3 questions about proofs

Question

I am fairly new to proofs and I have a test coming up. I had 3 questions that I can't seem to find the answers for. I would appreciate any help. Thank you!

$(1)$

Let's say I'm proving a direct proof ( $p \implies q$ )... if $p$ is false, can the statement be vacously true therefore proven? (I hope this makes sense)

$(2)$

When proving a direct statement, can you begin with the conclusion and work your way to prove the statement? Or do you have to always start with assumptions that end up becoming your "goal" statement?

$(3)$

If you have OR's in your conclusion, do you have to prove all or just one?

Answer 1

For some reason 2) is something the majority of students end up doing wrong in their first semester even though it is emphasizes over and over that you have to be careful with that. So it is very good that you ask this questions by yourself. The answer is "no" with a caveat. No because I could prove $2=1$ this way: $$\begin{align} &2=1\\ \implies & 0\cdot 2=0\cdot 1\\ \implies &0=0 \end{align}$$ The caveat is, that you can of course try to find a proof this way. But then you have to ask yourself if you can write it the other way. I.e. am I allowed to write? $$\begin{align} &2=1\\ \impliedby & 0\cdot 2=0\cdot 1\\ \impliedby &0=0 \end{align}$$ And the answer is no. Because I can't just divide by $0$ on both sides. You might think this is an obvious edge case and this is pedantry. But this is not the case. Because in mathematics you often deal with variables. And it is easy to forget that a complicated term might be zero. This is how many "proofs of 2=1" work. You obscure the fact that you are dividing by zero somewhere by doing a bunch of transformations hiding the time where you divide by zero. Like many magic tricks, distraction is the main ingredient. E.g. (from https://www.math.toronto.edu/mathnet/falseProofs/first1eq2.html):

$$\begin{align} &a=b\\ \implies & a^2=ab\\ \implies & a^2+a^2 = ab +a^2\\ \implies & 2a^2=a^2+ab\\ \implies & 2a^2-2ab=a^2+ab-2ab\\ \implies & 2a^2-2ab=a^2-ab\\ \implies & 2(a^2-ab)=a^2-ab\\ \implies &2=1 \end{align}$$

Answer 2

To complete on your other questions, as they can have rich answers in certain cases.

(1) - Yes : you can always prove $p \Rightarrow q$ if you can prove that $p$ is false. If you work with propositional logic, it is obvious, because you can just look at the truth table of $p \Rightarrow q$, but in more complex systems, it might not be all that obvious. In other (richer) systems like first order logic, or intuitionistic, it is also true, but might be a bit harder to see. In order to prove $p\Rightarrow q$, you start by assuming $p$, but in every system, from a wrong assumption, you can always deduce anything.

(3) - Yes and No : There are many ways of showing $P\vee Q$ classically. If you prove just $P$, then you can deduce $P\vee Q$, or if you prove just $Q$ you can also deduce $P\vee Q$, so you really need to prove only one of them. However, there are ways to prove $P\vee Q$ without knowing which one of them is right, for instance, you could suppose that $P\vee Q$ is false (that is both $P$ and $Q$ are false), and show that you get a contradiction. That way you know that $P$ and $Q$ cannot be simultaneously false, hence $P\vee Q$ is true. Note that by doing this I haven't proved $P$ nor $Q$, and I might not been able to tell which one is true and which one is false. This creates a discrepancy between the $\vee$ symbol and the intuitive meaning we have for "or" (and it gets even messier with undecidable statements). If you ever hear of inuitionistic logic, it is a logical system in which the reasoning I just presented is forbidden, and in which proving $P\vee Q$ is exactly the same as either proving $P$ or proving $Q$