Prove that the square root of 6 is irrational.
Sorry i don't know how to do the square root in here.
I would use the proof by contradiction method for this.
So let's assume that $\displaystyle {\sqrt {6}}$ is rational.
By definition, that means there are two integers a and b with no common divisors where:
$\displaystyle {\frac{a}{b}} = {\sqrt {6}}$
So let's multiply both sides by themselves:
$\displaystyle ({\frac{a}{b}})({\frac{a}{b}}) = ({\sqrt {6}})({\sqrt {6}}) $
$\displaystyle {\frac{a{^2}}{b{^2}}} = 6$
$\displaystyle a{^2} = 6b{^2}$
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. So $\displaystyle a{^2}$ must be even. But any odd number times itself is odd, so if $\displaystyle a{^2}$ is even, then a is even.
Since a is even, there is some integer c that is half of a, or in other words:
2c = a.
Now let's replace a with 2c:
$\displaystyle a{^2} = 6b{^2}$
$\displaystyle (2c){^2} = (2)(3)b{^2}$
$\displaystyle 2c{^2} = 3b{^2}$
But now we can argue the same thing for b, because the LHS is even, so the RHS must be even and that means b is even.
Now this is the contradiction: if a is even and b is even, then they have a common divisor (2). Then our initial assumption must be false, so $\displaystyle {\sqrt {6}}$ cannot be rational.