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

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- 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.

- zmudz

but what about proving for real numbers?

- 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.

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## More answers

- jtvatsim

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

- zmudz

I think I got it, thanks everyone

- jtvatsim

What was the key?

- 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.

- 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 \(f(x)\)

- jtvatsim

Convincing to me... QED.

- jtvatsim

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

- ganeshie8

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

- jtvatsim

Indeed... that is troubling. :/

- ganeshie8

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

- jtvatsim

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

- ganeshie8

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

- jtvatsim

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

- ganeshie8

true! i take my statement back :)

- 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.

- 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 \(f(x)\)

- zzr0ck3r

nice work

- 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! :)

- ganeshie8

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

- 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

- jtvatsim

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

- ganeshie8

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

- jtvatsim

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

- jtvatsim

I lack that topological property in general...

- 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 ?

- jtvatsim

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

- ganeshie8

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

- jtvatsim

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

- 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.

- ganeshie8

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

- 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.

- ganeshie8

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

- 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 \(f(x+\color{red}{\sqrt{2}}+\epsilon) = f(x+\color{red}{\sqrt{2}}) = f(x)\) . \(\blacksquare\)

- 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.

- ganeshie8

That looks really neat !

- jtvatsim

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

- ganeshie8

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

- 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! :)

- Empty

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

- 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

- ganeshie8

there is a reason for calling them "open" lmao

- zzr0ck3r

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

- jtvatsim

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

- 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

- zzr0ck3r

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

- Empty

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

- 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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