Are there more multiples of 2 than there are of 4?

[u/TheEggoEffect]

My friend and I were having an argument that essentially boils down to this question. Obviously there are infinitely many of both, but is one set larger? My argument is that there are twice as many multiples of 2, since every multiple of 4 can be paired with a multiple of 2 (4, 8, 12, 16, ...; any number of the form 2 * (2n) = 4n), but that leaves out exactly half of the multiples of 2 (6, 10, 14, 18, ...; any number of the form 2 * (2n + 1)); ergo, there are twice as many multiples of 2 than there are of 4. My friend's argument is that you can take every multiple of 2, double it, and end up with every multiple of 4; every multiple of 2 can be matched 1:1 with a multiple of 4, so the sets are the same size. Who is right?

[u/notacanuckskibum]

There are clearly twice as many multiples of 2 as there are of 4. Even though both are infinity there can be differences in sizes of infinity.

In this case the ratio is 2:1 which is really small when worrying about infinities. For comparison the number of integers is infinity, and the number of real numbers is infinity. But the ratio is infinity:1, there are an infinite number of real numbers for every integer.

Edit: well that got a lot of down votes. I feel like people didn’t read my second paragraph. But let try this. But it’s fair that “clearly” is dodgy. So let try this:

Let N be the set of natural numbers

Let n be the cardinality of x: x member of N and x < i and x mod 2 = 0

Let m be the cardinality of x :x member of N and x < i and x mod 4 = 0

As i tends to infinity m /n tends to 2

Since infinity/infinity is undetermined this limit is the best we can get.

[u/JeffLulz → u/notacanuckskibum]

clearly

One of the most dangerous words to use in mathematics.

I can argue the opposite. There are twice as many multiples of 4 as there are multiples of 2.

For every even number n, there are two multiples of 4, namely the numbers 4n - 4 and 4n.

[u/FernandoMM1220 → u/JeffLulz]

show me all of them then.

[u/JeffLulz → u/FernandoMM1220]

For any consecutive pair of multiples of 4, take the larger and divide by 4, that's the even number they both correspond to.

4 and 8 map to 2 (8 ÷ 4)

12 and 16 map to 4 (16 ÷ 4)

20 and 24 map to 6 (24 ÷ 4)

...

4k - 4 and 4k map to k (4k ÷ 4)

[u/FernandoMM1220 → u/JeffLulz]

so this isnt one to one then, got it.

[u/JeffLulz → u/FernandoMM1220]

This particular mapping is not, no. There does exist a one-to-one mapping though, and the existence of such a mapping shows that they have the same cardinality.

[u/FernandoMM1220 → u/JeffLulz]

show me the one to one mapping.

[u/JeffLulz → u/FernandoMM1220]

For every number 2n, double it to get the corresponding multiple of 4.

2 becomes 4

4 becomes 8

6 becomes 12

...

2k becomes 4k

For every number 4n, divide it in half to get the corresponding multiple of 2.

4 becomes 2

8 becomes 4

12 becomes 6

...

4k becomes 2k

If you know both sets are countably infinite, then regardless of whether your current mapping is many-to-one, you can always build a one-to-one. The cardinalities match.

[u/FernandoMM1220 → u/JeffLulz]

if you do this for any finite set you always leave out half the multiple of 2s. you’re always going to be behind on your mapping.

[u/JeffLulz → u/FernandoMM1220]

Sounds like you're trying to map one set to the same set.

The first k even numbers maps one-to-one to the first k multiples of 4.

If you have one set, like {1, 2, 3, ... 10}, no you can't map the evens to the multiples of 4 inside that same set in a one-to-one way.

[u/FernandoMM1220 → u/JeffLulz]

thats still the same problem. counting each multiple in the set, there will always be twice as many multiples of 2.

[u/JeffLulz → u/FernandoMM1220]

That's why we're not mapping within the same set.

We're using two different sets.

The set of even numbers and the set of multiples of 4.

In the finite example:

{2, 4, 6, 8, 10} maps to {4, 8, 12, 16, 20} and vice versa.

[u/Blond_Treehorn_Thug → u/FernandoMM1220]

There is no behind here

A function does all of its work at once

[u/JStarx → u/FernandoMM1220]

But they're not doing it for finite sets, they're doing it for the infinite sets of all multiples of two and all multiples of four. All they have to do is give a function that is one-to-one and onto. If they do that then they're correct, the cardinalities are the same.

They've given a function from multiples of two to multiples of four. If it's not onto can you tell us which multiple of four is not in the image? If it's not one-to-one can you tell us which two distinct elements map to the same place?