 So the topic of vector spaces is a good one for introducing proof. And again, it's worth pointing out a couple of important features about proof in mathematics. One of the reasons that we do proof, a proof result is unquestionably true, and this is actually the least important reason why we do proofs. Again, nobody bothers to find a proof for something they don't already believe to be true. More importantly, one of the things that proof does is it reminds us of things that we should know, and it clarifies our understanding of those things, and in particular, any place where our understanding is lacking, the process of proof will identify that. In other words, proof is a good way of studying mathematics. And then finally, the last important and possibly the most important thing about proof is that it suggests new topics for investigation. It suggests new questions. A good proof raises more questions than it answers. And this is good because by raising new questions, it guarantees that mathematicians will continue to have things to do, and there will be continued employment for mathematicians. Well, the most important thing to remember about proof, it's all about the journey. It's not about the destination. So let's take a nice simple problem. We want to prove, or possibly disprove, we don't yet know whether we should believe this or not, that the set L of linear expressions of the form AX plus B, where A and B are real numbers, and we use the ordinary rules of polynomial arithmetic, we want to prove or disprove that the set forms a vector space. And again, here's where the study aspect of proof comes in. I can't prove or disprove this unless I know what a vector space is. And so that means we want to go back to those 10 requirements for what makes up a vector space and verify that either the set meets all 10 requirements or that there's one of these requirements that the set does not meet. All right, so let's take a look at that. So we'll run through our checklist. So one of the first things that we need to make sure of is we need to make sure that this set is closed under addition and also under scalar multiplication. So let's verify that if I have two things that are in my set, so F is AX plus B, A and B and R, G is CX plus D, also those things are in R, and my addition or ordinary rules of polynomial arithmetic, I'll add them, and what I'd like to do is I'd like to verify that their sum is going to be in L. Well, let's go ahead and do that. So if F is something in L and G is something else in L, then, well, let's see, F plus G, I get to use the ordinary rules of polynomial arithmetic, so F plus G is going to be A plus CX plus B plus D, and I claim that this is going to be in L. Now, note that we have just simply claimed that this is going to be in L. We should at least elaborate or at least think about elaborating on this. A, B, C and D are real numbers, and I have the ordinary rules of arithmetic A plus C. If I add two real numbers, that's a real number. If I add two real numbers, that's a real number. So this is a real number, this is a real number. So I have real number X plus real number, and my set L consists of the linear expressions of the form real number X plus real number. Now, we don't have to elaborate on this, provided we are willing to offer the guarantee that the claim is true. And we're offering that guarantee. We have no choice of that when we write this down. When you write this, you are offering that guarantee that the claim is in fact true, and before you claim that something is true, try as hard as possible to find a case where it's false, because if you don't, somebody else will, and they won't be as nice about it as you could be. So let's see, what's our second requirement for being a vector space? So we've proved closure under addition. We want to prove closure under scalar multiplication. So I'm going to take something that's in L, I'm going to take some real number, and I want to check to see if C times F is an element of L. So C times F, well again, I know what F is, it's AX plus B, C times F, ordinary rules of polynomial arithmetic, CF is CAX plus CB, then CF is going to be in L. And again, here we have that implied assumption, C times A is a real number, C times B is a real number, so this expression is real number times X plus real number. Now we want to check the remaining properties. So let's first of all show commutativity. We want to show that F plus G equals G plus F. Well, let's go ahead and keep our reference functions, F and G, here. And here's a useful strategy and proofs. If I want to show that something is equal to something else, you can always write down one side of the equals. If I want to show F plus G equals G plus F, I can always write down the first part. What I have to be careful with is getting to the second part. But what I can do is I can leave a little bit of space and write down where I want to get to. And my goal is to try and fill in the gaps from here to here. If it's possible to do that, I've completed the proof. Alternatively, I may find that this is not going to be equal to this and so I can just put a cross through here, F plus G not equal to G plus F and I will have disproven the statement. So again, another useful strategy and proofs. Remember the definitions. In this particular case, I've defined F to be AX plus B. I've defined G to be CX plus D. So I can drop those in here. And definitions do go both ways, which means I can also use definitions to work my way backwards one step. Here, I have G. Well, I know what that is. I have F. I know what that is, so I can substitute those in. And that seems to be a reasonable thing for me to do. It seems that I can go from this line to this line because I'm using the ordinary rules of polynomial arithmetic and these two lines give me exactly the same thing. And again, you do want to ask yourself, is there any possibility that this is false? Is there any possibility that I can't go from here to here? And in this case, can I go from here to there? And again, because I'm using the ordinary rules of polynomial arithmetic, I can do that. So I've completed my proof of commutativity. So let's see. So far, we have done closure under addition. That's good. We've done closure under scalar multiplication. That's good. We've done commutativity of addition. That's good. So far, now we have to check associativity of addition. So let's go check that. So now, again, I want to prove that F plus G plus H is the same as F plus G plus H. And again, I want to prove something equals something so I can start by writing down the left-hand side. Now, I do need to introduce a new function H equals EX plus F. But now I have F plus G plus H. I know what F is. I know what G is. I know what H is. And again, what I want to do is I want to work my way down towards this expression here. So here's where I start. Here's where I end. And I want to fill in the gaps. So let's see if we can do that. So I know F, G, and H. I can write them here, F, G, and H. And also, I know F, G, and H. And I can write them from here. And again, my ordinary rules of polynomial arithmetic do allow me to go from this line to this and to regroup. And so there's my proof of associativity. And so I can check off associativity. Now, here's where things start to get a little bit challenging. We need to look for a zero vector. We need to find a zero vector such that for anything we choose, if I add it, I get the thing that I started with. And then once I find the zero vector, then I need to find the additive inverse. Now, this is where we have to start to be a little bit creative. So let's think about that. So I want to find an expression, which we're going to designate as zero, even though this isn't, should not be read as a number zero. This is the zero vector. Or if I add it to an expression, I get the same thing. And after a little bit of thought, again, we have to be creative here. There's no way around this. We realize, oh, how about this expression, zero x plus zero? It is in L because it is of the form A x plus B where A and B are real numbers. And this is the zero vector. If I add this to this under the ordinary rules of polynomial arithmetic, I do get A x plus B. Now that I know what the zero vector is, I can see if every element has an additive inverse. And so what we can do is to see if we can try and fill in the gaps. In order for this to have an additive inverse, I have to take an expression. I have to be able to find something that I can add to it that's going to give me the zero vector. So let's see, ordinary rules of polynomial arithmetic. Jeopardy theme song plays for a little bit. Da, da, da. You get the idea. And lo and behold, we come up with this as our possibility minus A x plus minus B. And under the ordinary rules of polynomial arithmetic, if I add these together, I do in fact get zero. And so that tells me that minus A x plus minus B is in L because those are both real numbers and that's going to be our additive inverse. So we'll check off a zero vector and our additive inverse. So what's next? Well, we have this multiplication by one property. One times our vector times any vector should give us the same thing. So let's check that out. So let's see, we'll check multiplication by one. And so I'll take something that's in our vector space and I'll find one times that. And again, we're using the ordinary rules of polynomial arithmetic. So when I do that, I do in fact get A x plus B. I get exactly what I started with. So multiplication by one is satisfied. Assessivity of scalar multiplication. So if I take two real numbers A and B, A times Bv is going to be the same as AB times V. So again, we'll go ahead and check for that. We'll write down where we start and where we want to end. I want to start P times Qf. I want to end PQ times F. And then I'll see if I can fill in the gaps. So again, F is A x plus B. And so this is P times Q times A x plus B and I had the ordinary rules of polynomial arithmetic. And the ordinary rules do allow me to regroup PQ times A x plus B and that gets me PQ times F. Assessivity of scalar multiplication also holds. And now we have our two distributive property. The distributive property of scalars over vectors. I want a real number times the sum to be the real number times the individual vectors. So let's go ahead and write that down. So again, I want the real number times the sum to be the real number times each of the vectors individually. Here's where I start. Here's where I want to end. I want to fill in the gaps. So again, under my assumption, F is A x plus B, G is C x plus D. That's going to be this. And again, under the ordinary rules of polynomial arithmetic, I do have that distributive property. I can distribute the P among the two factors there. And so now this is P times F plus P times G and this is where I do want to end up. So my distributive property of scalars over vectors holds. And then finally, the last thing I want to check is P plus Q times F. That's the distributive property of vectors over scalars. If I have the same vector multiplied by two different scalars and add it, I can combine the two scalars. And so I have P plus Q times F. I would like that to be P F plus Q F. So again, I want to start here, want to end there and I'll see if I can fill in the gaps. So again, F is A x plus B. And so this is P plus Q times A x plus B. And again, the distributive property under the ordinary rules of polynomial arithmetic, which I am using, which I'm allowed to use by the definition of what this space holds. And that gives me my final result. And it's always a good idea to summarize our results because we had quite a bit of information there. And so we might say something like the following. The preceding shows that L under the ordinary rules of polynomial arithmetic satisfies all the required properties for a vector space. Now, looking forward, at some point, you're going to graduate and you're going to get your diploma. And that diploma will actually say something very similar to this. It'll say something to the effect of let everybody who cares know that this person, name of person, has satisfied all of the requirements by the trustees, regent, et cetera, et cetera, and is therefore accorded the properties of Bachelor of Science, Bachelor of Arts, Master of Science, whatever the degree is going to be in. So this is very much like a diploma. Once you've satisfied all the requirements, you cannot be denied the accolade that comes with it.