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While reading, I came across uniqueness proofs, in which the existence of a unique element with a particular property is proven by taking two steps: prove existence and prove uniqueness.

For real numbers $a$ and $b$ with $a ≠ 0,$ show that there is a unique real number $r$ such that $ar + b = 0.$

Okay, I can do the Existence part: $r = -\frac{b}{a}.$

Suppose that $s$ is a real number such that $as + b = 0.$ Then $ar + b = as + b,$ where $r = -\frac{b}{a}.$ You subtract $b$ from both sides and divide both sides by $a$ to get $r = s.$ This means that if $s ≠ r$ then $as + b ≠ 0 ,$ and this establishes uniqueness.

How exactly does this part prove Uniqueness? When would placing some random variable in the same spot as the previous not end with the two variables being the same?

Consider $n^2 = 4.$ Following the previous example, suppose that $s$ is another real number such that $s^2 = 4.$ Then $n^2 = s^2,$ so $n = s.$ Also following the previous example, this means that if $s ≠ n$ then $s^2 ≠ 4.$ But of course this example does not have a unique solution, but instead two solutions $-2$ and $2.$ Where is my misunderstanding?

ryang
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user2205293
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4 Answers4

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Your proof concerning uniqueness is actually correct. You can prove it in a different way, more clearly:

Suppose $\exists s,r \in \mathbb{R}$, where $s\neq r$, but $as+b=0=ar+b$. Your computation establishes that nonetheless, $r=s$, which contradicts our assumption that $r\neq s$. This means our assumption must have been wrong: we conclude that there do not exist $s,r \in \mathbb{R}$ where $s \neq r$ but $as+b=0=as+r$. But given that $r = \frac{-b}{a}$ works, we conclude that we always have at least one $r$, but we never have two or more such $r$s: so if the number of admissible $r$s is always at least one and always less than two, it must be one: hence, uniqueness.

Newb
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  • Could you provide the same explanation but for the case of $n^2 = 4$ I'm missing what happens when the solution is not unique. – user2205293 Jan 26 '14 at 02:44
  • How exactly does the case of $n^2 = 4$ relate to $ar+b=0$? – Newb Jan 26 '14 at 02:46
  • It has nothing to do with the example of ar + b = 0 It's more I'm trying to understand when using the method of proving uniqueness what happens when you use it on something that is not unique. I want to see an example of it not being unique and the uniqueness proof failing. I don't quite understand that portion so I don't feel I have a full understanding of the uniqueness proof. – user2205293 Jan 26 '14 at 02:48
  • Just to try to explain what I am asking, if I use the same steps as in the $ar+b = 0$ example it would lead me to believe that $n^2 = 4$ has a unique solution as well. But wouldn't -2 and 2 both satisfy making the solution not unique? So I feel like I'm missing something and would like to see an example of it failing to fill in the gap. – user2205293 Jan 26 '14 at 03:02
  • In a uniqueness proof, we try to establish that two elements have to necessarily be the same. Square root is a funny operation that obviously doesn't really work for this purpose, because it restricts to non-negative values. Try to think of an example that doesn't involve taking roots. – Newb Jan 26 '14 at 03:05
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    The problem with the example $n^2=4$ is that we don't know what assumptions are placed on $n$. If $n$ is assumed to be a positive integer then there is only one solution to $n^2=4.$ If $n$ can be any real number then there is more than one solution, as is demonstrated by choosing $n_1=2$ and $n_2=-2$ and verifying that both are solutions where $n_1\neq n_2$. – R R Jan 26 '14 at 03:19
  • This actually explained my issue, I was forgetting that sqaure root changes the domain to positive real numbers. Meaning the proof was still correct for $n^2 = 4$ example. The only other example I could think where there would be non-unique would be with inequalities. Like 3 >= x + 3 < 4 would be unique because you can only add by 0 but x + 3 < 5 wouldn't be unique. I'm not sure how proofs would be used with this? – user2205293 Jan 26 '14 at 03:24
  • It's not clear to me what you're asking here. – R R Jan 26 '14 at 03:40
  • The last comment was put before I saw your answer. No need to worry. – user2205293 Jan 26 '14 at 04:09
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Consider $n^2 = 4.$ Following the previous example, suppose that $s$ is another real number such that $s^2 = 4.$ Then $n^2 = s^2,$ so $n = s.$ But of course this example does not have a unique solution, but instead two solutions $-2$ and $2.$ Where is my misunderstanding?

$n^2=s^2\implies n=\pm s\kern.6em\not\kern-.6em\implies n=s.$

For real numbers $a$ and $b$ with $a ≠ 0,$ show that there is a unique real number $r$ such that $ar + b = 0.$

Okay, I can do the Existence portion: $r = -\frac{b}{a}.$

Suppose that $s$ is a real number such that $as + b = 0.$ Then $ar + b = as + b,$ where $r = -\frac{b}{a}.$ You subtract $b$ from both sides and divide both sides by $a$ to get $r = s.$ This means that if $s ≠ r$ then $as + b ≠ 0 ,$ and this establishes uniqueness.

How exactly does this prove Uniqueness? When would placing some random variable in the same spot as the previous not end with the two variables being the same?

To prove that a given property $\boldsymbol P$ is satisfied by just one object:

  1. Show that $$\exists x\,P(x),$$ that is, there exists some object, $c,$ that satisfies $P$

    (in other words: at least one object satisfies $P);$

  2. show that $$∀x{,}y\;(P(x)∧P(y)\implies x=y),\tag2$$ that is, whenever two objects satisfy $P,$ they turn out to be the same object

    (in other words: up to one object satisfies $P);$

    note that if nothing satisfies $P,$ statement $(2)$ is vacuously true.

To be clear: uniqueness (which is identical in meaning to "unique existence") is not actually proven in part 2 alone. It is the combination of parts 1 and 2 that finally establishes that object $\boldsymbol c$ is uniquely identified by property $\boldsymbol P.$

ryang
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To show an object is unique one approach (the one taken here) is to assume that there is a second object that satisfies the given conditions. Then if you can show that this second object is actually the first object then you've shown that all objects that satisfy the condition are identical. In particular, by supposing that $s$ and $r$ are both arbitrary solutions to $ax+b=0$ and showing that $s=r$, you've shown that every solution to $ax+b=0$ is identical, i.e. the solution is unique (if it exists). Note that you can show uniqueness without showing existence.

R R
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That fact (that if you divide equal real numbers by equal real nonzero numbers, you get equal real numbers) may have been earlier in your book, or they may have assumed you know it. It's certainly true for the basic arithmetic operations (addition, subtraction, multiplication, and division), and comes from the fact that the operations are invertible (with zero a special case for multiplication/division).

tabstop
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