How to prove the uniqueness of the solution of $ax+b=0$? I have no background in mathematical analysis or the like, but I am interested to know how to prove the uniqueness of the solution of $ax+b=0$? Perhaps your answers will help me to prove other uniqueness problems.
 A: A standard way of showing that a certain object is unique is two assume that you have two objects that satisfy the desired properties, and deduce that they must be equal (when we say "two objects", we mean two "names", but which may refer to the same object).
In the case of the solutions to the equation $ax+b=0$, you have to distinguish two cases: if $a=0$, then the equation either has no solutions (if $b\neq 0$), or it has infinitely many solutions (if $b=0$). 
So uniqueness really only exists when $a\neq 0$. The uniqueness is based on the following fact about real numbers:

For any real numbers $r$ and $s$, if $rs=0$, then $r=0$ or $s=0$.

Once you have that:
Claim. If $a\neq 0$, then there is at most one solution to $ax+b=0$.
Proof. Suppose that both $x$ and $y$ are solutions. We aim to show that $x=y$. Since $x$ is a solution, $ax+b=0$. Since $y$ is a solution as well, $ay+b=0$. That means that $ax+b=ay+b$. Adding $-b$ to both sides we conclude that $ax=ay$. Adding $-ay$ to both sides, we obtain $ax-ay = 0$. factoring out $a$, we have $a(x-y)=0$. Since the product is $0$, then $a=0$ or $x-y=0$. Since $a\neq 0$ by assumption, we conclude that $x-y=0$, so $x=y$. Thus, if $x$ and $y$ are both solutions, then $x=y$, so there is at most one solution. $\Box$
Note that this argument works in the context of the real numbers, or other kinds of "numbers" where $rs=0$ implies $r=0$ or $s=0$. There are other situations where this is not the case. For example, if you work with "integers modulo 12" ("clock arithmetic", where $11+3 = 2$), then $2x+8 = 0$ has many different solutions $0\leq x\lt 12$: one solution is $x=2$ (since $2(2)+8 = 4+8=12=0$ in clock arithmetic), and another solution is $x=8$ since $2(8)+8 = 16+8=24 = 0$ in clock arithmetic).  
A: Suppose $x$ and $y$ are both solutions to your equation. Then we have:
$ax + b = 0$
and
$ay + b = 0$
Subtracting the two equations gives
$a(x - y) = 0$
So if $a$ is non-zero, then $x-y$ must be zero, i.e. $x = y$.
A: If you're working on a system with a unique inverse for every (or some) elements, then
$$ax+b=0\Longleftrightarrow x=-ba^{-1}=-\frac{b}{a}$$
is equivalent to "a has a unique inverse". For example, fields (like $\,\mathbb Q\,,\,\Bbb R\,,\,\Bbb C\,$ , etc.) , where the unique inverse element axiom is true for anyn non-zero element.
A: As Tony wrote, if it had two different solutions $\rm\,r\ne s,\,$ then  $\rm\: a\, x = 0\,$ would have root $\rm\, x = r-s\ne 0,\:$ contra $\rm\:x,y\ne0\:\Rightarrow\:xy\ne 0.\:$  This implication ("no zero-divisors") is equivalent to the fact that polynomials have no more roots than their degree (so a linear polynomial has at most one root). 
A: Would the following be valid?
if $a=0$ and $b\neq 0$ then there is no solution and if $a=0$ and $b=0$ then there are infinite solutions. (So we won't be using these conditions in our proof because the do not give us a unique solution)
but
if $a\neq 0$ then there is a unique solution to $ax+b=0$
Proof
let $x,y \in \Bbb R: ax+b=0$ and $ay+b=0$ meaning $x=y$
$$\begin{align}
&ax+b=ay+b\\
\implies &(ax+b)+(-b)=(ay+b)+ (-b)\\   
\implies &ax+(b+(-b))=ay+(b+(-b))\tag {A2}\\
\implies &ax+0=ay+0\tag{A4}\\
\implies &ax=ay\\
\implies &ax(a^{-1})=ay(a^{-1})\\
\implies &x(a(a^{-1}))=y(a(a^{-1}))\tag{M2}\\
\implies &x(1)=y(1)\tag{M4}\\
\implies &x=y
\end{align}$$
Proving that the solution is true
