1. ## Prove that integers are closed under addition

Prove that the sum of two integers is also an integer.

This is from an advanced calculus book. These are the facts presented earlier in the book which I assume are supposed to be sufficient to prove this:

$\mathbb{Z}=\text{the natural numbers, their negations, and zero.}$
$\text{If }m,n\in\mathbb{N}\text{, then }m+n\in\mathbb{N}\text{ and }mn\in\mathbb{N}$.
$\text{If }n\in\mathbb{N}\text{ and }n>1\text{, then }n-1\in\mathbb{N}$.
$\text{If }m,n\in\mathbb{N}\text{ and }n>m\text{, then }n-m\in\mathbb{N}$.

And what we want to show is that if $a,b\in\mathbb{Z}\text{, then }a+b\in\mathbb{Z}$.

So I'm not sure how to do it. Do you have to do cases where $a$ and $b$ are positive, negative, or zero?

2. ## Re: Prove that integers are closed under addition

Originally Posted by Ragnarok
Prove that the sum of two integers is also an integer.
This is from an advanced calculus book. These are the facts presented earlier in the book which I assume are supposed to be sufficient to prove this:
$\mathbb{Z}=\text{the natural numbers, their negations, and zero.}$
$\text{If }m,n\in\mathbb{N}\text{, then }m+n\in\mathbb{N}\text{ and }mn\in\mathbb{N}$.
$\text{If }n\in\mathbb{N}\text{ and }n>1\text{, then }n-1\in\mathbb{N}$.
$\text{If }m,n\in\mathbb{N}\text{ and }n>m\text{, then }n-m\in\mathbb{N}$.
And what we want to show is that if $a,b\in\mathbb{Z}\text{, then }a+b\in\mathbb{Z}$.
So I'm not sure how to do it. Do you have to do cases where $a$ and $b$ are positive, negative, or zero?
This is just a guess. (Unless you name the text)

Clearly if both are positive, the first axiom gives.

If one is positive and one is zero then what?

If both are negative then $-m-n\in\mathbb{N}$ so $m+n\in\mathbb{Z}$, HOW?

If one is negative and one is zero now then what?

3. ## Re: Prove that integers are closed under addition

Thanks so much! I figured it out with your hints.