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 is rational.
By definition, that means there are two integers a and b with no common divisors where:
So let's multiply both sides by themselves:
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 must be even. But any odd number times itself is odd, so if 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:
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 cannot be rational.