"Too simple to be true"

Question

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...

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

There is simple "proof" of four color theorem:

https://superliminal.com/4color/

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