What is the most useless theorem in mathematics?

Anything Cauchy,Weierstrass.

tao seems to let his cock spurt loads over it though

Well the hard part is usually translating the language of a problem into the language of pigeons and holes. It's useful to be able to say, "by the pigeonhole principle" and have people know what you're doing.

>"if n > m, n / m > 1"

wow

>not the generalised version

True, but it's used a lot, relatively speaking. So it's at least useful for mathematicians.

Basically any theorem which essentially goes

>The Existence Theorem
>A = A

Which I've seen in a half-dozen textbooks in various fields of math.

What in the world kind of math are you doing? Example of one of these theorems?

Symbolic logic has p -> p, that's the first thing that comes to mind.

I don't have any other examples off the top of my head. They're usually stuck in the middle of a table filled with other tiny little identities and theorems.

Probably independence results relating to the axiom of regularity. For example, in ZFC - {Regularity} it is consistent that {x | x = {x}} is not only nonempty, but a proper freaking class.

Interesting as it may be, this is useless to mathematics because literally everything can be done with regularity assumed. Every algebraic structure is homomorphic to a well founded set, every topology is homemorphic to a well founded set, etc... It would take something highly pathological to need a non-well founded set, (i.e. something that contradicts regularity.)

Yes, for any equivalence relation you have to show that A == A.

10 > -10
-1 > 1
wow

Archimède's theorem

You have to flip the inequality when you multiply by 1/-10 buddy

What if [math] n = | \aleph_0 | [/math] and [math] m = | \aleph_1 | [/math]?

The Jordan Curve Theorem:
"A continuous , closed loop divot gay doesn't intersect itself divides R^2 into 2 regions: inside and outside the loop"

This is literally a theorem that someone spent hundreds of hours of a professional career proving and was praised for it

10 > -10
-1 ^ 1
wow

if you thing it's useless you haven't done complex analysis. it's a VERY important theorem, and deceivingly hard to prove.

Banach-Tarski theorem, because you can't reassemble an object into two identical copies without completely abtracting yourself from any possible hope of real-world relevance

its you op

your math
and usless

>continuous
>continuous

It tells you the answer in the premise !

If if the line is genuinely continuous then of course it is going to divide space without any gaps

Yes even if the line is a fucking fractal.

Green, it is only Gauss or Stokes on a flat curve and a 2d vector field. no real application.

What the fuck do you think continuous means? Because it's not "obvious" that for f:[0,1]->R^2 such that "f(0)=f(1)", the condition "for all x in [0,1] and e>0, there exists a d>0 such that ||x-y||>d implies ||f(x)-f(y)||

a=b
b=c
Therefore a=c

lol

Go tell that to “CS majors” please, because apparently it's beyond their ability to understand.

that's clear when you're working in R^2, it's not clear when you're working in something that isn't R^2.

>a=b
>b=c
>thereforea=c
lolwhat?no.a=bretard

What are you getting butthurt for, blathering about irrelevant nonsense.

Are they actually that dumb?

example taken from javascript

>1 == " 1" // true
>1 == "1 " // true
>" 1" == "1 " // false

Lrn2pythagoras fgt pls

>theorem
>real-world
fgt pls

>dynamically typed languages
>mfw
Though I'm disappointed that you didn't realize that you use two different relations in your example

>Though I'm disappointed that you didn't realize that you use two different relations in your example
Not sure what you're referring to. Do you mean because one invocation compiled down to string comparison and the other down to integer comparison?

If so, that's a meaningless implementation detail - there's only a single == relation in JavaShit, because it operates on a single universal type. (You can compare every value to every other value)

No, the == relation is an overloaded operator and is dynamically resolved into the typed version, it's basically dynamic dispatch, but with relations instead of methods. It's not an implementation detail, the equality is defined only for the typed relations in the language standard. If your example shows anything, then it's that dynamic typing is stupid and leads to stupid pitfalls

It's not overloaded; it's syntactically and semantically a single operator with a single implementation.

Syntactical definition: ecma-international.org/ecma-262/7.0/index.html#sec-equality-operators

Operational semantics: ecma-international.org/ecma-262/7.0/index.html#sec-abstract-equality-comparison
(See also ecma-international.org/ecma-262/7.0/index.html#sec-samevaluenonnumber)

Note however that this is a single algorithm in principle. There's nothing in here about there being multiple distinct == operators, nor does JavaShit have any sort of dynamic dispatch mechanism for overloaded operators.

If you were to translate this from JavaShit into the language of the well-typed (e.g. Haskell), it would be like this:

[code]
data JsVal = JsNull | JsNum Double | JsString String | ...

instance Eq JsVal where
JsNull == JsUndef = True
JsUndef == JsNull = True
JsNum NaN == JsNum _ = False
JsNum _ == JsNum NaN = False
...
-- all of the described rules
_ == _ = False
[/code]

And *not* like this:

[code]
data JsNull = JsNull
data JsNum = JsNum D ouble

instance Eq JsNum where
...
[/code]

lol you're dumb
many languages have a top down approach
to state
a=b
b=c
and say that a=c is false since a was assigned b before b was assigned c
the = operator means assignment
what you're looking for is ==

Lowenheim-Skolem, and ultimately the Power Set Axiom.

Fuck your cardinals, the universe is discrete

Noether's Theorem, just to say fuck you to that guy who posts that shit like weekly.

U can't know nuffing bout models

Löwenheim-Skolem, in addition to compactness and completeness, is one of the most important theorems in model theory. If you feel model theory is useless, then you are also wrong, since it's extremely applicable in algebra at least.

pajeetgramming != compsci

go back to bed Wildburger

I am provoked by this

zorn's lemma

Like 80% of them.

a = a for sure

>mfw all these math-illiterate undergrad brainlets who say x = x not knowing reflexivity is fundamental as fuck
The correct answer is either [math]\mathrm{card}\,\mathbf P\,=\,\aleph_0[/math] or the Banach-Tarski theorem.

dafuq is [math]\mathbf P[/math]

1 * A = 1 * A is more descriptive, in my humble opinion

Under certain conditions, 1 * A = A * 1 can also be used.