Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
4  
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
1  
@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
1  
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
1  
@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
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. –  Fixed Point Jan 18 '13 at 20:05
show 4 more comments

5 Answers

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 $pa \not\equiv 0$ 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.

share|improve this answer
add comment

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.

share|improve this answer
9  
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
add comment

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.

share|improve this answer
1  
I think its obvious to see the mistake. –  Sawarnik Jan 21 at 17:23
add comment

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.

share|improve this answer
add comment

There is simple "proof" of four color theorem:

http://www.superliminal.com/4color/4color.htm

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

share|improve this answer
2  
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. –  Henning Makholm Feb 4 '13 at 23:36
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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