Let's say I have an algebraic statement A that I need to prove or disprove. Then I assume A and manipulate it to get the statement B. So, I have A implies B. Then I prove that B is false. Therefore, A must be false for the implication to be true.
Is this just called contradiction?
This is not proof by contradiction. It is proof by ex falso quodlibet (EFQ) or explosion: you have assumed $A$, derived a contradiction (false) from that and concluded that $A$ is false. A proof by contradiction of $A$ would first assume that $A$ is false. The distinction is important: EFQ is acceptable in intuitionistic/constructive logic, but proof by contradiction is not.
No it's called proof by contrapositive. It's an instance of the following scheme : $(P\implies Q) \implies (\neg Q \implies \neg P)$
NO. This is called proof by contradiction, also known by the Latin phrase reductio ad absurdum (rough literal translation: 'reducing to an absurdity'). Calling it simply contradiction would suggest that a genuine mathematical inconsistency had been arrived at, which is not what is intended or what has actually been achieved.
@Rob Arthan - Actually, you do not need to assume A is false and then deduce a contradiction to show A is true; you can just as easily do it the other way around. If that makes you uncomfortable, reinterpret it as showing (NOT A) is true instead.