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zmudz
 one year ago
Let \(f:\mathbb R \to \mathbb R\) be a function such that for any irrational number \(r\), and any real number \(x\) we have \(f(x)=f(x+r)\). Show that \(f\) is a constant function.
zmudz
 one year ago
Let \(f:\mathbb R \to \mathbb R\) be a function such that for any irrational number \(r\), and any real number \(x\) we have \(f(x)=f(x+r)\). Show that \(f\) is a constant function.

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Empty
 one year ago
Best ResponseYou've already chosen the best response.1This seems to be the case just in general for any function that satisfies \[f(a)=f(a+b)\] then f(x) would have to be constant.

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1but what about proving for real numbers?

Empty
 one year ago
Best ResponseYou've already chosen the best response.1@jtvatsim sounded like he knew more about this than I did, I was just making a passing comment. I don't really study this sort of math so I have no idea about density of numbers or whatever.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1@empty "sounded" is the key word. :) I'm trying to make this more rigorous. lol

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1I think I got it, thanks everyone

zmudz
 one year ago
Best ResponseYou've already chosen the best response.1the question really only asks me to prove that whatever I put into the function, I will get out \(r\), and so I can do that and spin off something similar to what empty said.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4It is sufficient to show that \[f(x+\epsilon)  f(x) = 0\] By the given definition the left hand side is same as \[f(x+r+\epsilon)  f(x) \] for any irrational \(r\). let \(r,\epsilon\) be positive for simplicity since anything added to iraational gives an irrational, the first term evaluates to \(f(x)\)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Convincing to me... QED.

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1But wait... This depends on the fact that irrational + anything is also irrational... is this necessarily true?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4yeah below is bugging me a bit \[\sqrt{2}+ (\sqrt{2}) \]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Indeed... that is troubling. :/

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4thats not really a problem because \[f(x+\sqrt{2}+(\sqrt{2})) = f(x+0) = f(x)\] :P

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1But we can make it a problem easily... :P lol

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4the problem comes only when \(\epsilon\) and \(r\) are additive inverses

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Consider \[2.dec(\sqrt{2})  0.dec(\sqrt{2}) = 2\] where \[dec(\sqrt{2})\] uses the digits from \[\sqrt{2}\].

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4true! i take my statement back :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I think we can get around that issue by simply fixing the irrational, lets say, \(r=\sqrt{2}\) we won't be losing anything.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Let \(\epsilon\) be some positive real number. It is sufficient to show that \[f(x+\epsilon)  f(x) = 0\] By the given definition the left hand side is same as \[f(x+\color{red}{\sqrt{2}}+\epsilon)  f(x) \] since \(\sqrt{2}+\epsilon\) is irrational for all positive \(\epsilon\), the first term evaluates to \(f(x)\)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I'm going to have to keep thinking through this one. Your new approach is clever, but I feel there is a way to force sqrt(2) + irrational to still be rational. It all depends on if the sequence of digits (10d) where d is the decimal digit of sqrt(2) is repeating or nonrepeating... however, I don't think I have the firepower at this point to deal with that. Too much fun! :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Ahh you want to make \[\epsilon = \text{10'scompliment}(\sqrt{2}1)\] very interesting xD

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Basically, it's hard to formalize.... However, I have this suspicion that even this would work in our favor since x can be any real number, we might be able to pair up our x's and epsilons in order to cover all real number possibilities.... I may be overthinking this xP

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Are we assuming here that f is continuous? That would be helpful I imagine to simplify all this crazy infinitesimal stuff...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4I thought earlier you're saying something like below \[\begin{align} \sqrt{2}+\epsilon &= \sqrt{2} + \text{9'scompliment}(\sqrt{2}1)\\~\\ &=1.413213\ldots + 0.586786\ldots \\~\\ &=1.999999\ldots\\~\\ &=2 \end{align}\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Yes, precisely. Sorry, my train of thought wasn't necessarily connected. :)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I lack that topological property in general...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4lol this is indeed very interesting, never thought of this trick before this observation essentially implies that sum of two positive irrational numbers need not be irrational ?

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Indeed, which is what I was afraid of. It is a very disturbing thought. :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4nvm i wasn't even thinking, see : \[(5\sqrt{2})+( 5+\sqrt{2}) = 10\]

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Oh duh... I can think complicated, just can't come up with simple counter examples. :P lol

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1There is an interesting proof out there that if x and y are irrational, then precisely one of x + y or x  y is also irrational.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that invalidates the proof, so we do need to consider two cases

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1We definitely know that the function evaluated at all the irrationals is the same. Let x = 0. Then f(0) = f(r) for all irrational r.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4that proves the function is constant for all irrational numbers, r

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Let \(\epsilon\) be some positive real number. It is sufficient to show that \[f(x+\epsilon)  f(x) = 0\] By the given definition the left hand side is same as \[f(x+\color{red}{\sqrt{2}}+\epsilon)  f(x) \] If \(\epsilon\) is rational, then \(\sqrt{2}+\epsilon\) is irrational, so the first term evaluates to \(f(x)\) If \(\epsilon\) is irrational, then \(f(x+\color{red}{\sqrt{2}}+\epsilon) = f(x+\color{red}{\sqrt{2}}) = f(x)\) . \(\blacksquare\)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Ooh ooh! How about this! Let x = p  r where p is rational. Now we DO KNOW that rational + irrational is irrational. Thus, x is an irrational. Then, f(x) = f(p  r) = f(p) for all rationals p. But all irrationals are equal to f(0). Hence, f(r) = f(p) = f(0) as desired.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4That looks really neat !

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Finally, we have something that works.... that was a real pain. Don't know why it was so elusive...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Haha idk from where i got the idea that irrational+irrational = irrational

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1Well, I thought it was true at first too! I guess it seems so obvious that "really weird number + really weird number = really weird number"... but, not so, math thwarts our intuition once again! :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Cool sounds like you guys are ready to solve one of these open questions: https://en.wikipedia.org/wiki/Irrational_number#Open_questions

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1I'm not going to even look... I will not be able to sleep tonight if I do.... :P But, I'll check it out at some point tomorrow. :D

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4there is a reason for calling them "open" lmao

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1\(\sqrt{2}+(2\sqrt{2})\in \mathbb{Q}\)

jtvatsim
 one year ago
Best ResponseYou've already chosen the best response.1seriously... props to everyone who stuck around! :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Hey if there's anything I learned from set theory it's that something can be both open and closed at the same time @ganeshie8 :P

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1Actually it has nothing to do with set theory. Topology my man :)

Empty
 one year ago
Best ResponseYou've already chosen the best response.1What weird kind of definition are you using for topological space?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.1The most general one... Set theory, in general, has nothing to do with open/closed. Topology by definition is the analytic notion of open and closed sets. Show me one theorem in set theory about open closed. Then google "topology".
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