$a+b=0$ implies $a=b=0$ Let $N$ be a set of non-negative integers. Of course we know that $a+b=0$ implies that $a=b=0$ for $a, b \in N$.
How do (or can) we prove this fact if we don't know the subtraction or order?
In other words, we can only use the addition and multiplication.
Please give me advise.
EDIT
The addition law mean that for $a, b \in N$, there is an element $a+b$ in $N$ and this operation is associative. 
The multiplication law means that for $a, b \in N$, there is an element $ab$ in $N$ and this operation is associative. 
Also the distribution laws hold.

EDIT2
Let me rephrase the question since I don't want arguments on orders.
Let $N$ be a set with operation $+$ and $\times$.
$N$ is a monoid with the operation $+$ and $\times$ respectively. There is an unit element $0\in N$.
The distribution laws hold as in the case of the set of integers.
Can we prove the fact above with this assumption?
 A: Please set out what you mean by the addition law.  You need the axiom that there is no number whose successor is $0$ or this fails.  That is what distinguishes the integers from the naturals.  It allows you to define order as $x \le y \leftrightarrow \exists (z) x+z=y$
A: What is the addition law?
If it the one from Peano arithmetic, 
it is 
$x+0=x$ and $x+S(y) = S(x+y)$,
where $S(x)$ is the successor of $x$.
If $x+y=0$, 
suppose $y$ was not zero.
Then there is a $z$ such that
$S(z) = y$.
Then $0 = x+y = x+S(z) = S(x+z)$
which is a contradiction, since
$0$ is not the successor of anything.
Therefore $y=0$.
Substituting this in $x+y=0$
and using $x+0=x$, we get $x=0$.
Is this what you wanted?
A: There are three axioms that you need to prove this:


*

*For all non-negative integers $n$, $n+1\neq 0$

*For all non-negative integers $a,b$, $a+(b+1)= (a+b)+1$

*Induction


Theorem: For all $b$, either $b=0$ or $a+b\neq 0$.
Proof by induction: If $b=0$ we are done.
Now assume it is true for $b$. Then $a+(b+1)=(a+b)+1$ by (2).  But by $1$, $(a+b)+1\neq 0$. So we get our result.
Now, if $a+b=0$ then $b=0$ and then $a+b=a+0=a$ so $a=0$ as well.
You can't do it just from addition law and multiplication law because those laws are true for all the integers, and it is not true for all the integers.
A: Use the Mazur swindle! Namely, if $a+b=0$ then
\begin{align*}
0 &= 0 + 0 + 0 + \cdots\\
  &= (a+b)+(a+b)+(a+b)+\cdots\\
  &= a+(b+a)+(b+a)+\cdots \\
  &= a + 0 + 0 + \cdots\\
  &= a.
\end{align*}
Regrouping the infinite sum is justified because everything is nonnegative. I leave it as an exercise to identify exactly which axioms of arithmetic we've used.
A: You don't need $+$ or $\times$ as forming monoids on the natural numbers, nor the distributive property.  You can get by with less as follows:
Addition has the property that $(0+b)\leq(a+b)$ and $(a+0)\leq(a+b)$.  This is NOT presupposing the order of the natural numbers, or that no number of the naturals has 0 as a successor, but only a monotonicity property of addition.
We'll also assume the existence of an additive identity.
Consequently, if $(a+b)=0$, then $b=(0+b)\leq(a+b)=0.$  So, $b\leq 0.$  Also, $0=(0+0)\leq (0+b)=b$ by substitution of 0 for a in $(a+0)\leq (a+b)$ and the identity rule.  Thus, $0\leq b.$  So, $0 \leq b\leq 0.$  Consequently, $b=0.$
Similarly, $a=(a+0)\leq(a+b)=0.$  Thus, $a\leq 0$.  Also, $0=(0+0)\leq(a+0)=a.$  So, $0\leq a.$  Thus, $0 \leq a\leq 0.$  Consequently, $a=0.$
Therefore, for an algebraic structure with an identity element "$0$", binary operation "$+$" and where $(0+b)\leq (a+b),$ and $(a+0)\leq (a+b),$ and $a \leq x \leq a \implies x=a,$ it holds that "if $(a+b)=0,$ then $a=b=0$".
As an example of an algebriac structure where this holds, and "$+$" is not natural number addition, let "$+$" denote the maximum of two numbers, and consider $(\{0, 1\}, +)$.  Both suppositions used in the proof above can get verified.
