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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.

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  1. Empty
    • one year ago
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    This 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.

  2. zmudz
    • one year ago
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    but what about proving for real numbers?

  3. Empty
    • one year ago
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    @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.

  4. jtvatsim
    • one year ago
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    @empty "sounded" is the key word. :) I'm trying to make this more rigorous. lol

  5. zmudz
    • one year ago
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    I think I got it, thanks everyone

  6. jtvatsim
    • one year ago
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    What was the key?

  7. zmudz
    • one year ago
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    the 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.

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

  9. jtvatsim
    • one year ago
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    Convincing to me... QED.

  10. jtvatsim
    • one year ago
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    But wait... This depends on the fact that irrational + anything is also irrational... is this necessarily true?

  11. ganeshie8
    • one year ago
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    yeah below is bugging me a bit \[\sqrt{2}+ (-\sqrt{2}) \]

  12. jtvatsim
    • one year ago
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    Indeed... that is troubling. :/

  13. ganeshie8
    • one year ago
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    thats not really a problem because \[f(x+\sqrt{2}+(-\sqrt{2})) = f(x+0) = f(x)\] :P

  14. jtvatsim
    • one year ago
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    But we can make it a problem easily... :P lol

  15. ganeshie8
    • one year ago
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    the problem comes only when \(\epsilon\) and \(r\) are additive inverses

  16. jtvatsim
    • one year ago
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    Consider \[2.dec(\sqrt{2}) - 0.dec(\sqrt{2}) = 2\] where \[dec(\sqrt{2})\] uses the digits from \[\sqrt{2}\].

  17. ganeshie8
    • one year ago
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    true! i take my statement back :)

  18. ganeshie8
    • one year ago
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    I think we can get around that issue by simply fixing the irrational, lets say, \(r=\sqrt{2}\) we won't be losing anything.

  19. ganeshie8
    • one year ago
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    Let \(\epsilon\) 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 \(f(x)\)

  20. zzr0ck3r
    • one year ago
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    nice work

  21. jtvatsim
    • one year ago
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    I'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 (10-d) 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! :)

  22. ganeshie8
    • one year ago
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    Ahh you want to make \[\epsilon = \text{10'scompliment}(\sqrt{2}-1)\] very interesting xD

  23. jtvatsim
    • one year ago
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    Basically, 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

  24. jtvatsim
    • one year ago
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    Are we assuming here that f is continuous? That would be helpful I imagine to simplify all this crazy infinitesimal stuff...

  25. ganeshie8
    • one year ago
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    I thought earlier you're saying something like below \[\begin{align} \sqrt{2}+\epsilon &= \sqrt{2} + \text{9'scompliment}(\sqrt{2}-1)\\~\\ &=1.413213\ldots + 0.586786\ldots \\~\\ &=1.999999\ldots\\~\\ &=2 \end{align}\]

  26. jtvatsim
    • one year ago
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    Yes, precisely. Sorry, my train of thought wasn't necessarily connected. :)

  27. jtvatsim
    • one year ago
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    I lack that topological property in general...

  28. ganeshie8
    • one year ago
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    lol 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 ?

  29. jtvatsim
    • one year ago
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    Indeed, which is what I was afraid of. It is a very disturbing thought. :)

  30. ganeshie8
    • one year ago
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    nvm i wasn't even thinking, see : \[(5-\sqrt{2})+( 5+\sqrt{2}) = 10\]

  31. jtvatsim
    • one year ago
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    Oh duh... I can think complicated, just can't come up with simple counter examples. :P lol

  32. jtvatsim
    • one year ago
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    There 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.

  33. ganeshie8
    • one year ago
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    that invalidates the proof, so we do need to consider two cases

  34. jtvatsim
    • one year ago
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    We 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.

  35. ganeshie8
    • one year ago
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    that proves the function is constant for all irrational numbers, r

  36. ganeshie8
    • one year ago
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    Let \(\epsilon\) 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 \(f(x+\color{red}{\sqrt{2}}+\epsilon) = f(x+\color{red}{\sqrt{2}}) = f(x)\) . \(\blacksquare\)

  37. jtvatsim
    • one year ago
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    Ooh 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.

  38. ganeshie8
    • one year ago
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    That looks really neat !

  39. jtvatsim
    • one year ago
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    Finally, we have something that works.... that was a real pain. Don't know why it was so elusive...

  40. ganeshie8
    • one year ago
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    Haha idk from where i got the idea that irrational+irrational = irrational

  41. jtvatsim
    • one year ago
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    Well, 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! :)

  42. Empty
    • one year ago
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    Cool sounds like you guys are ready to solve one of these open questions: https://en.wikipedia.org/wiki/Irrational_number#Open_questions

  43. jtvatsim
    • one year ago
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    I'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

  44. ganeshie8
    • one year ago
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    there is a reason for calling them "open" lmao

  45. zzr0ck3r
    • one year ago
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    \(\sqrt{2}+(2-\sqrt{2})\in \mathbb{Q}\)

  46. jtvatsim
    • one year ago
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    seriously... props to everyone who stuck around! :)

  47. Empty
    • one year ago
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    Hey 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

  48. zzr0ck3r
    • one year ago
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    Actually it has nothing to do with set theory. Topology my man :)

  49. Empty
    • one year ago
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    What weird kind of definition are you using for topological space?

  50. zzr0ck3r
    • one year ago
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    The 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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