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

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.

• It depends on how you define integers. But generally it is said that $+$ is an operation on $\Bbb Z$, which implies it must map pairs of elements of $\Bbb Z$ to an element of $\Bbb Z$. – Wojowu Sep 23 '15 at 20:34
• This ultimately will follow from the definition of addition of integers. To define addition, of course, we must first define the integers. However, in most introductory proof classes the integers are not rigorously defined and the most basic properties of the integers (such as closure under addition) are taken for granted. So for your case I think it is ok to state it without any sort of justification. – Seth Sep 23 '15 at 20:46
• To rigorously define the integers one can start by defining the natural numbers using peano's axioms (en.wikipedia.org/wiki/Peano_axioms), and then formally adjoin negatives, as is done in "Foundations of Analysis" by Landau. There are also other ways to define the natural numbers and integers that are more set theoretic in nature. – Seth Sep 23 '15 at 20:47

## 3 Answers

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.

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 $\mathbb{Z} \subseteq J$.
• You know guys, if you don't upvote, or at least leave a comment to the effect: "yes, this is correct," the OP often cannot know in advance that an answer is correct, and he probably won't invest any time trying to understand the relevant details unless he already knows that the answer is correct. Trust, to some extent, is a requisite for understanding, unfortunately. – goblin Sep 23 '15 at 23:18

Unless you are really going to go down the rabbit hole of defining the integers and addition formally, you can just appeal to the fact that the integers are "closed under addition", which is a fancy way of saying that adding two integers gives an integer.

I would say there is no problem taking it for granted, and working in the details is mostly pointless (though people on this site will disagree). Also, it will likely take the focus away from whatever interesting result you are actually proving.