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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I found this question a while ago now, but haven't made much progress with it.

(Or prove that such a function doesn't exist)

Can anyone shed light on this?

Edit: Just remembered, the question doesn't actually require you to find an example of such a function. Proving its existence would be enough.

Why did the vector cross the road?

It wanted to be normal.

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

Do we also need to assume the opposite, i.e. that

If not the question is trivial: just let f(x) = c where c is any irrational number.

*Last edited by TheDude (2009-01-20 07:57:12)*

Wrap it in bacon

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

What do you think means?

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

Did you read my post carefully?

Wrap it in bacon

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Look.

I wrote:

What do you think means?

Now stop asking stupid questions!

*Last edited by JaneFairfax (2009-01-20 08:12:52)*

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

Your statement comes out of my one by using contrapositions.

so

Replacing those right-arrows with left or double arrows is allowed, so by using double arrows we have that my condition is equivalent to yours.

Unfortunately, that means your trivial solution won't work.

Why did the vector cross the road?

It wanted to be normal.

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

I figured it couldn't be that easy.

Wrap it in bacon

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**TheDude****Member**- Registered: 2007-10-23
- Posts: 361

Perhaps my first thought will be more helpful. The irrationals are uncountably infinite, and f(x) will have to map them only to the rationals, which are countably infinite. Someone more educated on the subject will have to either verify or refute this, but it seems to me that such a mapping must include at least one member of the set of rationals which has an uncountably infinite number of irrational numbers mapped to it. Let's call this number m.

If that is incorrect then ignore the rest of this post. But if it is true then I don't think that any such function could be continuous. Further guesswork leads me to believe that there would be at least one irrational x where f(x) = m that would not have a limit because f(x) would oscillate infinitely many times between m and various irrational numbers no matter how close you approached x. Basically it would behave like sin(pi/x) near x = 0.

I'm sorry if my explanation isn't clear, if something I wrote doesn't make sense I can try to explain it better. Of course, I'm probably wrong anyway, but maybe it can lead you in the right direction.

*Last edited by TheDude (2009-01-20 08:40:07)*

Wrap it in bacon

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**Muggleton****Member**- Registered: 2009-01-15
- Posts: 65

Such a function may not exist, but if it does, then most likely it won't be expressible in closed form. Indeed I think such a function (if it exists) will not be differentiable (like the function discovered by Weierstrass).

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

The Weierstraß function is a pathological case (as described by the Wikipedia article on it). I highly doubt that the function were looking for (if it exists) is going to be anything as pathological otherwise people would have known about it much earlier and wouldnt have had to wait until the late 19[sup]th[/sup] century for an everywhere-continuous nowhere-differentiable function to be discovered.

So either it doesnt exist, or, if it does, it wont be totally undifferentiable. However, I agree that most likely it wont be expressible as a nice and neat formula (if it exists). Thats my theory.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Answer: http://planetmath.org/encyclopedia/ThereIsNoContinuousFunctionThatSwitchesTheRationalNumbersWithTheIrrationalNumbers.html

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

<That link's too long for my browser to show it all, so, Link>

Anyway, thanks Ricky! I'm not sure I'll understand all of that proof, but it's good that we have an answer.

Why did the vector cross the road?

It wanted to be normal.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

I love the proof. Instead of using analysis methods (as you would immediately think of doing at first) it uses topology methods. And I <3 topology more than analyss.

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