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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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beep-boop
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    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. – beep-boop Aug 27 '14 at 17:14
  • 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. – Adam Hughes Aug 27 '14 at 17:31
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    @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). – Steven Gubkin Aug 27 '14 at 17:52
  • 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.) –  Aug 27 '14 at 17:54
  • @StevenGubkin yes, but then one needs to justify why we care about that identity or what use it has. One can prove a ton of things from alternative definitions, and then prove trivial identities from them, but without a proper motivation, I don't think it's terribly interesting. – Adam Hughes Aug 27 '14 at 17:59
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    @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. – Steven Gubkin Aug 27 '14 at 18:04
  • @StevenGubkin I'm not discussing the validity of the proof. Only the utility. Why do we care that this relation holds between solutions to some differential equation? There are many objects which can be defined well and for which some trivial-from-definitions relations hold. But no one would ever bother to write down the result, because it is useless. With this definition, there is no motivation to care about the result. – Adam Hughes Aug 27 '14 at 18:10
  • @AdamHughes This is not the place for an extended discussion, so I will be done after this comment. The point is that these are not just any old differential equations, but the differential equations which define sine and cosine! I care about the proof because I care about trig functions. I can imagine that something like this proof is actually in undergraduate analysis texts where you want to develop trig carefully. Maybe it is in baby Rudin? – Steven Gubkin Aug 27 '14 at 18:33

2 Answers2

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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.

mathlove
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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.

fixedp
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