Which is your favourite mathematical prof/teorem?
Mine:
Newton–Leibniz
http://en.wikipedia.org/wiki/Fundamenta ... f_calculus
Idea of linkind definite and indefinite integral

or (Lagrange's) Mean Value Theorem:
http://upload.wikimedia.org/wikipedia/c ... heorem.svg
http://en.wikibooks.org/wiki/Calculus/S ... t_Theorems
We use it all the time:P

Favorite theorem: P(A|B) = P(B|A)P(A)/P(B)

I don't know if it is technically a theorem, but I love l'hopitals rule.

if lim x-> a (f(x)) is zero or infinity, and if lim x->a (g(x)) is zero or infinity

lim x->a (f(x)/g(x)) = lim x ->a (f'(x)/g'(x))

edit: Just to be clear, there are some other conditions, but what I wrote above is the jist of it

http://en.wikipedia.org/wiki/Lhopitals_rule

I've always like Euclid's proof of the infinitude of primes, and Cantor's diagonal argument that there are more real numbers than there are integers.

But my favourite would have to be:

_________________
Poka King of the south east.

In college, we encountered this particular wording by our professor
and it has remained my classmates' and my all-time favorite:

Picard's theorem --
In any neighborhood of an essential singularity,
a function takes on every possible value,
except perhaps one, an infinite number of times.

(my emphasis in red)

Ptolemy's theorem:

Select any four points on a circle.
Call one of them A, the next one B, and following this
direction call the next one C and the last one D.

Measure the sides of that quadrilateral
a = AB
b = BC
c = CD
d = DA

and measure the two diagonals
e = AC
f = BD

Magically it turns out that a*c + b*d = e*f

Cheers,
Rainer

I like this proof, beacuse it uses a nuclear weapon to crack a nut:

Theorem
2^(1/n) is irrational for n>=3.

Proof
Suppose otherwise. Then there exists integers n (which is 3 or greater), p and q such that:
(2^(1/n))^n=(p/q)^n

Hence q^n+q^n=p^n.

However, this contradicts Fermat's Last Theorem.
QED

A separate argument is needed for n=2 - Fermat's Last Theorem is not strong enough!

(Unfortunately, the proof above actually turns out to be a circular argument if one follows the details through, and so it's not actually valid.)

@(1): me too, thanks for sharing!

@(2): that's a fancy way to put it, but "not actually valid"

@(3): I can't detect a circulus vitiosus here.
Let me repeat - slowly and for the sake of simplicity for n=3.

Proposition: There are no natural numbers p and q
such that (p/q)^3 = 2.

Proof: Assuming the opposite we would have natural numbers
p and q such that (p/q)^3 = 2.
Then p^3 / q^3 = 2,
i.e. p^3 = 2 * q^3 = q^3 + q^3.
Let a = q, b = q, c = p.
Then a^3 + b^3 = c^3 for some natural numbers a, b and c.
That contradicts Fermat.
Thus the assumption led to a contradiction.
q.e.d.

Cheers,
Rainer

I like the proof that the base angles of an isosceles triangle are equal.

The triangle is congruent to its own reflection. QED.

_________________
The Adkins Principle:
At some point, doesn't thinking have to go on?
— Winona Adkins

Visualize whirled peas.

Everything with love. Stay safe.

Rainer

The circularity is non-trivial.

Assume we use Wiles' proof of FLT - as far as I know, it's the only known proof.

The proof shows equivalance of integer a^n+b^n=c^n with n>=3 to a "Frey curve" of the form y^2=x(x-a^n)(x+b^n). At this step of the proof, certain restrictions are placed on (a,b,c) such as them being pairwise coprime. Consider the equation we had p^n=q^n+q^n. The pairwise co-prime restiction applied to (p,q,q)is equivalent to a straightforward Euler proof that 2^(1/n) is irrational.

Seems so
But the circularity is not only non-trivial, but non-existent, too.

Pardon me: I don't have to go into the details of the proof
of any theorem I use for my proof. If the proposition, I am going
to prove, is already a trivial subcase of the theorem, then so
what? That is no problem and there is no circulus vitiosus.

If you want to convince me of having used a circular argument
you should point to a certain line in my proof and say:
"Ha, here you are - that is something you use as if already
proved!"

For your convenience and easier pointing-out here are the
lines of my proof, numbered:

1. Assume natural numbers p and q such that (p/q)^3 = 2.
2. Then p^3 / q^3 = 2,
3. i.e. p^3 = 2 * q^3 = q^3 + q^3.
4. Let a = q, b = q, c = p.
5. Then a^3 + b^3 = c^3 for some natural numbers a, b and c.
6. That contradicts Fermat.
7. Thus the assumption led to a contradiction.
8. q.e.d.

The first error is in line ....?

Cheers,
Rainer

6. (Or are you being pedantic about the fact that no step in the proof assumes Fermat, so 6 -> 7 is non sequitur?)

a similar thread on this forum: http://www.lifein19x19.com/forum/viewtopic.php?f=8&t=4885h
Cantors proof is also mentionned

_________________
In theory, there is no difference between theory and practice. In practice, there is.

I always liked the integral of a gaussian over the reals.

Integral(e^(-x^2)) from -Inf to Inf

1) Square the integral,
2) Then change from cartesian to polar,
3) Solve integral
4) Take the square root
5) Square again
6) Multiply by 2 EDIT: BAD, ignore my silly half-infinites
7) Have dessert

_________________
Tactics yes, Tact no...

love that one too

_________________
In theory, there is no difference between theory and practice. In practice, there is.

Being pedantic is not so bad in mathematical context, no?
"Fermat" is short for the (meanwhile proven) fact that there
are no natural numbers a, b, c such that a^n + b^n = c^n for
natural numbers n > 2.
Thus step 5. proposes something which is impossible.
That is, what my lines #6 and #7 are telling.

I really cannot detect a circular reasoning. That would be the case
if I wrote propositions p_1, p_2, p_3 etc. such that not all p_i
could be proven from the p_j with j < i, but at least one of the
p_i needed some p_j to be true, where j > i.

I completely agree that the unsolvability of (p/q)^3 = 2 is such
an easy thing to prove that it is part of "Fermat". But I doubt
you can call that a circulus vitiosus. At least I see your
point: you are trying to see "Fermat" not as a monolithic "fact",
but you glance into the how-and-why of this proven truth.

Greetings,
Rainer

Mine would either have to be Cantor's diagonalization proof for uncountable sets, or Gödel's first incompleteness theorem.

_________________
We don't know who we are; we don't know where we are.
Each of us woke up one moment and here we were in the darkness.
We're nameless things with no memory; no knowledge of what went before,
No understanding of what is now, no knowledge of what will be.

If you do not assume that Fermat's theorem is true, then our proof is false by non sequitur. If, on the other hand, you treat step 6 as an enthymeme which implies "... and we have a proof that Fermat's theorem is true," then our proof is false by petitio principii, since the proof of the theorem itself assumes our demonstrandum.

Fermat's theorem *is* true.

I see. I thought we were talking about mathematics, not rhetoric.
In rhetoric you may be right, in mathematics you are not.

Cheers,
Rainer