Hello Sylvia.
I would use the proof by contradiction method for this.
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.
So let's multiply both sides by themselves:
(a/b)(a/b) = (square root of 6)(square root of 6)
a^{2}/b^{2} = 6
a^{2} = 6b^{2}
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. So a^{2} must be even. But any odd number times itself is odd, so if 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:
a^{2} = 6b^{2}
(2c)^{2} = (2)(3)b^{2}
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 the square root of 6 cannot be rational.
There you have it: a rational proof of irrationality.
Stephen La Rocque.
