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I just started real analysis. I don't have a background in proofs or logic, simply calculus. So I'm trying to learn more about proofs--so forgive the basic question, please.

How do you go about proving this theorem: If $a$ and $b$ are any numbers, then there is one and only one number $x$ such that $a + x=b$. This number is given by $x=b+(-a)$.

First part of the proof I understand--it's simple. We simply use some axioms and do the following:

$a+b-a=b$, which of course is true. This is just true from plugging in the $b+(-a)$ for $x$.

But how about the uniqueness issue? In my text, it says:

$$(a+x)+(-a)=b+(-a) x=b+(-a)$$

How does this prove uniqueness? Is this equation essentially saying that the ONLY possible value of $x$ is expressed by $b+(-a)$?

How would I have known to set this up as part of the proof if I was asked to prove something is unique?

Sorry for the basic question--have to start somewhere.

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    Yes, it is essentially saying that the only possible value is b+(-a).2010-12-23
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    [Relevant SMBC comic](http://www.smbc-comics.com/?id=3227#comic) (I think)2014-01-05

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Usually when you want to prove something is unique you start by assuming there are two things that satisfy the given property, and then you show that they are the same. In this case, if you want to prove that there is exactly one $x$ such that $a + x = b$, you could start by assuming there are two numbers, say $x$ and $y$ that satisfy the identity, that is

$$a + x = b \quad \text{and} \quad a + y = b$$

but then this implies that $a + x = a + y$ since both are equal to $b$. Now you can add $-a$ to both sides of the equation to get $$-a + (a + x) = -a + (a + y) \implies (-a + a) + x = (-a + a) + y $$

$$\implies 0 + x = 0 + y \implies x = y$$

This takes care of the uniqueness of the number and since you already know one such $x$ then you're done. I suppose you're doing this in the real number field and that you know which axioms take place in the argument. Hope this helps a little.

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    So proof by contradiction? But what about the given argument, is it really air tight?2010-12-23
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    Yes, it is correct. What I did is to assume two numbers $x, y$ satisfy the identity and proved that $x = y$. What your text does is to show what $x$ should be, so it is another way of proving uniqueness.2010-12-23
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    án: But it's redundant to give such a uniqueness proof - see my answer.2010-12-24
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    @Bill Yes, you're right. I just wanted to give a different argument than the one the OP already had. I suppose I wanted to show a possible way to attack problems in which one has to prove uniqueness and there's no explicit form for the object that will show uniqueness immediately, since the OP asked for such a method. For instance when proving uniqueness of the inverse element in an arbitrary group.2010-12-24
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    án: But it's still redundant even for inverses in an arbitrary group - that's merely the special case $b = 0$ above. Perhaps you meant in an arbitrary *semigroup*, i.e. where an inverse operation need not exist.2010-12-24
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Simply use the (additive group) axioms to prove $\rm\ a + x = b\ \Rightarrow\ x = b + (-a)\:.\ $ That yields both existence and uniqueness. That the uniqueness doesn't require further proof is a subtlety that sometimes confuses students. This is discussed at length in my posts in Uniqueness of solution of x+a=b from field, sci.math, May 5, 2003 regarding a slick proof in Max Rosenlicht's Introduction to Analysis.