What are examples of things in mathematics that appear to be "obviously" true but in fact aren't?
>>8119265
the equation [math]x^n + y^n = z^n[/math] has no whole number solutions for integers n>2.
(Seems obvious, but isn't)
>>8119279
uh wasnt this proved to be true, though
>>8119279
Almost surely the "but in fact they aren't" was not refering to the "obviously" but to the "true".
That's what you get if yoy're not communicating in first order logic.
>>8119286
the negation of
>obviously true
is
>not obviously true
>>8119286
He means it's not obviously true, rather it's true following a ridiculously long proof
>>8119265
it seems obvious (and is very natural to assume) that given a set of axioms, all statements can either be proven correct or disproven.
Goedel's incompleteness theorem says that this is in fact false, for any axiomatic system powerful enough to encode integer arithmetic
>>8119265
any kind of intuition on area/volume is annihilated if you allow manipulations with non-measurable sets.
Like the Banach Tarski paradox: it seems obviously true that taken a sphere apart and reassembling it, its volume should stay constant.
>>8119265
>"simple" differential equations will result in "simple" behaviour
They won't. See chaos theory. (pic related is the Lorenz equations)
>>8119265
it seems intuitive to assume that quadratic integers (like Z[phi] for example, where phi is the golden ratio) should have the fundamental theorem of arithmetic, that is, they can be factored in primes uniquely.
This, in general, is not true
>>8119265
Continuous functions have to be differentiable somewhere.
An infinitely differentiable function is analytic.
The Borel set of R is P(R).
There can't exist a continuous curve filling the entirely filling the square [0,1]x[0,1]
Linear imply continuous.
>>8119306
>>8119308
These 2 have my interests peaked.
Please do go on.
[citation needed]
>One can design a clever algorithm which can decide whether a general Diophantine equation is solvable or not
a.k.a Hilbert's 10th problem. Hint: such an algorithm is impossible.
>>8119310
Discontinuous linear functions require the axiom of choice AFAIK, so it's Banach-Tarski-tier shit.
>You can have a continuous function that takes a 3D vector, and produces a perpendicular vector to that.
You can't. See the hairy ball theorem.
>>8119314
discontinuous w.r.t. what topology? Surely this doesnt work in Euclidean space with standard topology?
In fact, differentiation is linearization of a function, so a linear functions shouldnt just be continuous, but even differentiable (everywhere)
>>8119321
that is not the statement of the hairy ball theorem.
>>8119310
>peaked
You fucking illiterate mongoloid, please stop posting.
>>8119325
It's a corollary.
>>8119327
peeked? piqued?
Please note that for many posters, English is not their first language (like Spanish guys, Germans, Americans etc.)
>>8119310
1)https://en.wikipedia.org/wiki/Cantor_function
2)f(x)=0 on ]-inf;0]
f(x)=exp(-1/x) on ]0; +inf]
3) https://en.wikipedia.org/wiki/Borel_set#Non-Borel_sets
gonna need some latex for the other 2, give me some time.
>>8119333
>https://en.wikipedia.org/wiki/Cantor_function
oops I mean https://fr.wikipedia.org/wiki/Fonction_de_Weierstrass
my bad.
>>8119331
but the HBT is about cont. vectorfield on spheres. For example it's false on doughnuts.
I mean, isn't the cross product trivially continuous if you just fix one argument? So given any vector v, just take the function to be cross product with v
>>8119336
Okay. Write that function.
f(v) = v x ??
What do you take the cross-product with? You can't have a constant vector, because that'll result in a zero vector when they're colinear.
>>8119340
no the function is f(w)=vxw which is continuous in w and perpendicular to v at every point
>>8119342
Let me clarify: we need a function that takes a nonzero vector v, and returns a nonzero vector that's perpendicular to v. It's impossible to have such a function that's continuous.
>>8119310
4) http://math.stackexchange.com/questions/141958/why-does-the-hilbert-curve-fill-the-whole-square
5)https://en.wikipedia.org/wiki/Discontinuous_linear_map
Also I remembered one who fucked with my mind at an exam when I was still a student:
The inverse function of a continous function is not always continuous.
>>8119314
>>8119324
No, even with usual topology, on infinite dimension space, linear doesn't imply continuous.
>>8119359
>The inverse function of a continous function is not always continuous.
Wouldn't trig functions and their inverses make that obvious
>>8119482
Trigonometric functions on R are not invertible. Restrict them to a principal branch and they become invertible, with continuous inverses.
>>8119359
The inverse function not being continuous thing may not be logically intuitive, but finding an example is extremely easy. x -> 1/x.
>>8119495
What? f(x) = 1/x is not continuous at x=0.
The theorem was: there's a continuous function whose inverse is discontinuous.
>>8119482
Except you only define the inverse trig functions in a certain interval to begin with, on which they are continuous?
Where did you get that pic op
I like it
>>8119303
It's only "obvious" to a first year that it will have simple behaviour.
Past first term even you already know that it's very easy to make nasty ODEs that do not have an easy to find solution.
>>8119304
>This, in general, is not true
No one claimed that the FTA worked in other domains.
>>8119306
>Continuous functions have to be differentiable somewhere.
Said no one ever. In general differentiable doesn't even imply continuous.
>>8119359
>The inverse function of a continous function is not always continuous.
You don't need to much to know that this isn't true... homeomorphisms require a continuous inverse for a reason.
If you map a line to a circle then you can easily see that it does not have a continuous inverse.
>>8120208
Wow your autism is showing, the thread is about intuitively true statement that are false. Stop pretending you knew this statments were false the first time you saw them especially when you say something as retarded as :
>In general differentiable doesn't even imply continuous.
>>8119495
no you need to go in higher dimension than 1 to find a counter exemple, that's why you may think it's true if you only think in dim 1.
t->exp(it) t in [0;2pi[
is the simplest example, the inverse is not continuous at 1. (because the neighborhood as images as low as 0 and as high as 2pi)
>>8119321
Why not? Say v=(v1,v2,v3). Set w=(-v2,v1,0). Done.
>>8120479
After your first term class in analysis you should already know to be very careful with things in a more generalised setting. If being autistic means that I'm not as retarded as you to keep making the same intuitive mistake, then sure, I'll take it.
>In general differentiable doesn't even imply continuous.
And this is retarded why? Do you really still believe that this is true?
>>8119265
Parallel postulate
Axiom of choice
Continuum hypothesis
>>8120503
For v=(0,0,1), this fails.
>>8121143
>Parallel postulate
whut
>>8120479
I agree completely with this.
Great mathematicians of the past have fallen for these subtly "obvious" truths that turned out to be false.
For example, there is a quick way to prove Fermats last theorem if we assume that integers of number fields generally have the fundamental theorem of arithmetic. It is generally believed that this was the flawed proof Fermat found.
Math is very subtle and it's just arrogant to say all of these statements are immediately obvious.
>>8121143
>parallel postulate
is obviously true
>axiom of choice
is an axiom, so neither true nor false.
>Continuum hypothesis
hm, not sure how the word "obvious" applies to reasonings about infinities that are larger than other infinities
>>8121164
>Great mathematicians of the past have fallen for these subtly "obvious" truths that turned out to be false.
This doesn't make you great or even close to those mathematicians.
>great mathematicians took a shit, used thihng toilet paper and smelt their hands afterwards before going to work on their latest work
>I do this too so I must be worthy of spitting out something great
>>8121172
>mad
You are an unpleasant person to communicate with, and I wish you would leave.
>>8119265
>What are examples of things in mathematics that appear to be "obviously" true but in fact aren't?
The convergent series of continuous functions is continuous. Cauchy will never live that one down.
The intermediate value theorem. Seems to be true, but in fact requires an infinite amount of work to find the said point.
>>8121203
>theorem
>not true
ummm...
>>8121184
>mommy mommy he said something I didn't like
>>8121193
but that is obviously false if you know about the fourier series of the rectangle function, for example
>>8121138
Because being differentiable is a property way stronger than continuous. Differentiable imply continuous.
>>8119359
>The inverse function of a continous function is not always continuous.
yeah and this stems from the fact that inverse set functions do not commute with the conjunctions operator.
>>8121550
Did you read?
>In general
You realise that there are many different kinds of differentiability right? Not just the baby one variable one that you learn in babby's first real analysis class?
Learn to read properly.
>>8120208
>No one claimed that the FTA worked in other domains.
Not true at all. People tried to use this to prove Fermat's Last Theorem, thus prompting the work of Dedekind.
>>8120503
As I've said, it's impossible, but keep trying.
Hint: your solution fails for every (0,0,x) vector.
>>8121203
Hello Wildberger!
>>8121674
Back then yes, but nowadays I don't think so.
>>8120208
Do you even understand the thread? Are you an idiot?
>>8119295
>sphere
>nonmeasurable
>>8121729
>What are examples of things in mathematics that appear to be "obviously" true but in fact aren't?
None of those things appear "obviously" true so don't even belong in the thread.
>>8121706
I don't think you get this thread topic. Just because it is currently taught to people who understand the theory doesn't make something not intuitive. Do you know why the people who tried proving FLT long ago thought FTA would apply to things like Z[zeta_3]? It's because it appears perfectly logical at first glance. Only by digging deeper do we see that these things are not inherently true,
>>8121657
There is only one definition of diffentiability, F is differentiable at x if :
F(x+h)=F(x)+L(h)+o(h)
With L linear and continuous.
It's obvious from the definition that F is continous at x. Dimensionality has nothing to do with the definition.
>>8121755
There are examples of functions that are Gateaux differentiable but not continuous, so yes, dimensionality matters.
For example, the function [math]f: mathbb{R}^2 \rightarrow \mathbb{R} [/math] defined by:
[eqn]f(x,y) = \begin{cases} \frac{x^2y}{x^4+y^2}, & \text{if } (x,y) \neq (0,0)\\ 0, & \text{if } (x,y) = (0,0) \end{cases} [/eqn]
is Gateaux differentiable at the origin but is discontinuous there.
To think there is only one definition of differentiability is naive. The condition that you stated is something that an ideal derivative would satisfy, but not all of them do.
>>8121764
>Gateaux differentiable
Do you have any idea why we call it Gateaux Differentiable and not differentiable?
Because it's not differentiable. Differentiable umply continuous, always.
>>8121783
Because it was named after René Gâteaux and for no other reason.
The "better" derivative, the Fréchet derivative, does satisfy the "differentiable implies continuous" that we want, but no one calls it just "derivative" because there are too many different kinds of derivatives that you could be talking about.
>>8121170
>is an axiom, so neither true nor false.
I can make an axiom that everything is made of earth, wind, fire, and water, or that the earth is 6000 years old.
Those axioms are obviously false.
>>8121802
You've misunderstood axioms then. You can't state an axiom and then appeal to reality to "disprove" it.
>>8121802
>I can make an axiom that everything is made of earth, wind, fire, and water, or that the earth is 6000 years old.
Yes you could, and those that believe to choose this axiom would build on it and consider results with respect to this axiom.
Those that disagree will pick another axiom.
But as >>8121807 has said, you've misunderstood what an axiom is.
This is similar to those that say "you can make theories so why can I make a theory about god and adam and eve?"
>>8121755
>retarded highschooler detected
>>8121657
Do you understand the thread? Nobody is sitting around saying "Oh! Gateau differentiable functions need not be continuous. I bet there is a function that is nowhere differentiable in the sense of single variable calculus!"
>>8121855
Nobody is sitting around saying that differentiable implies continuous in a general setting either.
>>8121170
you realize that in their respective fields, these are each taken as axioms, and are independent of the other "naive" axioms (C and CH are independent of ZF, PP is independent of other euclidean axioms)
>>8121734
You are literally stupid.
He knows they aren't true. They seem like they would be true.
As for the continuous implies differentiable somewhere: for a long time people thought this you fucking hick. It was revolutionary when it was shown to be untrue.
>>8121896
>They seem like they would be true.
What the actual fuck. Perhaps it would seem true to the uneducated regulars of /sci/.
>>8119359
>The inverse function of a continuous function is not always continuous.
Can you give an example of an invertible [math]f: \mathbb{R} \rightarrow \mathbb{R} [/math] with this property?
>>8121913
http://math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous
>>8121919
I saw that, and not one example is a function [math]f: \mathbb{R} \rightarrow \mathbb{R} [/math].
>>8121926
Oh, I thought the first example would have been good enough for you. You might have a hard time finding something which goes from all of [math]\mathbb{R}[/math] to [math]\mathbb{R}[/math].
>>8119293
assuming arithmetic is consistent!
>>8121950
I want an example with my stated property because, when the topologies can be different, it's actually quite obvious (e.g. biject [0,1) with a circle in the obvious way).
But when both domain and range are the standard [math]\mathbb{R}[/math], that is when it goes against intuition.
It is clear that any example of a continuous bijection [math] f: \mathbb{R} \rightarrow \mathbb{R} [/math]that is not bicontinuous cannot be monotone, so it must look pretty funny. I'll see if I can either think of an example or prove there is none.
As >>8121966, I realize it's obvious that there is none.
This follows from my comment together with https://proofwiki.org/wiki/Continuous_Function_on_Closed_Interval_is_Bijective_iff_Strictly_Monotone
So the inverse of any continuous bijection [math]f: \mathbb{R} \rightarrow \mathbb{R} [/math] is also continuous.
>>8119265
The sum of all natural numbers is -1/12
>>8119279
how is that obvious?
>>8119495
>but finding an example is extremely easy. x -> 1/x.
1/x is continous everywhere. but its not the inverse function of x -> x, because x->x is defined for x = 0 as well
>>8122653
>because x->x is defined for x = 0 as well
The inverse of x -> x is itself..
>>8119332
>Americans
hearty out loud lol
That the multiplicative group of the complex unit circle is not isomorphic to the multiplicative group [math]\mathbb{C}^\ast[/math].
The two actually are isomorphic.
>>8119314
differentiation operators are linear and not continuous (you can't just swap limits with differentiation most of the time)
they don't require axiom of choice
linear + bounded implies continuous, which is a given for finite dimensional linear operators
Here are some things that are "obviously true", but which can fail in the absence of the axiom of choice:
1) Any two algebraic closures of [math]\mathbb{Q}[/math] are isomorphic.
2) Every field has an algebraic closure.
3) Every vector space has a basis.
4) If a real number is in the closure of a set [math]X \subset \mathbb{R} [/math], then it is the limit of a sequence from [math]X [/math].
5) The reals are not a countable union of countable sets.
6) If a set is infinite, it has a countably infinite subset.
7) If you have an equivalence relation on [math]\mathbb{R}[/math], then the number of equivalence classes is no larger than the cardinality of [math]\mathbb{R} [/math].
>>8124601
do you have proofs of 1, 4, 6 and 7 failing? 5 is always true: the reals are never countable
>>8119279
1^3+0^3=1^3
0^5+0^5=0^5
>>8124609
Hah! I was hoping someone would assert that 5 is always true.
No, 5 is not always true. The reason is that, in the absence of AC, a countable union of countable sets need [math]not[/math] be countable. The reals actually can be a countable union of countable sets. Therefore, in particular, the theory of Lebesgue measure can fail totally.
Proofs of the consistencies of the negations of what you asked require sophisticated forcing and inner model theory arguments. Refer to Jech's two textbooks "Set Theory" and "The Axiom of Choice", among others. If you'll accept an argument from authority, see http://mathoverflow.net/questions/20882/most-unintuitive-application-of-the-axiom-of-choice/70435#70435.
Here's one that I sort of implicitly assumed for a couple years in my visualization until I realized it to be false, even in the most trivial case:
"If [math]N[/math] is a transitive model of ZFC, [math]\kappa = \text{ORD}^N[/math], [math]M[/math], and [math]M \supset N [/math] is minimal such that [math]\kappa \in M[/math], then [math]\kappa[/math] is inaccessible in [math]M[/math]."
More informally, if one has a model [math]N[/math] whose class of ordinals is [math]\kappa[/math], then if one "keeps climbing" up the ordinal ladder past [math]\kappa[/math], adding precisely those necessary sets mandated by ZFC along the way, then when one "catches up" to a model of ZFC, [math]\kappa[/math] will be inaccessible.
_____
In fact, even if one deals strictly with models of V=L, it can be the case that [math]\kappa[/math] is forced to be not only not inaccessible, but countable. So sets one must constructively add higher up can collapse the entire model.
I happened to type up a proof here >>8124100.
>>8124601
no surprise at all.
In fact, the following seems intuitively true
>Given non-empty sets, their cartesian product will again be non-empty
but guess what..
>>8124601
>If a set is infinite, it has a countably infinite subset.
Can't we go further ? and say you can't be sure it has non trivial subsets (null and himself) without AC.
>>8124707
Well, the assertion that the cartesian product of nonempty sets is nonempty is clearly equivalent to AC; >>8124601 lists predominantly very non-trivial possibilities in the absence of AC.
>>8124900
No.
However, of note is that without AC there can exist an infinite Dedekind-finite sets. A set [math]X[/math] is "Dedekind finite" if there is no proper injection [math]f: X \rightarrow X [/math]. Intuitively, and in the presence of AC, the Dedekind-finite sets are precisely the finite sets. However, in the absence of AC, it is possible for there to exist infinite Dedekind-finite sets.
>>8124917
How do you prove an infinite set has finite subsets without AC ?
>>8124947
Your confusion is from the naive (read:non-formal) belief that containment of an element requires choosing an element. This is not the case. If [math]X[/math] is infinite, then in particular [math]X[/math] is nonempty, so there exists [math]x \in X[/math]. By the pairing axiom, [math]\{x\}[/math] is a set. By the definition of subset, [math] \{x\} \subset X [/math]. Therefore [math]X[/math] has a finite subset.
>>8124917
>Well, the assertion that the cartesian product of nonempty sets is nonempty is clearly equivalent to AC; >>8124601 lists predominantly very non-trivial possibilities in the absence of AC.
For all of those they either aren't that intuitive or the equivalence to AC is fairly obvious.
>2) Every field has an algebraic closure.
I'm not sure I believe you that this requires AC. Now if you mean to say "The algebraic closure of a fied is closed" then maybe, but you shouldn't need AC to just add all roots of all polynomials in at once.
>>8119306
weierstrass function senpai
>all gay ass analysis
Every topological manifold can be given a PL structure
Every PL manifold can be given a smooth structure
Any differential structure on R^n is diffeomorphic to the standard one
Any diagram of the unknot can be unknotted by Reidemeister moves without increasing the number of crossings
Poincare conjecture in dimensions 7 and higher
And many things that are unknown and very difficult to prove either way despite the fact that they are to some extent "obvious":
-There exist one way functions
-P != NP
-smooth poincare conjecture in dimension 4
-knot crossing number is additive
>>8125080
..an algebraic closure is closed in both the obvious field sense (closure under field operations) and in the algebraic sense that all polynomials have roots. So, regardless of what you meant, an algebraic closure is closed.
As for where AC comes in, the process of constructing an algebraic closure requires iteratively choosing an irreducible polynomial and quotienting the polynomial ring by the maximal ideal it generates. Key word choosing.
More formally, as you can see in https://en.wikipedia.org/wiki/Algebraic_closure#Existence_of_an_algebraic_closure_and_splitting_fields, Zorn's lemma (which is equivalent to AC) is required.
>>8125072
>a core class in multivariable differentiation is autism
>>8125131
I see, you're forced to do it one at a time if you aren't already contained in some larger field so you end up using Zorns lemma.
But if you are already contained in a larger (closed) field though it is easy, since then you can just take the set of all elements that are roots of some polynomial with coefficients in the field you are taking the closure of.
>>8125145
Failing to understand what people are talking about and using it to try to show off what you know about some boring ass analysis shit is autism.
>>8125152
>Failing to understand what people are talking about
maybe autism
>using it to try to show off what you know about some boring ass analysis shit
>analysis is shit
the fuck
>>8125152
>Banach spaces, the IFT, manifolds and optimizaton are boring
woah
Jordan curve theorem is very intuitive but pretty difficult to prove.
>>8119265
More numbers between 1 and 1000000 than there are between 1 and 2
>>8124601
>3) Every vector space has a basis.
Weird for me it always seemed obviously false even with AC, unless you can show me a basis of continuous function from R to R, or a basis of R viewed as Q vector space I will never believe it.
>>8126430
>a basis of R viewed as Q vector space
Huh? Just use a Hamel basis or a Schauder basis.
>>8126519
And what do you think guarantees the existens of a hamel basis?
a schauder basis is not really a basis in the sense that the guy you are replying to would like to see.
>>8126595
The AC of course - he said the statement seems obviously false even with the AC, and I was giving an example (inexplicitly) of a basis that he wanted.
What basis does he want to see? What are the limitations on a basis that would satisfy this guy?
>>8124601
also, this is just a trivial restatement of AC, but without it, a surjection need not have a section, which is unsettling.
>>8126634
oh shit, I didn't read the >even with AC part
nevermind then
