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In mathematics, it’s common to encounter statements that feel intuitively obvious but turn out to be extremely subtle, difficult to prove, or even false upon closer inspection. Examples include:

The statement “every continuous function is differentiable” obviously false, but intuition suggests smoothness.

“All continuous functions on the real line are monotone somewhere” false, but intuition nudges you otherwise.

The Banach-Tarski paradox counterintuitive and “obviously impossible” yet mathematically true.

The Four Color Theorem seemingly straightforward but requiring complex proofs and computers.

Many “folk” conjectures or naive assumptions in number theory or combinatorics that took decades or centuries to settle.

Why does intuition so often fail in mathematics, causing seemingly obvious claims to be false or require highly nontrivial proofs?

J. W. Tanner
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F. A. Mala
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  • I feel this question is too subjective for this site, and so I have voted to close, sorry. But for the record, my opinion is that to a considerable extent, this phenemon can be explained by selection bias. The overwhelming majority of mathematical statements (in a fixed formal language) are totally uninteresting, and probably not even comprehensible to humans. – Joe Jun 09 '25 at 16:44
  • In fact, even among the theorems that humans do study, e.g. those appearing in an undergraduate textbook, the majority are not especially surprising or interesting. It is a tiny slither of theorems which are nontrivial, counter-intuitive, and interesting, but humans spend a widlly disproportionate amount of time thinking about them. – Joe Jun 09 '25 at 16:45
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    Whose intuition? – ancient mathematician Jun 09 '25 at 16:48
  • You might get more responses if you try asking this question on a different site, like reddit. – Joe Jun 09 '25 at 18:19

1 Answers1

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Edit: I interpreted the question as asking for further examples. I will address the question at the bottom about why intuition fails. A related question: Why should we prove obvious things?

Some further examples

My undergraduate real analysis professor gave an example of this. The following theorem should be "obviously false" (from here):

Theorem. Let $I$ be an interval and $f : I \to \mathbb{R}$ strictly monotone. Then the inverse function $f^{-1} : f (I) \to I$ is continuous.

It would seem that because there are "disconnects" in the function going in one direction, that those disconnects would obviously appear in the inverse function as well. Yet, the theorem is true!

Another example is the well-ordering theorem, which states that every set can be well-ordered. Some people find this statement to be dubious (after all, most ordered sets are not total orders!) but it turns out that this can be shown to be equivalent to both the axiom of choice and Zorn's lemma, and therefore provable in set theory. As the old quip goes: "The axiom of choice is obviously true, the well-ordering principle is obviously false, and who knows about Zorn's lemma?"

Why intuition fails

The question asks:

Why does intuition so often fail in mathematics, causing seemingly obvious claims to be false or require highly nontrivial proofs?

I think the simple answer is that this is a part of mathematics. Mathematics is about proof, and intuition is (by nature) a heuristic which uses patterns we have seen before to try to identify the truth or falsehood of a statement in a new context. Hence, it is by nature possible for this intuition to fail.

It is worth noting that trained mathematicians in a given area typically have much stronger intuition about what statements are true compared to untrained students, so this intuition does "get better" over time as one is exposed to more statements in an area. I'm not sure if it converges at some point so that one is able to correctly guess the truth or falsehood of any statement, or at least never make a guess that is wrong. That would be interesting if true, and a bit surprising.

Caleb Stanford
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  • Not to be pedantic, but if by "set theory" you mean ZFC, then the well-ordering theorem is also equivalent to the assertion that $0=0$ (both are theorems). It's only over a set theory like ZF which doesn't prove the axiom of choice in which it is interesting that the axiom of choice is equivalent to the well-ordering theorem. [Side note: I am implicitly assuming that ZF is consistent here. If ZF is inconsistent then it proves ever statement; in particular, it proves the axiom of choice.] – Joe Jun 09 '25 at 16:52
  • @Joe I think it's fine as written? The statement can be shown to be equivalent to the axiom of choice either over ZF or over ZFC. Just because the statement of equivalence is between true theorems (of ZFC) does not make it a trivial statement. – Caleb Stanford Jun 09 '25 at 16:59
  • Well, working in ZFC, the assertion that the well-ordering theorem implies the axiom of choice is trivial, since an implication with a true consequent is always true. The assertion that the axiom of choice implies the well-ordering theorem is less trivial – but its nontriviality arises solely because it is difficult to prove the well-ordering theorem full stop. In ZF, the situation is different: assuming that ZF is consistent, neither the axiom of choice nor the well-ordering theorem can be proven; however, we can establish that they have the same truth value. – Joe Jun 09 '25 at 17:56
  • [Continued...] Said differently, any model of ZF where choice holds, the well-ordering theorem also holds (and vice versa). But every model of ZFC satisfies both choice and the well-ordering theorem (which is a less interesting situation). – Joe Jun 09 '25 at 18:01
  • [Sorry to comment again.] Saying that the axiom of choice is equivalent to the well-ordering theorem in ZFC is a bit like saying that in every ring $R$, the assertions $\forall x,y\in R((x+y)^2=x^2+2xy+y^2)$ and $\forall x,y\in R(x+y=y+x)$ are equivalent. I mean, they are, but only for the trivial reason that both statements hold in every ring. For an equivalence in ring theory to be "interesting", it usually needs to be the case that there are rings in which both conditions hold, and rings in which both conditions fail. There is a clear analogy with models of ZF(C). – Joe Jun 09 '25 at 18:13
  • Yes agree with all of the above, but don't think it negates my point. Note that the answer does not state equivalence over ZFC or over ZF, "over ZF" is the more plausible inferred meaning given it is a statement of equivalence and the other interpretation is (as you say) trivial in one direction. – Caleb Stanford Jun 09 '25 at 18:32