# Thread: Unique Solution

1. ## Unique Solution

Suppose a,b e G where G is a group under addition.

Then, the equation x + a = b has a unique solution in G. The solution is
x = b - a.

Proof:

Suppose the equation x + a = b has two solutions. Let the solutions be
x = b - a and y = b - a.

Now, by transitivity of equality, x = y. THus there is a unique solution in G.

How does this look?

2. Originally Posted by jzellt
Suppose a,b e G where G is a group under addition.

Then, the equation x + a = b has a unique solution in G. The solution is
x = b - a.

Proof:

Suppose the equation x + a = b has two solutions. Let the solutions be
x = b - a and y = b - a.

Now, by transitivity of equality, x = y. THus there is a unique solution in G.

How does this look?
Hi jzellt.

To prove uniqueness, you need to assume that there are two solutions, and show that they are equal. In this case, let the two solutions are $\displaystyle x$ and $\displaystyle y,$ so $\displaystyle x+a=b$ and $\displaystyle y+a=b.$ Your task is then to show that $\displaystyle x=y.$ Can you do that?