Is the set of natural numbers closed under subtraction? I was reading my textbook right now on elementary number theory, and this question came up. It got me wondering: How can subtraction be performed on two elements $x,y\in\mathbb{N}$ if subtraction is defined by $x-y=x+(-y)$, and $\forall a\in\mathbb{N}, 0<a<\infty$?
I realize that $1-10=-9$, which is not in $\mathbb{N}$. But the question just bothered me, because I wonder if you can technically subtract two elements of $\mathbb{N}$?
 A: When one subtracts a natural number from another, say $1-2=-1$, one may not get a natural number. However, this does not forbid us from defining substraction. Substraction would be closed if we enlarge the set to the set of all integers.
A: Subtraction is partially defined in $\Bbb N$, i.e. it is defined only for some pairs of natural numbers. You cannot define it as $x-y=x+(-y)$, though, since $-y\notin \Bbb N$.
You can define $x-y=z$ for $x,y\in \Bbb N$ just as the $z\in \Bbb N$, if it exists, such that $y+z=x$.
A: No, subtraction is not closed on the set of natural numbers. 
One can define the difference between $a$ and $b$, $a, b \in \mathbb N\,$ in terms of the magnitude of the difference: taking the absolute value: $|a - b|$ for $a, b \in \mathbb N$, but the problem with "normal subtraction" is that $\,a - b = a + (-b)$. And here, $-b$ is the additive inverse of $b$: and since here we have $b \in \mathbb N$, unless $b = 0$ (if $\mathbb N$ includes $0$), $-b \notin \mathbb N$.


*

*The additive inverse of an integer $n$ is the number such that for any $n \in \mathbb Z$,  $\,n + -n = -n + n = 0,\;$ where $\,0\,$ is the additive identity. 

*Hence, we have the integers, which are closed under subtraction, (or rather closed under inverses), and hence defining subtraction on the integers presents no problems.

*However, for all $n \in \mathbb N$, n\neq 0,\; -n  \notin \mathbb N$. That, essentially, is what is meant when we say that the set of natural numbers is not closed under subtraction (...because it is not closed under inverses).
A: When we write for the natural numbers $a-b=c$ we mean that the number $c$ is the natural number (if it exists) that verify $c+b=a$ so the subtraction is defined by the sum operation. Now if we write for example $1-10=c$, the question is are there a natural number $c$ s.t $c+10=1$? The answer is obviously NO! so the set of natural number is not closed under subtraction.
A: You can prove that for any natural numbers $a,b$ there is at most one $c$ such that $b+c=a$. That means that we can define $z=a-b$ to mean $z+b=a$. This means that if $z=a-b$ and $w=a-b$ then $z=w$ - that is, subtraction is well-defined when it is defined.
You could also just define the integers based on the naturals, and define subtraction there for all pairs of integers.
A: Regular subtraction is not well-defined on the natural numbers.  In natural number contexts one often deals instead with truncated subtraction, which is defined: $$a\dot-b = \begin{cases}0,&\text{if $a\le b$}\\ a-b&\text{if $a\ge b$}\end{cases}$$
For example, one can define a truncated subtraction in Peano arithmetic as follows:
$$\begin{array}{rcrl}
0 & \dot- & n & = 0 \\
Sn & \dot- &  0 & = Sn \\
Sn & \dot- & Sm & = n\dot- m
\end{array}
$$
One can similarly define it in the context of Church numerals, or in the context of total recursive functions.
This is often sufficient for whatever purposes one needs subtraction.
