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.
$\endgroup$ 34 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ 9 $\begingroup$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).
$\endgroup$ 1 $\begingroup$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.
$\endgroup$
