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Inspired by this very concise answer, which proves that $$\sin^2(\theta)+\cos^2(\theta) \equiv 1 $$ as follows:

$f(\theta)=\cos^2\theta+\sin^2\theta \quad;$

then it's simple to see that $$f'(\theta)=0,$$ then $$f(\theta)=f(0)=1,$$

I'm looking for theorems (and corollaries and lemmata) with one-line proofs.

Rules:

1) The proof of the theorem has to actually deduce something; it can't be, for example, a definition.

2) No vacuous truths; e.g. "If the sky is green, then the Riemann Hypothesis is true" is not allowed.

3) The words "trivially true" are banned!

4) The proof (when rendered in $\LaTeX$), must be under 100 characters, and must be no longer than one sentence.

If you could state the theorem along with its short proof, that would be great!


Obviously, there are lots of theorems with proofs of varying lengths-- I just care about the short ones!

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closed as too broad by mrf, Jean-Claude Arbaut, Adam Hughes, user147263, Tomás Aug 27 '14 at 18:20

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ I'm not sure why this is considered "too broad a question". I thought that was the idea behind the "big list" tag: that there are lots of examples, of which I would like to see a few. I'm obviously not asking for a complete list or anything; just some examples. $\endgroup$ – beep-boop Aug 27 '14 at 17:14
  • $\begingroup$ I don't understand your original example. The derivation of the derivatives of $\sin$ and $\cos$ use that identity to prove, so that "proof" is cyclical reasoning. $\endgroup$ – Adam Hughes Aug 27 '14 at 17:31
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    $\begingroup$ @AdamHughes you can define sine and cosine as solutions of differential equations, so that this becomes a valid proof of the identity in that context. This is my preferred definition of sine and cosine (and the exponential). $\endgroup$ – Steven Gubkin Aug 27 '14 at 17:52
  • $\begingroup$ What one considers "short" can depend on context. I assume you want this to be done with no previous machinery developed? (I can think of a very nice, very short proof of finiteness of the class number of a number ring - that uses at least 30 pages of machinery.) $\endgroup$ – user98602 Aug 27 '14 at 17:54
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    $\begingroup$ @AdamHughes most textbooks "prove" this formula using the unit circle definition and the pythagorean theorem. On the other hand, most real analysis books either define sine and cosine as solutions to differential equations, or as power series, since the unit circle definition is kind of horrible theoretically. The proof above is great if those are the definitions you are working from. I do not see why this has to be justified further. $\endgroup$ – Steven Gubkin Aug 27 '14 at 18:04
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A famous one : Irrational number to an irrational power can be rational.

Proof : If $\sqrt 2^{\sqrt 2}$ is rational, we are happy. If $\sqrt 2^\sqrt 2$ is irrational, then $(\sqrt 2^{\sqrt 2})^\sqrt 2=2$ is rational.

P.S. $\sqrt 2^\sqrt 2$ is actually irrational because it is transcendental by Gelfond–Schneider theorem, but we don't need to know this theorem to prove the above statement.

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Hairy Ball theorem (for $n=2$): There are no non-vanishing continuous tangent vector field for $S^2$

Proof:: If such a vector field did exist, let $v_x$ be the vector at $x$. The function $H:S^2 \times [0,1] \to S^2$ mapping $(x,t)$ to the point $t\pi$ radians away from $x$ along the great circle defined by $v_x$ is a homotopy between the identity and the antipodal map on $S^2$, which is impossible.

Well I wrote two sentences, but essentially it is one.

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  • $\begingroup$ I think you might need 3 so that you can justify why the identity and antipodal maps on $S^2$ aren't homotopic. $\endgroup$ – user98602 Aug 27 '14 at 17:48
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    $\begingroup$ Yeah I suppose so. It really depends if the reader knows about degrees of maps, which shouldn't be too much to ask for if they're looking at a proof of the hairy ball theorem. $\endgroup$ – fixedp Aug 27 '14 at 17:51

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