Hello everyone. In this screencast, we're going to talk about uniqueness. So I have two examples here that are usually found inside Group Theory and Ring Theory, but I'm going to use them just to emphasize the purpose of uniqueness, and I'll do this over the real numbers so that we have an idea.
Question $1$ Show that there's a unique real number $y$ such that $xy=x$ for all real numbers $x$.
So this example, you can kind of intuitively tell yourself, "Okay this is trying to show you that there's a unique $1$ element" right? $y$ here's supposed to take the value $1$ basically.
Question $2$ Show there is a unique real number $y$ such that $x+y=x$ for all real numbers $x$.
This is sort of saying there's a unique $0$ element in the real numbers. That's intuitively what you should read these as but the proof of these things is going to be a little different.
So, how are we proving these things? It doesn't say anything with existence so I mean, we know that they exist, but it says about uniqueness. Well okay, so I guess existence is sort of implied but again, we already talked about $1$ and $0$ being there.
Notice that $y=1$ is a possible value for $y$. So let's just write that sentence in just so that we can talk about the more important part of this pencast, which is uniqueness.
Now, suppose there are two values say, $y$ and $z$ such that $xy=x$ and $xz=x$ for all real values $x$. Well now they're both equal to $x$, so I should be able to subtract the two equations.
Subtracting yields $xy-xz=0$, and factoring gives $x(y-z)=0$, which holds for all real $x$.
So if we take $x$ to be non-zero, then, what does this mean? Well it must mean that $y-z=0$. Since this has to hold for every single value of $x$, you could plug in, let's say, $x=1$ or $x=2$ it has to work for everything, right? So we really do know that $y-z=0$, and hence $y=z$.
Again, this is valid because we have to have this hold for every single real value of $x$ so it has to hold for, in particular, non-zero values of $x$.
So the second one you're going to just go the same way, right, so what's the key idea here with uniqueness? Now suppose there are two values, if you want to prove something's unique, suppose two exist, and then show that that can't happen, or show that if you have two of the values then they must be equal, okay?
So that's what we did there. I'm going to probably write this one the same way, but again, you can also derive a contradiction by assuming that $y$ and $z$ aren't equal and using that somehow.
So again, notice that $y=0$ is a possible solution. Now suppose that there exists a $y,z\in\mathbb{R}$ such that $x+y=x$ and $x+z=x$. Subtracting these yields $y-z=0$ and hence, $y=z$.
So the second example is a lot easier, but that's okay. Again, it's the point of these questions that's important, right? So in the second example, again, $y=0$ is a possible solution. Now suppose that we have two values, you subtract them, and then you get $y-z=0$, and hence $y=z$.
Again, I think these are just the types of examples for uniqueness. Like I said at the beginning of the screencast, these will make more sense in Group Theory and Ring Theory, if you ever take these classes, but because they hear that proofs aren't quite as easy as they are here, but nonetheless this gives you an idea of how to show that something is unique. Assume that there are two of them, and then show either that they must be equal, or you assume that they're not equal and derive a contradiction. Hopefully that gives you an idea, thank you very much for listening.