Prove √6 as an irrational number
Prove √6 as an irrational number
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 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 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 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.
