Dividing Two Infinities I am Curious if the following is mathematically correct:
Let $a$ be the infinite set of all nonnegative integers $0,1,2,3...$.
Let $b$ be the infinite set of all nonnegative EVEN integers $0,2,4,6...$.
If I take the sum of $a$ and divide it by the sum of $b$ do I get 2?
Is this correct? Can one divide infinities like this?
If yes, does this mean that SUM $a$ > SUM $b$ (despite that both are infinite)?
(if this is not possible can one do some kind of equivalent division using mapping of sets?)
 A: It is mathematically incorrect, but the idea is intuitively good in certain contexts. The right way to say things is that it all depends on how you take the limits. Since both series sum to infinity, you need to make your result "$2$" precise. For instance,
$$
\lim_{n \to \infty} \frac{\sum_{i=0}^{n} 2i }{\sum_{i=0}^n i}
$$
is probably what you had in mind when you thought of the result being "two". Of course, this limit is worth two, but we cannot say 
$$
\lim_{n \to \infty} \frac{\sum_{i=0}^{n} 2i }{\sum_{i=0}^n i} = \frac{\lim_{n \to \infty} \sum_{i=0}^n 2i}{\lim_{n \to \infty} \sum_{i=0}^n i}
$$
because the two "things" on the numerator and denominator are not numbers, so their ratio is not a priori defined. To understand what I mean better, notice that if we tweak things a little bit :P , it is not hard to show that 
$$
\lim_{n \to \infty} \frac{\sum_{i=0}^{n} 2i }{\sum_{i=0}^{n^2} i}= 0, \qquad \lim_{n \to \infty} \frac{\sum_{i=0}^{n^2} 2i }{\sum_{i=0}^{n} i}= \infty.
$$
Hope that helps,
A: As pointed out in the comments, your series are divergent, so their sums are not defined.
However, if you have a convergent infinite series, say $b_1 + b_2 + b_3 \cdots$ converges to $S$, and $c$ is a constant, then $cb_1 + cb_2 + cb_3 + \cdots$ converges to $cS$.
Moreover, given a second series $a_1 + a_2 +a_3 \cdots$, if $0 \le a_n \le b_n$ for all $n$, then the second series $a_1 + a_2 + a_3 + \cdots$ converges to a value less than or equal to $S$.  This is known as the comparison test for infinite series.
A: You cannot divide the sum of the elements of $a$ by the sum of the elements of $b$. The reason is that what we expect out of our "normal" rules of addition, subtraction, multiplication and division become very tricky when infinity becomes involved. 
This is seen very commonly in a number of misleading but seemingly correct manipulations. Consider, for instance, the infinite sum $$S = 1 + 2+ 4 + 8 + 16 + \cdots$$
Now multiply $S$ by 2 to find $$2S = 2 + 4 + 8 + 16 + \cdots = S - 1$$
This would imply that $S = -1$, which is an absurd conclusion indeed! 
Math is rife with examples like this where infinities are tricky to manage and must be handled with exceptional care. For more examples like these, the excellent notes of Dr. Tao are quite informative:
http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/week1.pdf
A: I assume you don't know very much about number theory so I am well suited to answer you, also being an amateur.
You cannot sum up infinitely many natural numbers. You can take the sum of the first n numbers, S1(n) and divide one with the other, S2(n) and then divide to get S1(n)/S2(n) and the find a limit as n goes to eternity, which means it doesn't change very much as n goes bigger and bigger. In some cases you can do this, but the series to be summed have to consist of smaller and smaller terms, and only some of those will work. 1+1/2+1/4 +1/8 can be summed up but 1+1/2+1/3+1/4... cannot.
