SOLUTION: Using proof by contradiction, prove that 7√2 is irrational.

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Question 1130704: Using proof by contradiction, prove that 7√2 is irrational.
Answer by MathLover1(20849) About Me  (Show Source):
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suppose that 7sqrt%282%29 were rational:
Then we can write 7sqrt%282%29 as a fraction+a%2Fb+where+a and b are integers with no common factors.
Since 7sqrt%282%29+=+a%2Fb, square both sides=> 7%5E2%2A2+=+a%5E2%2Fb%5E2
So, 7%5E2%2A2b%5E2+=+a%5E2
By the definition of even, this means a%5E2 is even. But then a must be even if a%5E2 is even .
So a+=+2n for some integer n.
If+a+=+2n and 7%5E2%2A2b%5E2+=+a%5E2, then
7%5E2%2A2b%5E2+=+4n%5E2.
So 7%5E2%2Ab%5E2+=+2n%5E2. This means that b%5E2 is even, so b must be even.
We now have a contradiction. a and b were chosen not to have any common factors.
But they are both even, i.e. they are both divisible by 2.
Because assuming that 7sqrt%282%29 was rational led to a contradiction, it must be the case that 7sqrt%282%29 is irrational.