Prove $b+x=a$ has one and only one solution

Question

Suppose $a$ and $b$ in $R$

The equation $b+x = a$ has one and only one solution.

How do i prove this theorem ? I assume first $x$'s existence has to be shown then its uniqueness

Answer 1

Assume two arbitrary solutions to the equation for now, say $α$ and $β$. then both $α$ and $β$ satisfy the equation, so it follows that,

$b + α = a$ and $b + β = a$

which implies that,

$α = a - b$ and $β = a - b$

which immediately tells us that,

$α = β$

This exercise thus shows/proves the uniqueness of the solution of the equation.

Answer 2

I'm assuming that $R$ is an abstract ring, since the question was tagged as ring theory.

But that doesn't make the algebra more abstract. How would you solve the equation in the integers? Add $-b$ to both sides of the equation.

In a general ring (or even, as Hagen points out, a general abelian group), the addition operation is commutative and has inverses. So all the properties you need are available to do the same thing. You just need to justify each step and point out which axiom you are using.