When does a proof need to be proved both ways?

Question

I am very sorry I understand that the title is confusing, but I am not totally sure how to reword it because I am completely lost on this subject. I am taking a discrete mathematics course right now, and one subject that is confusing me is proofs.

I understand the basic methods of proving usually start with an implies statement or an if...then statement, but in class there are times when it seems we need to prove it going both ways. Like we start with "Prove: If A, then B" and we need then need to prove both "If A, then B" and "If B, then A". I am completely lost because I don't know when this style of proof is needed. I have also heard the term "If and only if Proofs" getting thrown around and this confuses me even more because I don't know if they are using direct proofs or contradiction.

I understand if this gets downvoted because I have no idea what I am talking about right now. But, any honestly any advice on how to get started with understanding this topic would be great.

Answer 1

"If and only if" proofs are exactly what you say. Really, it is a shorthand for writing two different proofs of two different theorems: A implies B and B implies A. Sometimes this is referred to as proving both "directions" of an if and only if statement. The A implies B direction is referred to as sufficiency and the B implies A direction is referred to as necessity.

With each of those proofs, you can choose to prove directly or by contrapositive or whatever. You don't have to pick the same method for both proofs.

Answer 2

If you're only asked to prove "if $A$ then $B$", that's one way, and make sure you don't instead attempt a proof of "if $B$ then $A$". If you're asked to prove "$A$ iff $B$", that's two directions. In some cases, the two directions' proofs are so different you have to write them separately. But sometimes you can insert $n$ intermediate statements viz. $A\iff C_1\cdots\iff C_n\iff B$. That only needs to be one argument.

Answer 3

The "both ways" proofs comes as a result of some typical strategies for proving statements of various forms. In particular:

• If you want to prove a statement of the form "$P$ implies $Q$", you can assume $P$ and then prove $Q$ using that assumption.

• If you want to prove a statement of the form "$R$ and $T$" (i.e. both are true), you can first prove $R$ then, separately, prove $T$.

These aren't the only ways to prove such statements, but they are, in a sense, the most fundamental way to prove such statements.

You generally see these "both ways" proofs when you try to prove $$P \text{ if and only if }Q$$ which is defined to mean $$(P \text{ implies }Q)\text{ and }(Q\text{ implies }P).$$ So, if you just follow the strategies above, you see that you're trying to prove two statements: $P \text{ implies }Q$ and $Q \text{ implies } P$. You can prove either of these statements however you want - though the common way is to, for the former, assume $P$ and prove $Q$ and then, for the latter, to separately assume $Q$ and prove $P$.

The utility of the "if and only if" construction is that it essentially says that two propositions are equivalent: if one is true, so is the other and if one is false, so is the other. For instance, you could prove that "$n$ is the square of an integer" implies "$n$ is not negative", but the statements aren't equivalent because there are non-negative numbers that are not perfect squares - thus only one direction works.

Answer 4

As you mention in your comment to Alexander Gruber's answer, "both ways proofs" also occur when you have to prove an equality between sets, say $E = F$. Quite often, you first prove $E \subseteq F$ and then $F \subseteq E$. A similar idea is also used when you want to prove that two functions, or numbers, are equal. Suppose you want to prove that $f(x) = g(x)$. A possibility is to first prove $f(x) \leqslant g(x)$ and then $g(x) \leqslant f(x)$.

The "if and only if" proofs are just a logical instance of the same idea, which is to define an equivalence relation (the $\iff$ relation, the equality relation, etc.) as the intersection of a preorder relation (the $\Rightarrow$ relation, the inclusion relation, the relation $\leqslant$, etc.) and its opposite (the $\Leftarrow$ relation, the relation $\supseteq$, the relation $\geqslant$, etc.).

Answer 5

what happens when it does not outright say "$P$ implies $Q$" or vice versa, and I need to prove an identity like $$A-(B-C) = (A-B)\cup(A\cap C)\;?$$

When proving an equality is there are way to accomplish it by proving one way, or does it need to be proved with an iff?

The former. If statement $P$ is true, and $P$ implies statement $Q,$ then by modus ponens, $Q$ must be true; thus, to prove an equality $Q$ like the above, it is valid to just show that $$P\implies Q,$$ where $P$ is some true statement (for example, a known identity).