Prove that the sum of two positive integers is positive? On a practice final exam for my Discrete Math class, I've been asked to prove that the sum of two positive integers is positive. I've been pulling my hair out over how to prove this, as it seems so obvious. I even asked some of my friends taking proof-based calculus, and they said that this was the type of thing that is just assumed to be true. Anyone have an idea of how to prove this? Any help is appreciated.
 A: What's important about the Integers is that they are an totally ordered commutative ring with identity, though more precisely they form a totally ordered integral domain.
There are two components here: being a commutative ring with identity, i.e. an integral domain refers to the algebraic properties of $\mathbb{Z}$: commutativity and associativity of addition and multiplication, distributivity, cancellation for multiplication, etc. These don't really matter to the question. You can look them up if you're interested (you should).
What does it mean for $\mathbb{Z}$ to be totally ordered? It means that the following four axioms of order hold for $\mathbb{Z}$:

*

*Trichotomy: if $a,b \in \mathbb{Z}$, then one and only one of the following holds: $a<b$, $a=b$, $a>b$.

*Transitivity: If $a,b,c \in \mathbb{Z}$ with $a<b$ and $b<c$, then $a<c$.

*Addition: If $a,b,c \in \mathbb{Z}$ and $a<b$, then $a+c<b+c$.

*Multiplication by Positive Elements: If $a,b,c \in \mathbb{Z}$, $a<b$ and $c>0$, then $ac<bc$.

You need to use the addition rule to complete your proof. Take a positive integer $a>0$. Take a positive integer $b>0$. So $a+b > b+0$. So $a+b > b$. And $b>0$. So by transitivity, $a+b>0$.
A: If you have the more or less usual definitions and operations:
$$a,b\;\;\text{positive}\iff a\stackrel 1>0\,,\,\,b\stackrel 2>0$$
and from here
$$a+b\;\;\text{non-positive}\;\iff a+b\le 0\iff a\le -b\stackrel 2<0\implies a<0\;\;\text{contradiction to}\;\;1$$
Thus is must be $\;a+b>0\implies\;\text{the sum is positive}\;$
