It might be a silly question but I was just wondering if this was unprovable or false...
It's a poorly worded question. How many natural numbers are there, and how do you divide that by half?
As a start up to this kind of stuff, consider the function
f(n) = 2n
You now have a bijection, or a 1-1 and onto function, from the set of natural numbers to the set of even numbers. So for every even number you give me, I can give you a natural number, and for every natural number, I can give you an even number. Hence there must be the same number of even numbers as there are natural numbers.
This is more related to set theory than number theory (counting the size of different sets)
There are some concepts of "density" of a set of natural numbers. The set of even numbers has density 1/2.
It's not silly but it depends on exactly what you mean by "half" of an infinite set. As g. edgar points out, there are several ways of defining that.
The asymptotic density of the even numbers is indeed 1/2. But the Schnirelmann density is 0.
...and the Schnirelmann density of the odd numbers is 1/2
So the Schnirelmann density of the even numbers is 0 and the Schnirelmann density of the odd numbers is 1/2? Peculiar!
The Schnirelmann density relates to sumsets, and sets missing enough small numbers won't be able to sumset to given values even if iterated many times. In particular if the set lacks 1, it won't be able to sum to 1. Since the even don't have one, their density is 0.
thats basically the answer right there.
Questions like these are very tricky because you're operating with infinity. How Shredder said, the size of the set of even numbers is the same as the size of the set of natural numbers.
A famous paradox by David Hilbert shows why doing arithmetic with infinity is so tricky, and the last part about the odd rooms of his Grand Hotel shares some similarities with the quetsion on this thread.
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