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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.

Bill Dubuque
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rsteckly
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2 Answers2

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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.

Adrián Barquero
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  • So proof by contradiction? But what about the given argument, is it really air tight? – rsteckly Dec 23 '10 at 23:13
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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. – Adrián Barquero Dec 23 '10 at 23:16
  • án: But it's redundant to give such a uniqueness proof - see my answer. – Bill Dubuque Dec 24 '10 at 19:57
  • @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. – Adrián Barquero Dec 24 '10 at 20:39
  • 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. – Bill Dubuque Dec 24 '10 at 21:16
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Simply use the (additive group) axioms to prove $\, a + x = b\iff 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) about a proof of this form in Max Rosenlicht's Introduction to Analysis. See also the further discussion in this answer.

AF = Alan E. Feldman spamsi...@yahoo.com wrote:

BD = Bill Dubuque w...@nestle.ai.mit.edu wrote:

LB = Leonard Blackburn blac...@math.umn.edu wrote:

AF = Alan E. Feldman spamsi...@yahoo.com 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 also 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: that solutions $\,x\,$ of $\,x + a = b\,$ are unique means precisely this: $ $ if $\,r, r'\,$ are two solutions for $\,x,\,$ then $\,r = r'.\,$ Rosenlicht's proof shows: $ $ if addition is associative with some 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 both equal $\,b + -a,\,$ so are equal, i.e. solutions 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 choices while constructing $\,e,\,$ since to equalize the set $\,S\,$ all we require is the existence of one such $\,e.$

Bill Dubuque
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