which is your favourite mathematical prof/teorem?

[marvin]

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

[Solomon]

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

[speedchase]

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

[TheBigH]

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:

Code: Select all

　πix
e     =  cos(x) + i sin(x)

[EdLee]

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)

[GoRo]

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

[drmwc]

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.)

[GoRo]

(1)... I like this proof
(2)... Fermat's Last Theorem is not strong enough!
(3)... circular argument ... 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

[Bill Spight]

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

The triangle is congruent to its own reflection. QED.

[drmwc]

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.

[GoRo]

The circularity is non-trivial.

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

[Fedya]

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:

Code: Select all

　πix
e     =  cos(x) + i sin(x)

http://xkcd.com/179/

[jts]

The circularity is non-trivial.

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?)

[perceval]

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

[shapenaji]

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