# B Natural Numbers and Odd Numbers

1. Jun 11, 2017

### davidge

Given any finite set of natural numbers, it seems evident that the odd numbers form a subset of the natural numbers. But what happens "at infinity"? I mean, if we account for all infinitely many natural numbers, there would be also infinitely many odd numbers. In such case, is it still true that the odd ones form a subset of the natural ones?

<Title edited along with this note about the title. fresh_42>

Last edited by a moderator: Jun 11, 2017
2. Jun 11, 2017

### PeroK

What's the definition of subset?

3. Jun 11, 2017

### davidge

Given two sets $A$ and $B$, $A$ is said to be a subset of $B$ if $A \cap B = A$ and $A \cap B \neq B$.

4. Jun 11, 2017

### PeroK

I wasn't expecting that!

I might prefer the following:

$A$ is a subset of $B$ if $x \in A \ \Rightarrow \ x \in B$

5. Jun 11, 2017

### davidge

So, since by definition any positive odd number is also a natural number, we conclude that the odd positive numbers form a subset of the natural numbers even when we have infinitely many numbers?

6. Jun 11, 2017

### Staff: Mentor

7. Jun 11, 2017

### PeroK

Yes.

8. Jun 11, 2017

### davidge

9. Jun 11, 2017

### Staff: Mentor

Usually a subset can be the full set: $A \subseteq A$. In this case the second condition is not satisfied.

If the second condition is satisfied, it is a proper subset.

Edit: Better symbol.

Last edited: Jun 11, 2017
10. Jun 11, 2017

### davidge

This is a bit nonsense. But definitions are definitions.

11. Jun 11, 2017

### Staff: Mentor

The only nonsense is that he should have written $A \subseteq A$ instead of $A \subset A$, but the rest and the main idea is correct. $A \subseteq A$ is a subset. $A \subsetneq B$ is a proper subset. The additional condition $A \cap B \neq B$ is unusual as long as one doesn't define a proper subset. To exclude equality makes the entire topic only unnecessarily complicated, IMO.

12. Jun 11, 2017

### Staff: Mentor

Many mathematical statements would have to use "a subset of A or A itself" everywhere if you exclude the full set as subset.

It is like the convention that 1 is not a prime number. Otherwise you have "for every prime apart from 1" everywhere.

13. Jun 11, 2017

### PeroK

I've always liked this convention as I find something very satisfying about:

A = B iff A is a subset of B and B is a subset of A.

Having to say "subset or equal to" would spoil that.

14. Jun 11, 2017

### WWGD

I don't know if just a convention; if 1 were prime, every number would be composite as n=n(1).

15. Jun 11, 2017

### Staff: Mentor

If a unit was prime the entire concept would be meaningless. We have this (IMO senseless) discussion only because they learn at school "if only divisible by $1$ and itself". If they learnt it correctly, this wouldn't be necessary.

16. Jun 11, 2017

### WWGD

Yes, but I ( think I ) get mfb's point that , by strict definition, 1 is a(n) ( integer) prime, since it is divisible only by itself...and by 1.

17. Jun 11, 2017

### Staff: Mentor

But $1$ is neither irreducible (in $\mathbb{Z}$) nor does $1 \mid ab$ imply $1 \mid a$ or $1 \mid b$. It only happens both to be true as for every unit. Why did never ever ask anyone, why $-1$ isn't prime? It simply contradicts the idea behind it.

18. Jun 11, 2017

### WWGD

Still, I guess is the rype of thing that needs to be clarified just once, after which one can move on.

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