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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

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

but what about proving for real numbers?

3. Empty

@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

@empty "sounded" is the key word. :) I'm trying to make this more rigorous. lol

5. zmudz

I think I got it, thanks everyone

6. jtvatsim

What was the key?

7. zmudz

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

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

Convincing to me... QED.

10. jtvatsim

But wait... This depends on the fact that irrational + anything is also irrational... is this necessarily true?

11. ganeshie8

yeah below is bugging me a bit $\sqrt{2}+ (-\sqrt{2})$

12. jtvatsim

Indeed... that is troubling. :/

13. ganeshie8

thats not really a problem because $f(x+\sqrt{2}+(-\sqrt{2})) = f(x+0) = f(x)$ :P

14. jtvatsim

But we can make it a problem easily... :P lol

15. ganeshie8

the problem comes only when $$\epsilon$$ and $$r$$ are additive inverses

16. jtvatsim

Consider $2.dec(\sqrt{2}) - 0.dec(\sqrt{2}) = 2$ where $dec(\sqrt{2})$ uses the digits from $\sqrt{2}$.

17. ganeshie8

true! i take my statement back :)

18. ganeshie8

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

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

nice work

21. jtvatsim

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

Ahh you want to make $\epsilon = \text{10'scompliment}(\sqrt{2}-1)$ very interesting xD

23. jtvatsim

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

Are we assuming here that f is continuous? That would be helpful I imagine to simplify all this crazy infinitesimal stuff...

25. ganeshie8

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

Yes, precisely. Sorry, my train of thought wasn't necessarily connected. :)

27. jtvatsim

I lack that topological property in general...

28. ganeshie8

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

Indeed, which is what I was afraid of. It is a very disturbing thought. :)

30. ganeshie8

nvm i wasn't even thinking, see : $(5-\sqrt{2})+( 5+\sqrt{2}) = 10$

31. jtvatsim

Oh duh... I can think complicated, just can't come up with simple counter examples. :P lol

32. jtvatsim

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

that invalidates the proof, so we do need to consider two cases

34. jtvatsim

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

that proves the function is constant for all irrational numbers, r

36. ganeshie8

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

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

That looks really neat !

39. jtvatsim

Finally, we have something that works.... that was a real pain. Don't know why it was so elusive...

40. ganeshie8

Haha idk from where i got the idea that irrational+irrational = irrational

41. jtvatsim

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

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

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

there is a reason for calling them "open" lmao

45. zzr0ck3r

$$\sqrt{2}+(2-\sqrt{2})\in \mathbb{Q}$$

46. jtvatsim

seriously... props to everyone who stuck around! :)

47. Empty

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

Actually it has nothing to do with set theory. Topology my man :)

49. Empty

What weird kind of definition are you using for topological space?

50. zzr0ck3r

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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