Question

prove that
$\sqrt{6}$
is a irrational no?????

Answer 1

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)
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 a2must 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.

There you have it: a rational proof of irrationality.