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As indicates the title, this question is about "proofs" of true statements which are short and/or look elegant but are wrong.

I mean example like Cayley-Hamilton's theorem, which states that for a $n\times n$ matrix over $\Bbb C$, and $\chi$ its characteristic polynomial, then $\chi(A)=0$. The well-known fake proof consists of a substitution $\lambda=A$ in $\chi(\lambda)=\det(A-\lambda I)$, which is not allowed.

So, I think writing a big-list could be interesting, where each answer will contain:

  • the statement;
  • the fake proof;
  • an explanation of the gap in the proof;
  • if possible, a reference to a good proof.

Each one can concern any field of mathematics. It will be good to have an example in every field: real analysis, measure theory, etc...

Davide Giraudo
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    The fake proof is easy to fix though: change the base field from $\mathbb{C}$ to $\overline{\mathbb{C}(x_{1,1}, \ldots, x_{n,n})}$... – Zhen Lin Dec 27 '12 at 11:02
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    @ZhenLin I agree, but it's not so simple to think about such a fixation when we believed we could replace a scalar by a matrix. – Davide Giraudo Dec 27 '12 at 11:06
  • @ZhenLin Could you give a reference? (I'm interested in the detailed proof. I'm not an algebraist.) Thanks. – vesszabo Dec 27 '12 at 11:20
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    Isn't this more or less the same question as Pseudo Proofs that are intuitively reasonable? It seems that many of the answers there would qualify as answers to your question. – Martin Dec 27 '12 at 11:20
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    @Martin I've done a research and I didn't find it, and indeed it's related. However (it's my opinion, not necessarily true), the other thread gives "non rigorous" proofs, which are not necessarily wrong. – Davide Giraudo Dec 27 '12 at 11:24
  • The gap in the proof of Cayley-Hamilton is that "evaluation at $A$" would result in $\det(I \otimes A - A \otimes I )$, where $\otimes$ is the kronecker product, and the determinant is taken as an $n \times n$ matrix over $\mathbb{C}[A]$, not as an $n^2 \times n^2$ matrix over $\mathbb{C}$. I'm sure the proof could be finished in this manner, but alas I don't see an easy method. –  Dec 27 '12 at 14:57
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    There is a whole book about that ranging over various fields of mathematics. Mathematical Fallacies, Flaws and Flimflam by Edward J. Barbeau. Love it! And for all of the "proofs", it takes care of the first three of your bullets. – Fixed Point Jan 18 '13 at 20:05
  • Is this related? – Red Banana Jun 09 '13 at 20:21
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    @BandeiraGustavo No, because I want true statements with an attempt of proof which doesn't work. – Davide Giraudo Jun 10 '13 at 19:33
  • Possibly related: https://math.stackexchange.com/questions/8814, https://math.stackexchange.com/questions/743067 – Watson Dec 27 '19 at 21:34

5 Answers5

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My favorite is the following:

Let $\pi$ be rational, and write $\pi = a/b$ in lowest term. Let $p \neq 2$ be a prime not dividing $a$. Then in $\Bbb{Q}_{p}$, we have

$$ 0 = \sin(pb\pi) = \sin(pa) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)!}(pa)^{2n+1} \equiv pa \ (\mathrm{mod} \ p^2), $$

which is absurd since $a \not\equiv 0$ mod $p$. Therefore $\pi$ is irrational.

The essential gap in this too-good-to-be a proof is that a $p$-adic power series may not converge to the same value as in the real field case, even the series consists of only rational terms. Thus the value of $\sin x$ need not coincide in $\Bbb{R}$ and $\Bbb{Q}_{p}$.

This false proof appears in Neal Koblitz's p-adic Numbers, p-adic Analysis, and Zeta-Fnctions.

Cardboard Box
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Sangchul Lee
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Let $\displaystyle \int$ denote $\displaystyle \int_0^x (\cdot ) dx$. Consider solving the equation $$\int f = f-1.$$ Rearranging, we get that $$f - \int f = 1 \implies \left(1 -\int \right)f = 1$$ Hence, $$f = \dfrac1{1 - \displaystyle\int} = \left(1 + \int + \int \int + \int \int \int + \cdots \right)1\\ = 1 + \int_0^x 1 dx + \int_0^x \int_0^x 1 dx + \int_0^x \int_0^x \int_0^x 1 dx + \cdots = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots = e^x$$which indeed satisfies the equation.

Adapted from this post. The post has lot of other interesting answers as well.

Community
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    Is that proof really that false?

    As an example, on $(\mathcal C([0,1/2]),|\cdot|\infty)$ the operator $\int$ you defined has norm $|\int|{\mathcal L (\mathcal C([0,1/2]))}=1/2$ and all the operations you carry seem to be legitimate. Although it would be better to use two different letters for the identity on $\mathcal C([0,1/2])$ and the constant function equal to one.

     – Sebastien B Jan 24 '13 at 17:16
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What I have in my mind is Wilson's theorem, which says that if $p$ is a prime number, then $$(p-1)!\equiv -1 \pmod p.$$

The fake proof I have learned is the following: Since $p=p-1+1$, by taking factorial on both sides, we have $$p!=(p-1+1)!=(p-1)!+1!=(p-1)!+1.$$ Now taking mod $p$, we obtain $$0\equiv(p-1)!+1 \pmod p.$$

Of course the "proof" is wrong. The gap occurs because factorial is not distributing in the sense that $(a+b)!\neq a!+b!$ in general. In fact, same "proof" would work without assuming $p$ is prime.

Paul
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There is a whole book about that ranging over various fields of mathematics. Mathematical Fallacies, Flaws and Flimflam by Edward J. Barbeau. Love it! And for all of the "proofs", it takes care of the first three of your bullets.

Fixed Point
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2

There is simple "proof" of four color theorem:

https://superliminal.com/4color/

unfortunately I still can't see the gap in it.

user3840170
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tom
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    In figure 2b, the green-red and green-blue paths can cross through each other. Therefore, after we swap yellow/blue inside the green-red path, it may be that we thereby break the green-blue path. And since there are now two blue neighbors of $V$, the argument that a green-blue path must exist cannot be repeated. – hmakholm left over Monica Feb 04 '13 at 23:36
  • This link is now broken. Could anyone provide an updated URL? – BSplitter Aug 02 '22 at 21:20
  • @BSplitter: An updated URL is here. I'm not going to edit the (community wiki) Answer above because the content is not really "simple" in my view (but your view might differ). – hardmath Aug 24 '22 at 04:27