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Expert-verified Found in: Page 67 ### Abstract Algebra: An Introduction

Book edition 3rd
Author(s) Thomas W Hungerford, David Leep
Pages 608 pages
ISBN 9781111569624 # Let a and b be elements of a ring R. (a) Prove that the equation ${\mathbit{a}}{\mathbf{+}}{\mathbit{x}}{\mathbf{=}}{\mathbit{b}}$ has a unique solution in R. (You must prove that there is a solution and that this solution is the only one.) (b) If Ris a ring with identity and a is a unit, prove that the equation ${\mathbit{a}}{\mathbit{x}}{\mathbf{=}}{\mathbit{b}}$ has a unique solution in R.

(a). It is proved that the equation $a+x=b$ has a unique solution.

(b). It is proved that the equation $ax=b$ has a unique solution.

See the step by step solution

## Solution to Part (a)

The given equation is $a+x=b$, where we need to prove that the equation has a unique solution. Solve the given equation for x.

$x=\left(-a\right)+b$

By using theorem 3.3, there exists an additive inverse of $-a$, that is a.

$\begin{array}{rcl}x& =& a+\left(\left(-a\right)+b\right)\\ & =& a+\left(-a\right)+b\\ & =& {0}_{R}+b\\ & =& b\end{array}$

This implies that $\left(-a\right)+b$ is a solution of the given equation. Now, we have to prove that this solution is unique, for which we assume that there exists two different solutions of the given equation that are x and y, then:

$a+x=b=a+y\phantom{\rule{0ex}{0ex}}⇒x=y$

Which is a contradiction hence, there exists a unique solution, which is: $\left(-a\right)+b$.

## Solution to Part (b)

The given equation is $ax=b$.

It is given that a is a unit. Then, there exists ${a}^{-1}\in R$, so that $a{a}^{-1}={1}_{R}$, then for the given equation we have,

$\begin{array}{rcl}a\left({a}^{-1}b\right)& =& \left(a{a}^{-1}\right)b\\ & =& {1}_{R}b\\ & =& b\end{array}$

So, ${a}^{-1}b$ is the solution of the given equation. Now, we have to prove that it is unique. Let two different solutions are x and y.

$\begin{array}{rcl}{a}^{-1}/\left(ax& =& ay\right)\\ {a}^{-1}\left(ax\right)& =& {a}^{-1}\left(ay\right)\\ \left({a}^{-1}a\right)x& =& \left({a}^{-1}a\right)y\\ {1}_{R}x& =& {1}_{R}y\\ x& =& y\end{array}$

Which is a contradiction hence, there exists a unique solution, which is ${a}^{-1}b$. ### Want to see more solutions like these? 