Question

$2 + 3 \sqrt{6 \: } is irational$

Answer 1

to prove 2+3√6 is irrational , we have to prove that √6 is irrational .
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 = √6.

So let's multiply both sides by themselves:

(a/b)(a/b) = √6 * √6
a²/b² = 6
a² = 6b²

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 a² must be even. But any odd number times itself is odd, so if a² 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:

a² = 6b²
(2c)² = (2)(3)b²
2c² = 3b²

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.

so 2+3√6 is also irrational.