8
$\begingroup$

I will start off by saying that I am an elementary student of mathematics and do not possess the deep and rigorous knowledge of most members of this site. Nonetheless, whilst learning how to do a proof by contradiction, I had a fascinating thought that I would like clarification on.

I realise that, by Gödel's incompleteness theorems, mathematical axioms must be either consistent or complete but not both. If we ever encounter an inconsistency in our (correct) mathematical reasoning, then that is a sign that our mathematical axioms are complete but inconsistent.

This may be a really stupid question, but I would rather ask it anyway:

If one is attempting a proof by contradiction (say, at the most advanced levels of mathematics) and they encounter a contradiction, how would we know that it is the contradiction we were searching for and NOT an inconsistency of our axioms? I realise that, given the success of our axioms thus far, this is extremely unlikely, but if the unlikely did happen in such a situation, how would we know?

Again, I apologise if this is a stupid question, but I would greatly appreciate it if someone would entertain my thought.

$\endgroup$
  • 2
    $\begingroup$ The brief summary of the incompleteness theorem in the post is missing several important hypotheses. $\endgroup$ – Carl Mummert Feb 9 '17 at 12:40
  • 1
    $\begingroup$ @CarlMummert I know. I was just summarising the ones that I thought were relevant to my question. $\endgroup$ – The Pointer Feb 9 '17 at 12:41
  • 4
    $\begingroup$ Mathematicians are very sensitive to such things; it's important to list all the hypotheses, particularly for the incompleteness theorem. There are complete and consistent theories, after all, such as the theory of the field of real numbers. (Also, a system can be both incomplete and inconsistent.) $\endgroup$ – Carl Mummert Feb 9 '17 at 12:44
  • $\begingroup$ @CarlMummert Such as the theory of any model over any signature, by definition. $\endgroup$ – xavierm02 Feb 9 '17 at 13:47
  • 1
    $\begingroup$ @CarlMummert Wait, how can a theory be both incomplete and inconsistent? If you're inconsistent you prove everything! $\endgroup$ – Noah Schweber Feb 9 '17 at 18:58
12
$\begingroup$

how would we know that it is the contradiction we were searching for and NOT an inconsistency of our axioms?

We don't. But either way, the conclusion holds (since, if the axioms themselves are inconsistent, then all statements are true).

By that, I mean that

  • if the axioms are consistent, then $\neg P\implies \bot$ is proof that $\neg \neg P$ is true, and from that, most will conclude that $P$ is true.
  • If the axioms are inconsistent, then, because $\bot\implies P$ is true and $\bot$ is true, $P$ must also be true.

So, in both cases, the conclusion, $P$, is true (in ZFC).

$\endgroup$
  • 1
    $\begingroup$ Thanks for the response. You mean everything we thought to be true/false instantly becomes unknown, since the inconsistency means we could have proved it either way (true or false) by "mistake"? $\endgroup$ – The Pointer Feb 9 '17 at 11:48
  • 2
    $\begingroup$ @ThePointer Formally, yes. However, many of the proofs may be re-used in some other set of axioms, so it might not be a disaster. $\endgroup$ – 5xum Feb 9 '17 at 11:49
  • 2
    $\begingroup$ @ThePointer My advice is just to not think about these things too much. After all, math works in the real world, so whatever formalization we choose must be something darn close to what we have right now. $\endgroup$ – 5xum Feb 9 '17 at 11:53
  • 2
    $\begingroup$ @ThePointer See this answer on the topic "What if ZFC is inconsistent?": mathoverflow.net/a/41030/103918 $\endgroup$ – Wolfram Feb 9 '17 at 12:06
  • 2
    $\begingroup$ By true in a system, I mean "provable". I thought that much was clear. $\endgroup$ – 5xum Feb 9 '17 at 12:59
8
$\begingroup$

The short answer is that there is no simple way to distinguish an inconsistency that comes from a proof by contradiction from an inconsistency that comes from working with inconsistent axioms.

In some cases, if the axioms were the source of the inconsistency, an examination of the proof in question might show it. But if the proof is sufficiently subtle, this analysis may not be straightforward.

In many cases the proof we are working with uses very simple axioms. In these cases, it may be clear that the axioms are consistent. This is the case, for example, with the axioms for a field or a vector space.

In other cases, such as ZFC, there is a general belief in the consistency of the system, because we have been working with it for so long. Many people have tried to construct a contradiction in ZFC, and none has been found. This is not a proof that ZFC is consistent, of course (although there are some arguments that do aim to prove the consistency).

Even if ZFC was found to be inconsistent, it would not affect the vast majority of mathematical results, which do not really rely on special features of ZFC. We could reprove these results using many sets of axioms. If an inconsistency in ZFC was found, it would be interesting for logicians, but unlikely to affect most mathematicians.

$\endgroup$
  • 2
    $\begingroup$ "If an inconsistency in ZFC was found, it would be interesting for logicians, but unlikely to affect most mathematicians" which is why, dear OP, you shouldn't be too worried about inconsistencies in ZFC. $\endgroup$ – 5xum Feb 9 '17 at 13:01
3
$\begingroup$

When you're doing proof by contradiction, you explicitly assume some statement $A$ and after some logical process obtain a contradiction, that is, for some statement $B$ you prove both $B$ and $\neg B$. This only proves that you assumption was false, that is, it proves $\neg A$. The contradiction in ZFC would be entirely different matter - you doesn't assume anything but the axioms of ZFC and obtain the proof of both $B$ and $\neg B$. In some sense, when you prove by contradiction, you prove that ZFC, if you add $A$ to it, is inconsistent and therefore ZFC implies $\neg A$. However, inconsistency of $ZFC+A$ doesn't say anything about inconsistency of ZFC itself.

$\endgroup$
  • $\begingroup$ Thanks for the response. Excuse my ignorance, but it sounds like your answer disagrees with that of 5xum? Can you please elaborate on what you disagree with, if anything? $\endgroup$ – The Pointer Feb 9 '17 at 12:05
  • 1
    $\begingroup$ No, I agree with the answer, that any statement we have proved by contradiction is true in ZFC, no matter is ZFC consistent or not (because any statement is true in inconsistent system). $\endgroup$ – Wolfram Feb 9 '17 at 12:11
  • $\begingroup$ I understand. Thank you for the answer and clarification. $\endgroup$ – The Pointer Feb 9 '17 at 12:15
  • $\begingroup$ @Wolfram: "true in ZFC" is not usual terminology. Statements can only be true or false in a model; if ZFC is inconsistent it has no model. Normally we are concerned with whether statements are true in the standard model. Importantly for this question, it is not the case that "every statement is true in an inconsistent system". $\endgroup$ – Carl Mummert Feb 9 '17 at 12:43
  • $\begingroup$ @Wolfram: In the first part of your answer ("When you're doing ... it proves $\neg A$.") you should make clear if, besides the explicit assumption $A$, also axioms of ZFC are implicitly assumed. Because then, if you prove $B$ and $\neg B$, you know only that $A$ and axioms of ZFC cannot be simultaneously true in any model. I believe that the situation you had in mind was the first one, when no implicit assumptions were made. However, OP asks about a proof "say, at the most advanced levels of mathematics", where axioms of ZFC are almost certainly assumed. $\endgroup$ – Peter Elias Feb 9 '17 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.