# Proving closure of $\mathbb{Z}^+$ under multiplication

I understand that $\mathbb{Z}^+$ is said to be closed under multiplication because $a\in \mathbb{Z}^+, b\in \mathbb{Z}^+\implies a\cdot b\in \mathbb{Z}^+$ but my question is: is this enough to be considered a proof?

Clarification: I guess I am really wondering if it is necessary/possible to prove $a\cdot b\in \mathbb{Z}^+$

• Yes it is. What else would you expect? What is the definition of "being closed under an operation"? – amrsa Apr 16 '17 at 15:41
• I guess I would rather ask, what is the context? How much are you allowed to assume? – Maximal Ideal Apr 16 '17 at 15:56
• I have attempted to clarify a little better – ifearthenight Apr 16 '17 at 16:01
• The answer to that question (clarification), depends on how is the product defined: if you assume you have an operation on that set, then it is already closed under that operation, by definition; if you have some construction of the integers (says, starting with the natural numbers), then you have to prove that the definition of multiplication gives an operation, that is, that the set is closed under that map. – amrsa Apr 16 '17 at 16:21

It is important to understand that no proof exists in a vacuum: a proof only exists in the context of the axioms and definitions you are assuming are already true.

In your case, you need to specify what definitions and axioms you are using for the integers and natural numbers. Under Peano arithmetic, multiplication is defined via the relationship $$x\times 0 = 0$$ $$x\times S(y) = x + x\times y.$$ You can then prove that multiplication is well-defined for any pair of natural numbers using induction and the fact that addition is well-defined (which you can prove separately).

I've seen other axiomizations, where existence of the natural numbers as a well-ordered semiring contained in $\mathbb{Z}$ is itself an axiom of the integers. Then there is nothing to prove.

Proving closure only makes sense when you're considering something as a subset of a larger structure, like proving a subset $H$ of a finite group $G$ is closed and therefore a subgroup. If you just use the natural number construction, then the multiplication operation is by definition closed, and you just have to prove it's well-defined.

If you define $\mathbb{Z}^+$ as the intersection of all inductive subsets of $\mathbb{R}$, that is: $$\mathbb{Z}^+=\bigcap\{X: X\subseteq \mathbb{R},\,1\in X,\,n\in X \implies n+1\in X\}$$ you can use the definition of addition and multiplication from $\mathbb{R}$ to ask if $\mathbb{Z}^+$ is closed relative to those operations.

To prove closure of addition consider the set $$A=\{x : x\in\mathbb{Z}^+; y\in\mathbb{Z}^+\implies x+y\in\mathbb{Z}^+\}$$ Because $\mathbb{Z}^+$ itself is an inductive set $1\in\mathbb{Z}^+$, and $y\in\mathbb{Z}^+\implies 1+y=y+1\in\mathbb{Z}^+$, so $1\in A$. If $n\in A$ then $n+1\in\mathbb{Z}^+$ and $y\in\mathbb{Z}^+\implies (n+1)+y=n+(1+y)\in\mathbb{Z}^+$, so $n+1\in A$, and we see that $A$ equals all of $\mathbb{Z}^+$(because it's an inductive subset of the intersection of all inductive sets), which proves closure of addition.

Next to prove closure of multiplication consider the set$$B=\{x : x\in\mathbb{Z}^+; y\in\mathbb{Z}^+\implies x\cdot y\in\mathbb{Z}^+\}$$

Using again the fact that $\mathbb{Z}^+$ is inductive, we see $1\in\mathbb{Z}^+$ and $y\in\mathbb{Z}^+\implies 1\cdot y=y\in\mathbb{Z}^+$, so $1\in B$. Assuming $n\in B$, we have $n+1\in\mathbb{Z}^+$ and $y\in\mathbb{Z}^+\implies (n+1)\cdot y= n\cdot y + y\in\mathbb{Z}^+$, using the assumption and the fact we've already proved closure of addition, so $n+1\in B$, and we have proved closure of multiplication.