Is there a theorem or axiom stating that integers added to integers always yields integers?

Question

I have finished a small proof for school but I realize it relies on the statement that integers added to integers yields integers. I assumed this statement was pretty much just accepted, but considering how central it is to my proof I wanted to cite the theorem or law or something. So far no internet search has turned up anything useful.

Answer 1

It can be part of the definition of "integer". A natural number is an object inductively defined as follows: $0$ is a natural, then we define another object $S(0)$ (the "successor" of $0$) which by definition a natural, then $S(S(0))$, then…

We define a shorthand $1$ for $S(0)$, $2$ for $S(S(0))$, …

Let $S^n(0)$ be a further shorthand for $n$, and given natural $n$ write $n-1$ for the unique object $P$ which has $S(P) = S^{n}(0)$. (We define an axiom to say that if $S(n) = S(m)$ then $n=m$, so that gives us uniqueness.) We would then be justified in writing $P = S^{n-1}(0)$ - that is, in saying that $n-1$ is the unique solution to $(n-1)+1 = n$.

We then define the sum of two natural numbers by induction: $n + m := S(n) + (m-1)$, and $n+0 = n$.

Now define integers to be equivalence classes of pairs of naturals under the relation $(a, b) \sim (c, d)$ iff $a+d = b+c$. Informally, $[(a, b)]$ represents the integer $a-b$, so we're capturing $a-b = c-d$, or $a+d = b+c$. We can define addition to work in the right way on these equivalence classes.

Finally, we have the statement that "the sum of two integers is integer", which just falls out: inductively the sum of two naturals is natural, and this property gets inherited by the equivalence classes, so we get that the sum of two integers is integer.

Answer 2

Notation: write $\mathbb{Z}$ for the set of integers and $\mathbb{R}$ for the set of real numbers.

Case 0. Perhaps we're viewing $\mathbb{Z}$ as a standalone object, independent of its status as a subset of the real line.

Then by definition, it is a set $\mathbb{Z}$ equipped with functions $$+:\mathbb{Z}, \mathbb{Z} \rightarrow \mathbb{Z}, \qquad -:\mathbb{Z} \rightarrow \mathbb{Z}, \qquad 0:\mathbb{Z},\qquad \cdot : \mathbb{Z}, \mathbb{Z} \rightarrow \mathbb{Z}$$ satisfying certain conditions.

(The problem then remains of showing that this mathematical structure actually exists.)

Anyway, in this case, there is no need to prove that the sum of two integers is an integer; its just a consequence what functions are (and/or how they work).

Case 1. On the other hand, perhaps we're viewing $\mathbb{Z}$ as a subset of $\mathbb{R}$. In this case, in principle we need to prove that the sum of two integers is an integers. This is essentially immediate from the defintion, however.

Definition. $\mathbb{Z}$ is the least subset of $\mathbb{R}$ such that:

• $1 \in \mathbb{Z}$
• $-1 \in \mathbb{Z}$
• $x+y \in \mathbb{Z}$ whenever $x,y \in \mathbb{Z}$

Notice that the sum of two integers is itself always an integer, essentially by definition (again!)

It remains to show that this thing actually exists, in other words that $\mathbb{R}$ actually has a least subset satisfying these three conditions. We can prove this from accepted mathematical principles as follows.

• Let $\mathbb{Z}$ denote the intersection of all subsets of $\mathbb{R}$ satisfying the above three condtions.
• Show that $\mathbb{Z}$ must itself satisfy these conditions.
• Show that if a subset $J$ of $\mathbb{R}$ satisfies these conditions, then $\mathbb{Z} \subseteq J$.