Are all "Theorems" necessarily in the "if - then" form?

Question

When someone talks about "Theorems" in Mathematics, something of the sort below comes to my mind.

Theorem 1: For every two real numbers $a$ and $b$ with $a \lt b$, there exists a rational number $r$ satisfying $a \lt r \lt b$.

Basically, there's a Hypothesis (the "if" part) and a Conclusion (the "then" part). So if I were to, say, prove this theorem using contradiction, I would start by negating the conclusion and then proceeding logically until a contradiction is found with the hypothesis (contrapositive proof) or some other accepted fact. The hypothesis and conclusion can be seen very easily here. Let us take a different example.

Theorem 2: There exists no rational number $r$ whose square is $2$.

I am having a hard time seeing what the hypothesis is in this case. Maybe if this theorem is worded in a different way, it might be obvious but that's just my opinion.

And if you're familiar with the proof of this theorem, the contradiction comes from the fact that we had supposed a rational number $\frac pq$ in lowest terms but after proceeding logically, we find a factor of $2$ common between $p$ and $q$ which goes against our assumption that we started with and so our proof ends.

So my questions are:

1. Are all "Theorems" necessarily in the "if-then" form?
2. What is the Hypothesis in Theorem $2$? Is the proof for this theorem a contrapositive proof?

I heartily welcome any extra information in relation to this question.

Answer 1

First of all, we have to know what we mean with a theorem, for instance if we have a theorem such as

Statement All numbers are equal to $0$

Theorem 1 Statement is false

We could of course rewrite the two sentences as a new theorem

Theorem 2 There exists a number that is not $0$

But, what if wanted to represent Theorem 1 more generally (that is, without using anything said in the Statement), would we always be able to write a "Theorem 2".

Short answer, no.

TL;DR There are theorems that can't be written as "if, then".

If you want to read more about the topic, I'd suggest reading some First Order Logic (FOL), a subject where proofs and theorems get a new meaning.

Edit

Without giving any proofs, but as Salmon suggested, there are reasons as to why we can't reformulate everything, and it has to do with the axioms made in FOL, different sets of axioms yields different "types of logic", but i we restrict ourselfs to classical (what you're most likely used to) logic, we have "if-then" as an axiom, as well as falsum (that something is false) and the "for all"-statement. Not that it proves anything, but axioms are usually chosen such that they can't be derived from other axiom.

Answer 2

Regarding question 2.

When I read a theorem, let's say "If A and B then C", I find all its assumptions and conclusions and than I imagine they are written like this:

• Let A
• Let B
• Then C

If I wanted to prove it by contradiction, I would add an extra assumption,

• Let A
• Let B
• Let not C

and I would play with the statements until I find some contradicting.

Applying this to your Theorem 2, I would assume nothing - except axioms and to some point my knowledge - and state

• Then $\neg \exists_{r\in \mathbb{Q}} \; r^2=2$

Now for the contradiction I would assume the opposite.

Answer 3

Most systems of logic prove assertions of this type:

$$\Gamma \vdash a$$ which means that, starting from the assumptions in the set $\Gamma$, we can prove the statement $p$. For example, $$a\to b, a\vdash b$$means that, given the assumptions “$a\to b$” (“$a$ implies $b$”) and “$a$” we can prove “$b$”. Usually one can prove a meta-theorem called the deduction theorem that states that if $$\Gamma, a\vdash b$$ is provable, then so is $$\Gamma\vdash a\to b.$$

All these require the assumptions in the set $\Gamma$. But $\Gamma$ may be empty, and if it is then $\Gamma\vdash p$ is an assertion of $p$ that requires no assumptions. For example, in typical systems $$\vdash a\to(b\to a)$$ is provable, and one can understand this as a theorem with no if-then: “Whatever $a$ and $b$ are, it is provable that $a$ implies that $b$ implies $a$”.

You may object that this is just a rephrasing of “If $a$ is true then, if $b$ is true then $a$ is true”. Very well. Most logics will also prove $$\vdash \lnot(a\land \lnot a),\tag1$$ which says that, whatever $a$ might mean, it is not the case that $a$ is both true and false. And many will prove $$\vdash a\lor \lnot a\tag 2$$ which says that any $a$ is either true or false.

I don't know if this is what you were looking for, but I don't see anything in either $(1)$ or $(2)$ that resembles “if…then…”, so perhaps that will answer your question.