Are half of all natural numbers even?

[textbooks]

It might be a silly question but I was just wondering if this was unprovable or false...

Thanks guys

 

[Office_Shredder]

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)

 

[g_edgar]

There are some concepts of "density" of a set of natural numbers. The set of even numbers has density 1/2.
http://planetmath.org/encyclopedia/AsymptoticDensity.html"
http://mathworld.wolfram.com/SchnirelmannDensity.html"

 

[HallsofIvy]

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.

 

[CRGreathouse]

There are some concepts of "density" of a set of natural numbers. The set of even numbers has density 1/2.
http://planetmath.org/encyclopedia/AsymptoticDensity.html"
http://mathworld.wolfram.com/SchnirelmannDensity.html"

The asymptotic density of the even numbers is indeed 1/2. But the Schnirelmann density is 0.

 

[g_edgar]

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

 

[CRGreathouse]

...and the Schnirelmann density of the odd numbers is 1/2

Right.

 

[HallsofIvy]

So the Schnirelmann density of the even numbers is 0 and the Schnirelmann density of the odd numbers is 1/2? Peculiar!

 

[CRGreathouse]

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.

 

[camilus]

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)

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.

Consider a hypothetical hotel with infinitely many rooms, all of which are occupied – that is to say every room contains a guest. Suppose a new guest arrives and wishes to be accommodated in the hotel. If the hotel had only finitely many rooms, then it can be clearly seen that the request could not be fulfilled, but because the hotel has infinitely many rooms then if you move the guest occupying room 1 to room 2, the guest occupying room 2 to room 3 and so on, you can fit the newcomer into room 1. By extension it is possible to make room for a countably infinite number of new clients: just move the person occupying room 1 to room 2, the guest occupying room 2 to room 4, and in general room N to room 2*N, and all the odd-numbered rooms will be free for the new guests.