B Natural Numbers and Odd Numbers

davidge
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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>
 
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davidge said:
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?

Obs: there is a mistake in the thread title. It should be "natural numbers and odd numbers".

What's the definition of subset?
 
PeroK said:
What's the definition of subset?
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##.
 
davidge said:
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##.

I wasn't expecting that!

I might prefer the following:

##A## is a subset of ##B## if ##x \in A \ \Rightarrow \ x \in B##
 
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PeroK said:
I wasn't expecting that!

I might prefer the following:

##A## is a subset of ##B## if ##x \in A \ \Rightarrow \ x \in B##
:smile:
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?
 
davidge said:
:smile:
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?

Yes.
 
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davidge said:
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##.
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.
 
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  • #10
mfb said:
Usually a subset can be the full set: ##A \subset A##. In this case the second condition is not satisfied.

If the second condition is satisfied, it is a proper subset.
This is a bit nonsense. But definitions are definitions.
 
  • #11
davidge said:
This is a bit nonsense. But definitions are definitions.
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.
 
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  • #12
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.
 
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  • #13
mfb said:
Many mathematical statements would have to use "a subset of A or A itself" everywhere if you exclude the full set as subset.
.

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.
 
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  • #14
mfb said:
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.
I don't know if just a convention; if 1 were prime, every number would be composite as n=n(1).
 
  • #15
WWGD said:
I don't know if just a convention; if 1 were prime, every number would be composite as n=n(1).
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 learned it correctly, this wouldn't be necessary.
 
  • #16
fresh_42 said:
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 learned it correctly, this wouldn't be necessary.
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
WWGD said:
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.
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.
 
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  • #18
fresh_42 said:
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.
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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