On a uniqueness proof for solutions of $\ a + x = b\,$ in a group
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.
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 (excerpted below) regarding a slick proof in Max Rosenlicht's Introduction to Analysis. See also the further discission in this answer.
Alan E. Feldman [email protected] wrote:
Bill Dubuque [email protected] wrote:
Leonard Blackburn [email protected] wrote:
Alan E. Feldman [email protected] wrote:
[Now, ] F3: For any $\,a,b \in \Bbb R\,$ the equation $\,x + a = b\,$ has one and only one solution. For if $\,x \in \Bbb R\,$ is such that $\,x + a = b,\,$ then
$$\begin{align} x &= x + 0\\ &= x + (a + (-a))\\ &= (x + a) + (-a)\\ &= b + (-a)\end{align}$$
so $\,x = b + (-a)\,$ is the only possible solution; that this is indeed a solution is immediate. [OK so far, except how do we know that -a is unique?
LB: You've already gotten some useful replies. However, I am wondering why nobody remarked that you are correct and that the author has made an error (in my opinion it is very important to explicitly acknowledge this). Above, you state F3 as a theorem and give the author's proof. The author's proof is wrong since he did not first prove the uniqueness of $\,-a\,$ (the uniqueness of $\,0\,$ doesn't play a role in his proof).
BD: The author has not made an error -- the proof is correct as it is. The proof employs only the existence (not the uniqueness) of both an additive inverse of $\,a\,$ (denoted $\,-a)\,$ and neutral elt (denoted $0).\,$ Uniqueness is a corollary, the special cases $\,b = 0;\ b = a\,$ resp.
AF: This gets to the crux of the problem. How can $\,x = b + (-a)\,$ be known to be unique when $\,-a\,$ is not known to be unique? Sure, you can say the solution has to be $\,x = b + (-a),$ but if $\,-a\,$ is not unique, then how can $\,x\,$ be? Suppose $\,-a\,$ could be $3$ or $5.$ Then $\,x\,$ could be $\,b+ 3\,$ or $\,b + 5\,$ and would then not be unique. So please explain how a not-yet-known-or-shown-to-be-unique, i.e., a possibly multi-valued, $\,-a,\,$ could be added to $\,b\,$ and produce a unique value for $x.$
BD: I explained this in a prior post, but perhaps I was too terse, so I elaborate: To say that the solutions $\,x\,$ of $\,x + a = b\,$ are unique means precisely: if $\,r, r'\,$ are two solutions for $\,x,\,$ then $\,r = r'.\,$ Rosenlicht's proof shows: if addition is associative with neutral $\,0\,$ and $\,a\,$ has some additive inverse $\,-a,\,$ then any solution $\,x\,$ must satisfy $\,x = b + -a.\,$ So if $\,r, r'\,$ are two solutions, they must both be equal to $\,b + -a,\,$ hence $\,r = r',\,$ i.e. any solution are unique.
The structure of the proof is clarified if we abstract a little. Above we have a set $\,S\,$ (of solutions) which we wish to prove has no more more than one element. To prove this it suffices to prove there exists an "equalizer" $\,e\,$ for $\,S,\,$ i.e. $\,x \in S\Rightarrow x = e.\,$ Rosenlicht's proof shows that $\,e = b + -a\,$ is an equalizer for the set $\,S = \{x\ :\ x + a = b\}.\,$ It doesn't matter that the proof made some arbitrary choice while constructing $\,e,\,$ since to equalize the set $\,S\,$ all we require is the existence of one such $\,e.$