Why this proof $0=1$ is wrong?(breakfast joke)

Question

We have $$e^{2\pi i n}=1$$

So we have $$e^{2\pi in+1}=e$$

which implies $$(e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e$$ Thus we have $$e^{-4\pi^{2}n^{2}+4\pi in+1}=e$$

This implies $$e^{-4\pi^{2}n^{2}}=1$$

Taking the limit when $n\rightarrow \infty$ gives $0=1$.

I took the liberty of doing an edit that actually changed the meaning of your question, in the middle formula. It made no sense as it stood, but it does now, I think. — Harald Hanche-Olsen, Aug 04 '12 at 11:55
Do you want to undo my edit? You certainly may do so, if you insist. — Harald Hanche-Olsen, Aug 04 '12 at 12:02
You actually show that 1 is equal to infinitely many other numbers (if the steps of the argument are correct) — i. m. soloveichik, Aug 04 '12 at 12:08
This puzzle is due to Thomas Clausen). Have you perhaps found it in Nahin's "Story of $\sqrt{-1}$"? — Raskolnikov, Aug 04 '12 at 15:41

score 9 · Accepted Answer · answered Aug 04 '12 at 12:29

9

Your error is (as in most of those fake-proofs) in the step where you use the power law $(a^b)^c=a^{bc}$ without the conditions of that power law being fulfilled.

answered Aug 04 '12 at 12:29

celtschk

44,737

8

It could be useful to recall these conditions, or give a reference. – Vincenzo Tibullo Aug 04 '12 at 13:12
But such conditions are not easy to produce, except for some special cases: For example, having $a$ and $b$ real (with $a$ nonnegative) is sufficient, unless you base your powers on a highly unusual branch of the natural log function. – Harald Hanche-Olsen Aug 04 '12 at 14:15

score 8 · Answer 2 · answered Aug 04 '12 at 12:00

8

The proof is wrong because an expression of the form $x^y$ is actually ambiguous, when $x$ is a complex number: Rewrite it as $e^{y\ln x}$ and note the multivalued nature of the natural logarithm as used on complex numbers. For your proof to be correct, you would need $\ln e^{2\pi in+1}=2\pi in+1$, but that is not consistent with $\ln e=1$.

answered Aug 04 '12 at 12:00

Harald Hanche-Olsen

32,768

Would be useful to clarify why this doesn't mean complex analysis is an inconsistent theory. – akkkk Aug 04 '12 at 16:10
2

@Auke: No. That would lead to a paraphrase of half a chapter's worth (or more!) of some complex analysis book, talking about branch points, Riemann surfaces, and I don't know what. This is not the purpose of this site. Anyway, the point of all this is that the union of complex analysis and the blind application of certain rules learned from real analysis is an inconsistent theory. – Harald Hanche-Olsen Aug 04 '12 at 17:50

score 1 · Answer 3 · edited Aug 05 '12 at 10:14

1

From wikipedia on Euler identity

The identity is a special case of Euler's formula from complex analysis, which states that $e^{i x}=\cos x+i\sin x$. for any real number $x$.

Note $x$ should be real number.

$$e e^{i x} \neq e^{i x + 1} = e^{i(x-i)} = \text{undefined} $$

edited Aug 05 '12 at 10:14

Vincenzo Tibullo

11,397

answered Aug 04 '12 at 15:34

Ankush

239

3

Euler's formula also holds for complex $x$. In fact, it is a great trick to define cos and sin for arbitrary arguments. – akkkk Aug 04 '12 at 16:07
Oh! so $e*e^{ix} = e^{ix + 1}$ is true? – Ankush Aug 04 '12 at 17:15
2

@Ankush: Yes. But please, please don't use the asterisk $*$ for multiplication! In mathematics, it usually means convolution. It's an abomination coming from programming languages, where it arose due to the lack of a more appropriate symbol in ASCII (or EBCDIC, or …). Use \cdot instead: $e\cdot e^{ix}$. – Harald Hanche-Olsen Aug 04 '12 at 17:41
ok :) I'm new to writing equations in $ – Ankush Aug 04 '12 at 18:27
1

@Ankush: it is called LaTeX. – akkkk Aug 04 '12 at 19:22
@Auke I know that. But never got a chance to write equations into TeX system. Only used for resume n final report purpose :) – Ankush Aug 04 '12 at 19:41

Why this proof $0=1$ is wrong?(breakfast joke)

3 Answers3