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This question already has an answer here:

I am not sure whether the following statement is true: $ ℤ^+ = ℕ$

if not, why?

Thank you in advance! I appreciate your help!

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marked as duplicate by Andrés E. Caicedo, user91500, Asaf Karagila set-theory May 24 '14 at 16:57

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    $\begingroup$ It depends on the definition. $\endgroup$ – Git Gud May 24 '14 at 14:48
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    $\begingroup$ Peano says $0\in\mathbb{N}$, while - as far as I've always been told - $\mathbb{Z}^+$ denotes the set of strictly positive integers. $\endgroup$ – MattAllegro May 24 '14 at 14:49
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    $\begingroup$ So $ℤ^+$ always means {1, 2, 3, ...}, whereas $N$ means either { 0, 1, 2, 3, ...} or { 1, 2, 3, ...}. Right? $\endgroup$ – cherry8.8vanilla May 24 '14 at 14:50
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    $\begingroup$ @cherry8.8vanilla Any reasonable interpretation is one those two, yes. $\endgroup$ – Git Gud May 24 '14 at 14:51
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    $\begingroup$ It is very annoying that you always need to have or give a definition of $\mathbb N$ before you can really get to work. I am afraid that this annoyance will unfortunately never come to an end. Is $0$ natural? Yes, no, yes, no, yes, no,..... As @Git Gud says: it depends on the definition. This is true, but should not be the case, unless it is a commonly accepted definition. $\endgroup$ – drhab May 24 '14 at 15:01
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You need to know the definitions.

$\mathbf{N}$ usually includes zero and $\mathbf{Z}^+$ usually does not. But occasionally people define $\mathbf{N}$ to exclude zero or $\mathbf{Z}^+$ to include zero, or both.

I often use $\mathbf{Z}^>$ (respectively $\mathbf{Z}^{\geq}$) for excluding (resp. including) zero, as I've never seen any ambiguity with those.

(or sometimes the longer forms $\mathbf{Z}^{>0}$ and $\mathbf{Z}^{\geq 0})$

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  • $\begingroup$ Is $0$ an even or an odd number? $\endgroup$ – cherry8.8vanilla May 24 '14 at 15:47
  • $\begingroup$ @cherry: Even, because it is divisible by $2$. $\endgroup$ – Hurkyl May 24 '14 at 15:48
  • $\begingroup$ But it's divisible by any number, or not? (by $3,5,7,9,$ etc. as well, and those are odd numbers $(2k-1)$) $\endgroup$ – cherry8.8vanilla May 24 '14 at 15:50
  • $\begingroup$ @cherry8.8vanilla: When checking whether a number is even or not, we just care about its divisibility by 2. We don't care if it's divisible by 3, 4, etc... or not. $\endgroup$ – user49685 May 24 '14 at 15:56
  • $\begingroup$ @user49685: Thank you for explaining that to me. $\endgroup$ – cherry8.8vanilla May 24 '14 at 15:57
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If $\mathbb{N} \!\,$:= {$1,2,3,4,5,...$} and ℤ+ ={$1,2,3,4,5,...$}, then we can show that $\mathbb{N} \!\,$ is isomorphic to ℤ+.

So we can think of them as "equal".

Obviously, if you define $\mathbb{N} \!\,$ or ℤ+ otherwise then this won't hold.

Usually, $\mathbb{N} \!\,$ is defined not to include $0$ as originally stated.

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    $\begingroup$ Usually is defined not to include $0$? If anything, usually it is defined to include $0$. $\endgroup$ – Git Gud May 24 '14 at 15:43
  • $\begingroup$ From my experience, it is defined without $0$. $\endgroup$ – Mr Croutini May 26 '14 at 14:02
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It is a matter of convention. Always keep in mind that "positive" $\neq $ "non-negative". I like to consider $0 \in \mathbb{N}$, because of the notation $\mathbb{N}^*$, which means $\mathbb{N} \setminus \{ 0 \}$. And indeed $\mathbb{Z}^+$ and $\mathbb{N}^*$ are isomorphic.

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  • $\begingroup$ ... but also keep in mind that people will often say "positive" when they really mean "non-negative", so you have to be on guard to avoid confusion. $\endgroup$ – Hurkyl May 24 '14 at 16:00
  • $\begingroup$ @Hurkyl With 'people' you mean 'analysts'. $\endgroup$ – Git Gud May 24 '14 at 16:02

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