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Okay so here's what I tried: $ a + x = b\\\implies (-a) + (a + x) = (-a) +b \\\implies (-a+a)+x=b+(-a) \\\implies 0 + x = b + (-a) \\\implies x = b+(-a)$

But I'm not sure whether this shows that the solution is unique. I've only shown that $b+(-a)$ is one solution. Maybe there is some other combination of $b$ and $a$ that'll work. But I don't know how to prove that this is the only one.

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Assume there is some second solution $x_1$ such that $x \neq x_1$ and arrive at a contradiction. That's usually the way to show uniqueness. – Noble. Mar 2 '13 at 22:04
Each of your arrows can be replaced by a double arrow. – André Nicolas Mar 2 '13 at 22:04
provided that a and b are some given real/complex/whatever constants... – V-X Mar 2 '13 at 22:05
@Noble. So my proof doesn't show that $x$ is unique then, right? Or does it? I'm kind of confused whether my proof is sufficient or not. – Alraxite Mar 2 '13 at 22:06
Actually, given the direction your arrows are pointing, you have shown uniqueness. That is, you've shown that if you can find $x$ such that $a+x=b$, then $x=b+(-a)$. What you haven't shown is that $b+(-a)$ actually satisfies the equation (i.e., existence), but that shouldn't be too hard... – Micah Mar 2 '13 at 22:07

Assume there is more than one solution to a + x = b. So let us assume that a + x = b and a + y = b, x not equal to y. Then

x = b - a


y = b - a.

x = y

So x must equal y which disproves our assumption. Therefore, proved by contradiction, there is only one solution to a + x = b.

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Well, what you actually proved is: if $x$ satisfies $a+x=b$ then $x=b+(-a)$. For the other part, verify that it indeed satisfies $a+x=b$. Then, you're done.

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I have shown that $a+x=b$ implies $x=b+(-a)$. I've understood that this shows that the solution is unique, but I'm pretty certain that this also shows that $b+(-a)$ is the solution that will satisfy the equation. I don't get the point of substituting this expression into the original equation to verify this if I've already shown that if $a+x=b$ then $x$ must be $b+(-a)$. – Alraxite Mar 3 '13 at 0:21
Well, in this case it is obvious and seen readily that $b+(-a)$ is a solution. But, generally, this logic requires the verification as well. I can't think of but a poor example right now: Solve $\frac x{x-1}=\frac 1{x-1}$. Then, we can conlude that if there is solution, then $x=1$. But this is not a solution now.. – Berci Mar 3 '13 at 10:54
Sorry if it seems that I'm being too concerned about this but I want to understand when a proof is sufficient or not sufficient to a problem. I don't want to add unnecessary detail if I've already shown what needs to be. As for your example I would solve it like this: Clearly, the expression $\frac{x}{x-1}$ is undefined at $x=1$. So suppose $x\not=1$. Then we can multiply both sides by $(x-1)$ to obtain $x=1$. But $x\not=1$ by assumption, so there is no solution. – Alraxite Mar 3 '13 at 13:16

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