[/tex] hence [tex]p[/tex] is even, which can be written [tex]p=2p'[/tex]
Hence [tex]4p'^2=6q^2[/tex] thus [tex]2p'^2=3q^2[/tex]
Thus p,q are both even, which is absurd.
Hence sqrt(6) is irrational
Then [tex]p^2=6q^2[/tex] hence [tex]p[/tex] is even, which can be written [tex]p=2p'[/tex]
Hence [tex]4p'^2=6q^2[/tex] thus [tex]2p'^2=3q^2[/tex]
Thus p,q are both even, which is absurd.
Hence sqrt(6) is irrational
