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While reading I came across Uniqueness Proofs. Where a theorem asserts the existence of a unique element with a particular property. In order to prove this two steps are needed, Prove existence and Prove Uniqueness. The example given is Show that if $a$ and $b$ are real numbers and $a ≠ 0$, then there is a unique real number $r$ such that $ar + b = 0$.

I could do the existence portion. $r = -\frac{b}{a}$

I would like some clarification on proving the uniqueness part.

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 $r = s$. Then it says that this means if $s ≠ r$ then $as + b ≠ 0 $and that this establishes uniqueness.

I guess my issue is 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? One example I thought of where it wouldn't be unique I wasn't able to follow the same steps to disprove. Consider this $n^2 = 4$

following the previous example suppose that $s$ is another real number such that $s^2 = 4$

$\implies n^2 = s^2$

square root both sides $n = s$

now in the other one they just jumped to the conclusion that this means if $s ≠ n$ then $s^2 ≠ 4$

however... -2 and 2 could fill meaning this example does not have a unique solution. Could someone please give some clarification so I can better understand uniqueness proofs.

user2205293
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  • Do you mean $a\neq 0$ (as opposed to the factorial of $a$ being zero)? – tabstop Jan 26 '14 at 02:24
  • Yes that is what I meant, sorry I couldn't find how to get that formatting and instead used != but the suggested edits to include the examples in dollar signs moved the placement. It should be fixed now. Curious how did you go about getting the ≠ I just copied yours for the edits. – user2205293 Jan 26 '14 at 02:31
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    $\neq$ gives $\neq$. – tabstop Jan 26 '14 at 02:34

4 Answers4

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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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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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for the case of n^2=4, consider this:

Suppose there is a real number s such that s^2=4. Given n^2=4, it follows n^2=s^2. And you could not just take the square root here. Instead, n^2-s^2=0. (n+s)(n-s)=0. It implies n=-s or n=s. There are two unique solutions.

quantif
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TCKwok
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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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