>>19
k is integer => k^2 is integer.
This implication is always true. Let's try it out with k=2.
2 is integer => 2^2 is integer
≡ 2 is integer => 4 is integer
≡ T => T
≡ T
Let's try it out with k=pi.
pi is integer => pi^2 is integer
≡ F => F
≡ T
Let's try it out with k=sqrt(2).
sqrt(2) is integer => sqrt(2)^2 is integer
≡ sqrt(2) is integer => 2 is integer
≡ F => T
≡ T
As you can see, since the implication is a tautology, there is no way of making it take a false value. Theorems are usually formulated in the form of "assume blah1, blah2, blah3, blah4. then resultant", which really just means "(blah1 & blah2 & blah3 & blah4) => resultant". If you fail to make one of the "blah"s true, the theorem is still true even if "resultant" isn't true because you failed to keep your end of the deal and make sure that all the "blah"s are true.