Section1.6How to prove it
¶Below is a list of suggestions to consider when writing proofs.
- Figure out what you are trying to prove. This starts by determining the logical structure of the statement. From there, you can set up the basic structure of your proof. Part of this is also determining the assumptions (or hypotheses and antecedents) and the conclusions (or consequents).
- Fill in the boilerplate material. For those of you who may be unfamiliar with this term, material is said to be boilerplate if it is standard regardless of context. Boilerplate material consists of that material which provides the basic structure to your proof (e.g., “assume \(P\text{,}\)” in a direct proof of \(P \implies Q\)), as well as the insertion of definitions (e.g., inserting “\(m = 2k\text{,}\) for some \(k \in \mathbb{Z},\)” after the statement “\(m\) is even.”). It is also recommended that you fill in the boilerplate material that goes at the end of the proof because this will show you what is your ultimate goal.
- Play around. Once you've filled in the boilerplate and gotten down to the meat of the proof, just try things. That is, combine equations, look at a few examples, try and understand the concepts. It's this stage, before you've found the whole proof, where true understanding and insight can occur. We saw this in the proof of Proposition 1.4.8 during class where we tried a few examples before we understood the key idea. Once you think you understand why the result is true, try and turn that into a proof.
Now, different logical forms lend themselves to different methods of proof, which we review below.
- \(P \implies Q\text{.}\) Try a direct proof first. If you try it for a while and get nowhere, consider trying to prove the contrapositive. As an absolute last resort (rarely necessary), consider a proof by contradiction. Such a proof would start “assume \(\neg (P \implies Q)\text{,}\)” which is equivalent to “assume \(P \wedge (\neg Q)\text{.}\)”
- \((P \vee Q) \implies R\text{.}\) Use a proof by cases, where in the first case you assume \(P\) and in the second case you assume \(Q\text{.}\)
- \(P \iff Q\text{.}\) Start with a two-part proof (where you prove \(P \implies Q\) and \(Q \implies P\) separately) and then see if each step is reversible. If it is, turn the proof into a concise if-and-only-if proof.
- \((\forall x) P(x)\text{.}\) Generally the best way is to prove these by starting with an arbitrary element. Proofs by contradiction should be avoided.
- \((\exists x) P(x)\text{.}\) Most often you will want to play around or guess until you find an object satisfying \(P(x)\text{.}\) Then, just write down a proof which highlights that object and shows it has the specified property. However, if that doesn't work, you might try showing the object exists indirectly (i.e., without constructing it) by contrdiction. This is one of the places where a proof by contradiction can really shine. In this case, you would start by assuming \(\neg (\exists x) P(x)\text{,}\) which is equivalent to \((\forall x) \neg P(x)\) and then proceed to derive a contradiction.