# Set question - $ℤ^+ = ℕ$ [duplicate]

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

if not, why?

## marked as duplicate by Andrés E. Caicedo, user91500, Asaf Karagila♦ set-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 24 '14 at 16:57

• It depends on the definition. – Git Gud May 24 '14 at 14:48
• Peano 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
• So $ℤ^+$ 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
• @cherry8.8vanilla Any reasonable interpretation is one those two, yes. – Git Gud May 24 '14 at 14:51
• 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. – drhab May 24 '14 at 15:01

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 $\mathbf{Z}^{\geq 0})$

• Is $0$ an even or an odd number? – cherry8.8vanilla May 24 '14 at 15:47
• @cherry: Even, because it is divisible by $2$. – Hurkyl May 24 '14 at 15:48
• 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
• @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
• @user49685: Thank you for explaining that to me. – cherry8.8vanilla May 24 '14 at 15:57

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.

• Usually is defined not to include $0$? If anything, usually it is defined to include $0$. – Git Gud May 24 '14 at 15:43
• From my experience, it is defined without $0$. – Mr Croutini May 26 '14 at 14:02

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.

• ... 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. – Hurkyl May 24 '14 at 16:00
• @Hurkyl With 'people' you mean 'analysts'. – Git Gud May 24 '14 at 16:02