Math Is Fun Forum

A true limit exists only if the limits from both the left and the right sides exist and are equal.  As you pointed out the limits from the left and right sides in your example are not equal, which means that the limit as a whole does not exist.

I don't follow this part.  Why can't the discriminant be positive?

That's not what it comes out to though.  Using that substitution gives

http://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational

So are you saying that pi's digits do not go on infinitely, or that we haven't proven it?

I'm not sure what you're saying.  Are you saying that we can express as many digits of pi in binary as we can in decimal?  If so then you're right.  Although it will take many more digits we can express pi to the same accuracy in binary that we can in decimal.

However, countability is a whole different subject that deals with the size of infinite sets like the naturals, the rationals, and the reals.

You didn't do the induction step correctly.  It should be:

2 + 4 + 6 + 8 + ... + 2k + (2k + 2) > (k + 1)^2 = k^2 + 2k + 1

You already know that 2 + 4 + 6 + ... + 2k > k^2, so subtract both sides from the inequality.  The proof is trivial from there.

I could have said that, except that I didn't know about it .

Anyway, U is an odd one because it is typically drawn with a curved line.  Y is odd because it is a consonant.  A is odd because it is the only one that forms a word by itself (in English, anyway).  E is odd because it is (as far as I know) the only letter that has been explicitly avoided in a published novel.

Hydrogen is the only element in the list with a stable isotope that contains no neutrons.

Sorry, it's been too long since chemistry class for me to come up with anything for the other 3.

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.