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.

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    $\begingroup$ Theorem 2 may be written as "If $r$ is rational then $r^2\neq2$". $\endgroup$ – Arthur Nov 2 '18 at 17:31
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    $\begingroup$ At a basic level, all theorems are proved assuming some axioms (usually ZFC), so all theorems can be expressed as "if <axioms> then <theorem>" $\endgroup$ – TreFox Nov 2 '18 at 17:33
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    $\begingroup$ If $r^2=2$ then $r$ is not a rational number $\endgroup$ – Vasya Nov 2 '18 at 17:33
  • $\begingroup$ @Arthur So the hypothesis is that $r$ is rational and the conclusion is $r^2 \neq 2$. And we started our proof assuming that $r^2 = 2$ and $r$ is rational. I am not sure that this counts as a proof by contradiction because we started by negating the conclusion as well as assuming that the hypothesis is true. $\endgroup$ – Salman Qureshi Nov 2 '18 at 17:42
  • $\begingroup$ @Vasya This is the contrapositive of the above statement. I know they are equivalent but in relative terms, the hypothesis and conclusion are different for them. $\endgroup$ – Salman Qureshi Nov 2 '18 at 17:45

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.


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.

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    $\begingroup$ How about: "if by numbers we mean an element of (some nontrivial ring, for example), then statement is false", or "if we work in classical logic, then such and such is a theorem" $\endgroup$ – user10354138 Nov 2 '18 at 18:13
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    $\begingroup$ I sort of understand your point. Would you please elaborate a little on why all theorems can't be written as "if-then"? I don't have enough time on my hand right now to spare for a new subject. $\endgroup$ – Salman Qureshi Nov 2 '18 at 18:20
  • $\begingroup$ Of course every theorem.can be rewritten as an if/then by butchering the phrasing, e.g. "if 0=0 the [insert theorem here]". But this is not usually a natural way to express the theorem. $\endgroup$ – Carl Mummert Nov 3 '18 at 1:49

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.


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