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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4It depends on the definition. – Git Gud May 24 '14 at 14:48
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3Peano says $0\in\mathbb{N}$, while - as far as I've always been told - $\mathbb{Z}^+$ denotes the set of strictly positive integers. – MattAllegro May 24 '14 at 14:49
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1So $ℤ^+$ always means {1, 2, 3, ...}, whereas $N$ means either { 0, 1, 2, 3, ...} or { 1, 2, 3, ...}. Right? – cherry8.8vanilla May 24 '14 at 14:50
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1@cherry8.8vanilla Any reasonable interpretation is one those two, yes. – Git Gud May 24 '14 at 14:51
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Ok, thank you @GitGud. I have never heard that $0$ could be a natural number, which is weird. – cherry8.8vanilla May 24 '14 at 14:56
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1It 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. – drhab May 24 '14 at 15:01
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@حكيمالفيلسوفالضائع: It's weird because in all these years of educating nobody mentioned to me that $0$ could actually be treated as a natural number. – cherry8.8vanilla May 24 '14 at 15:04
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@drhab: It's exactly how I feel about this. I think that they should finally decide on whether $0$ is or isn't a natural number. This question is really causing a lot of confusion around people. – cherry8.8vanilla May 24 '14 at 15:07
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1@cherry8.8vanilla The reasonable thing is for $0$ to be a natural number. But since analysts are lazy, they choose to start $\mathbb N$ at $1$ so they don't have to exclude $0$ every time. – Git Gud May 24 '14 at 15:14
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@GitGud: Thank you, that explains a lot. – cherry8.8vanilla May 24 '14 at 15:17
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Also an exact duplicate of this. – Asaf Karagila May 24 '14 at 16:59
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@AsafKaragila: My apologies for not noticing that before. Thank you for notifying me and others about that. – cherry8.8vanilla May 24 '14 at 17:13
3 Answers
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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.
Ivo Terek
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... 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. – May 24 '14 at 16:00
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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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But it's divisible by any number, or not? (by $3,5,7,9,$ etc. as well, and those are odd numbers $(2k-1)$) – cherry8.8vanilla May 24 '14 at 15:50
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@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. – user49685 May 24 '14 at 15:56
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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.
Mr Croutini
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2Usually is defined not to include $0$? If anything, usually it is defined to include $0$. – Git Gud May 24 '14 at 15:43
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