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This is my first math problem set, and I am a little confused. This is my solution to the problem, however it think it might be a little too easy...

I have a Field $F$ and $ a, b \in F$. I am supposed to show that the equation $ a + x = b $ has a unique solution.

First I show existence:

$ a + x = b \Leftrightarrow a + x + (-a) = b + (-a) \Leftrightarrow a + (-a) + x = b + (-a) \Leftrightarrow 0 + x = b - a \Leftrightarrow x = b - a. $

Now I want to show that the solution is indeed unique: Suppose there is another number $y$ for which $y = b - a$. Then $ y = b - a = x \Rightarrow y = x$

But again, I am really not sure about the last part. Can you tell me if I am wrong, and if so, give me a hint as to why this wouldn't work?

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I would for no good reason prefer to add $-a$ on the left. So $a+x=b$ iff $(-a)+(a+x)=(-a)+b$ iff $((-a)+a)+x=(-a)+b$ iff $0+x=(-a)+b$ iff $x=(-a)+b$. The end. Small note: Your $b-a$ introduces a new operator, subtraction. That's unnecessary, and might even be downmarked by an excessively fussy person. – André Nicolas Oct 28 '12 at 17:40
@AndréNicolas thanks for the hint andré. in the question it says that I should show that the solution is of the form x = b - a, so subtraction seems to be legitimate (although we have not clearly introduced it in lecture) – padrino Oct 30 '12 at 17:13
The OP as given did not say that. So I was afraid that if you introduced subtraction without detailed definition and verification of properties, a grader in a sour mood might object. – André Nicolas Oct 30 '12 at 17:18
up vote 2 down vote accepted

I think your existence proof establishes uniqueness on its own: through a series of if and only if statements, you show (necessarily) that $x = b - a$.

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You are absolutely right. More generally, in every abelian group the substraction is unique.

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