Unraveling the Right Angle Proof of Derivatives of Inverse Trig Functions

[harvellt]

I am still a little early in my math education but we were talking about this the other day.
Mine is the right angle proof of the derivatives of inverse trig functions. The first time I went threw them I was just given the rule that the derivative of arcsin is such and such, with no real explanation. It was hard to remember and didn't make a lot of sense but after I took the time to look at the very simple proof it all beautifully clicked together.
What are yours?

 

[quasar987]

This one is fine. First, try to think up for a while of a function f:R-->R that is nowhere locally bounded. That is to say, such that for any point a and any interval I however small around a , f is unbounded on I.

Then consider the following solution.The function f given by

f(x)=0 if x is irrational and f(x)=n if x is rational and x=m/n where the fraction is reduced and n>0

has the desired property. To see this, consider a in R and e>0 and suppose f is bounded in (a-e,a+e). This would mean that the denominators (in reduced form) of all rational numbers in (a-e,a+e) are bounded. And since the interval (a-e,a+e) itself is bounded, this implies that the numerators are bounded too! But then this would mean that there are only finitely many rationals in (a-e,a+e), which is absurd.

 

[deiki]

Hölder inequality.

 

[Enjoicube]

I have just begun analysis, so all of my favorite proofs are very simple, but I love the proof that R is uncountable, and yet that any open interval within it is also uncountable. I am sure this will change with time though.

 

[uman]

The proof of Lagrange's theorem: If H is a subgroup of G, the order of G is divisible by the order of H. It seems to go from a lot of simple ideas arranged into good definitions to a deep result, almost by magic.

 

[Kurret]

The (two) proof(s) of Pythagoras Theorem which involves creating a square with sides equal to the hypothenuse, and then creating 4 equal triangles on the inside(outside) of the square with the sides being the hypothenuses. Very elegant, maybe not my favorite but the first that came to my mind right now.

 

[uman]

I also really like the proof that the Cantor set (which has measure zero) has the same number of elements as the set of all real numbers. Just more evidence that real numbers are counterintuitive and bizarre!

 

[slider142]

Some of my favourite theorems have already been listed. I'll add Stokes's Theorem for elegance and physical applications.

 

[uman]

Which are your favorites that have already been mentioned, slider?

 

[Enjoicube]

I always thought Greens theorem's proof was more beautiful than Stokes' theorem, even htough it is just a special case. Nevertheless, both are very beautiful, and Stokes is a close second in vector calc. The only proofs I can say I don't like are proof by cases when a direct proof would do, it's like whacking a lock off a door intead of using the key under the mat.

 

[Kurret]

Just had to post about these proofs, which is two of my favorites
Proof of Fermats little theorem, from "problem solving strategies" by Arthur Engel:
Consider necklaces of p pears in a different colors where p is a prime number. We want to count the number of necklaces that does not only have all pearls in one color. So we start with a string of p pearls. there are a^p such strings. Now we connect ends and throw away the one colored necklaces, and we are left with a^p-a. But each necklaces now come in p copies, so the number we are looking for is (a^p-a)/p. Now this must be an integer, thus p|a^p-a.

Another proof I think is very elegant is a proof of Ptolemys theorem. Can be found at page 37 in http://www.math.rochester.edu/people/faculty/dangeba/geom-080399.pdf

 

[quasar987]

I always had a sweet spot for the proof that \sqrt{2} is irrational:

First observe that if an integer k is such that k² is even, then it means that k is even. To see this, prove the contrapositive: k odd ==> k² odd. Indeed, to say that f k is odd is equivalent to saying that k is of the form k=2n+1, and thus k²=4n²+4n+1=2(2n²+2n)+1. That is to say, k² is odd too.

Now, for the proof that \sqrt{2} is irrational, let x be such that x²=2 and suppose x is rational: x=p/q with gcd(p,q)=1 (that is to say, p and q have no common divisor, that is to say, the fraction p/q is "reduced".) Squaring that equation yields p²=2q². That is to say, p² is even. So by the above lemma, p is even: p=2n. Thus q²=4n²/2=2n² and so q² and q are even too. But this is absurd because it means that gcd(p,q)\geq2 (that is to say, 2 is a common divisor of p and q, that is to say, the fraction p/q is not reduced.)! Thus x, and in particular, ±\sqrt{2} are irrational.

 

[Russell Berty]

Speaking of the square root of two… Here is a fun little nonconstructive existence proof.

Prove that there exists irrational numbers a and b such that a^b is rational.

Proof – Note that \sqrt{2}^{\sqrt{2}}

is either rational or irrational. We will argue by cases.

Case 1) If \sqrt{2}^{\sqrt{2}}
is rational then we have our irrational a and b, namely a=\sqrt{2} and b=\sqrt{2}. Thus, in this case, there exists irrational numbers a and b such that a^b is rational.

Case 2) If \sqrt{2}^{\sqrt{2}}
is irrational then notice
\left(\sqrt{2}^{\sqrt{2}}\right)^{\sqrt{2}}=\sqrt{2}^{\sqrt{2}\sqrt{2}}=\sqrt{2}^{2}=2

Hence, letting a=\sqrt{2}^{\sqrt{2}} and b=\sqrt{2}, we see in this case that there exists irrational numbers a and b such that a^b is rational.

In either case, there exists irrational a and b such that a^b is rational.

 

[Dadface]

An ancient one but I like it Circle area=pi*r squared by dividing the circle into smaller and smaller slices.There's something fascinating about pi.

 

[Jamma]

I saw this proof on the internet that 1=0, it was totally amazing.

 

[AUMathTutor]

I like the proof that the halting problem is undecidable.