Understanding Proof of Uniqueness

[Mr Davis 97]

I'm trying to really get a grasp on proofs of uniqueness.

Here is a model problem: Prove that $x=-b/a$ is the unique solution to $ax+b=0$.

First method:
First we show existence of a solution: If $x = -b/a$, then $a(-b/a)+b = -b+b = 0$.
Now, we show uniqueness: If $ax+b=0$, then $ax = -b$ and so $x = -b/a$.

Second method:
First we show existence of a solution: If $x = -b/a$, then $a(-b/a)+b = -b+b = 0$.
Now, we show uniqueness: Suppose that $y$ and $z$ are two different solutions to the equation. Then $ay+b=0$ and $az+b=0$, so then $ay+b=az+b$, which implies that $y=z$.

What is the difference between these two methods? Are they both valid? Which is preferable?

 

[member 587159]

Both are correct.

In the first method, you show that if there is a solution, it MUST be $x = -b/a$. So there is only one possibility for the solution, and your existence shows that it is a solution, so there is at least one solution. So, you can conclude that there is precisely one solution.

In the second method, you show that any two solutions must be the same. So again, there can be at most one solution.

Both methods are used frequently, none of the two should be preferred above the other imo.

 

[fresh_42]

I'm trying to really get a grasp on proofs of uniqueness.

Here is a model problem: Prove that $x=-b/a$ is the unique solution to $ax+b=0$.

First method:
First we show existence of a solution: If $x = -b/a$, then $a(-b/a)+b = -b+b = 0$.
Now, we show uniqueness: If $ax+b=0$, then $ax = -b$ and so $x = -b/a$.

Second method:
First we show existence of a solution: If $x = -b/a$, then $a(-b/a)+b = -b+b = 0$.
Now, we show uniqueness: Suppose that $y$ and $z$ are two different solutions to the equation. Then $ay+b=0$ and $az+b=0$, so then $ay+b=az+b$, which implies that $y=z$.

What is the difference between these two methods? Are they both valid? Which is preferable?

First of all, we don't need existence to show uniqueness. Uniqueness means: if exist, then unique. So it doesn't require the existence.

The difference between the two proofs is (second parts), that the first doesn't show uniqueness. It shows instead: If a solution exists, then it is necessary that $x=-b/a$. Theoretically we could still have two different values for $x$, say $x=y$ and $x=z$. Now we apply transitivity of equality and get by our necessary condition: $x=-b/a \text{ and } y=-b/a \Rightarrow x=y$. This step would formally be needed in the first version.

The second version shows directly that two solutions have to be the same by starting with transitivity: $ay+b=0\text{ and }az+b=0 \Rightarrow ay+b=az+b$.

 

[member 587159]

Theoretically we could still have two different values for $x$, say $x=y$ and $x=z$. Now we apply transitivity of equality and get by our necessary condition: $x=-b/a \text{ and } y=-b/a \Rightarrow x=y$. This step would formally be needed in the first version.

I disagree with what you say here. How would it be possible to obtain 2 values? That proof really says that "if there is a solution, it must be -b/a"

 

[fresh_42]

I disagree with what you say here. How would it be possible to obtain 2 values? That proof really says that "if there is a solution, it must be -b/a"

This is a necessary condition. You still have to show somehow, that it is a unique description of a number or at least mention it - which it isn't by the way, but that's another discussion. To me there is a formal step missing, as from necessity alone cannot be concluded uniqueness for elements which fulfill the condition. I mean the entire subject is about logic, and I think in logical terms such a step is needed. The more as the second version of the proof also uses it, namely (hidden) at the start. In real life the entire example is ridiculous, but in general, a necessary condition doesn't guarantee anything, as e.g. in $y=\sqrt{x}$ over $\mathbb{R}\quad x>0$ is a necessary condition, but it doesn't say anything about uniqueness. As mentioned: formally.

And by the way, the entire example neglects the case $a=b=0$ in which case uniqueness will be lost.

 

[Mark44]

Theoretically we could still have two different values for $x$, say $x=y$ and $x=z$. Now we apply transitivity of equality and get by our necessary condition: $x=-b/a \text{ and } y=-b/a \Rightarrow x=y$. This step would formally be needed in the first version.

I disagree with what you say here. How would it be possible to obtain 2 values? That proof really says that "if there is a solution, it must be -b/a"

It seems to me that @fresh_42 is doing a proof by contradiction here. At the start, he's assuming that there are two solutions, x = y and x = z. Subsequent work shows that there really is only one solution, which is a contradiction of the assumption that there are two solutions. Thus, along with the fact that a solution exists, that solution must be unique.

 

[member 587159]

It seems to me that @fresh_42 is doing a proof by contradiction here. At the start, he's assuming that there are two solutions, x = y and x = z. Subsequent work shows that there really is only one solution, which is a contradiction of the assumption that there are two solutions. Thus, along with the fact that a solution exists, that solution must be unique.

He never says that the solutions must be distinct, so there is no contradiction until this is explicitely mentioned.

 

[fresh_42]

Whether you use (without mentioning) the fact that $|\{\,-\frac{b}{a}\,\}|=1$ or (by mentioning) use transivity of equality doesn't make a difference. In either case you add another statement to $x=-\frac{b}{a}$. You must somehow explain, why this necessary condition already describes the solution completely. This is obvious in this case, but not in general. To deduce a necessary condition covers only one aspect of the solution, because it is merely necessary. That it covers the entire solution - in this case, not in general - needs another statement.

 

[Mark44]

He never says that the solutions must be distinct, so there is no contradiction until this is explicitely mentioned.

But @fresh_42 essentially said that, with his phrase

Theoretically we could still have two different values for x, say x = y and x = z

(underscore added by me).
"Different values" says "distinct" to me.

 

[fresh_42]

My main issue was, that a necessary condition is generally not sufficient to conclude uniqueness. It is just here in this trivial example the case, that the solution appears unique to everybody.

If we had e.g. $x^2=4$ we can deduce $|x|=2$ from that. This is a necessary condition. Now everybody sees, that this is not sufficient to claim uniqueness. So formally, there has to be another argument, why a solution is unique. To derive a necessary condition is always just that: necessary. E.g. we would also have to verify, that $x=-\frac{b}{a}$ is actually a solution, as both proofs in the OP do.

So, although this example wasn't actually difficult and thus not a good one, I want to sharpen the readers' minds, that a necessary condition alone is usually not enough. Often we use equivalent phrases between condition and a necessity, so sufficiency is already given. But as soon as examples get only a bit more complicated: possible zero factors, inequalities, squares etc. equivalences tend to get lost. And in those cases, one has to know, what the difference between a necessary condition and a solution is.

 

[WWGD]

I don't know if this is OT , but don't we need to assume we are either working on a group or with an elements that are units in a ring? 1/a , which is , I assume here, the multiplicative inverse of a, may not always exist, depending on the setting.

 

[Mark44]

I don't know if this is OT , but don't we need to assume we are either working on a group or with an elements that are units in a ring? 1/a , which is , I assume here, the multiplicative inverse of a, may not always exist, depending on the setting.

I think it's safe to assume that we're dealing with real numbers. In any case, the question is about existence and uniqueness of solutions. If this had been about groups or rings, I'm sure that would have been stated. So yes, OT.

 

[WWGD]

I think it's safe to assume that we're dealing with real numbers. In any case, the question is about existence and uniqueness of solutions. If this had been about groups or rings, I'm sure that would have been stated. So yes, OT.

What I meant is that it is a good idea, if one expects to do some serious ( grad-level) Math at some point, to get used to specify the setting: algebraic object, type of number, etc one is working with. But that is up to Mr. Davis.