 Let's take a look at some of the properties of vector arithmetic. So far, we know, well, none of the properties of vector arithmetic. So in the spirit of mathematics, let's play around with the properties of vector arithmetic and see if any interesting results show up. For example, one useful feature, which our underlying scalar field has, is that addition is commutative. So we might ask whether vector addition is commutative. And so we'll prove, or possibly disprove, that the vector u plus v is equal to the vector v plus u. So a good way to structure simple proofs like this is to think of them as a bridge. We have some starting point, in this case the sum of the vector u plus v, and we have some ending point, the vector v plus the vector u. What we want to determine is whether or not it's possible to go from our starting point, u plus v, to our ending point, v plus u. So we can structure it this way. We'll put down our starting point, and we'll put down our ending point, separated by some distance to give us some space to work. We then want to build a bridge from our starting point to our ending point. And if we can do that, that bridge is going to be our proof. Now remember, definitions are the whole of mathematics, all else is commentary, and so one of the most important things we can do is we can pull in any relevant definitions. And in this particular case, because we're looking at the sum of two vectors, we might want to pull in our definition for the sum of two vectors. And this tells us that the sum of two vectors is going to be found by adding the corresponding components of the two vectors. So if my vector u has components u1 through un, and my vector v has components v1 through vn, the sum u plus v is going to be the vector whose components are u1 plus v1, u2 plus v2, and so on. Now a useful thing to remember is that you can build a bridge by starting at both ends and working towards the middle. And in this case, at the end of the bridge, I have the vector sum vector v plus vector u. And my definition tells me that if vector v has components v1 through vn, and vector u has components u1 through un, then this sum is going to be the vector with components v1 plus u1, v2 plus u2, and so on, up to vn plus un. And now we have that final step. Can we bridge this gap that takes us from where we left off, coming from u plus v, and where we got to, starting at v plus u? And our vector components look very similar, and the only thing that we have to worry about is whether or not u1 plus v1 is the same as v1 plus u1, whether u2 plus v2 is the same as v2 plus u2. And we're in luck, because remember these vector components came from a scalar field, and in the scalar field, addition is commutative. And so u1 plus v1 is equal to v1 plus u1, u2 plus v2 is equal to v2 plus u2, and so on. And that means we are able to bridge the gap, which means we're able to go from our starting point to our ending point, and that completes our proof. Well that worked out really nicely. Let's see if we can do something else. How about scalar associativity? So suppose I have two scalars in R and some vector in Rn. So I could multiply our vector v by b, and then multiply the result by a. Or I could multiply a and b together, and then multiply the result by v. And the question is, do I get the same thing? Well, let's check it out. Again, since this proof seems to involve scalar multiplication, let's go ahead and pull in our definition of scalar multiplication. And again, we want to build a bridge from our starting point to our ending point. So I'll write down a times the quantity b times the vector. And a definition of scalar multiplication says this is going to be a times the vector where b is multiplied by every component. And that's going to give me a times b times every component. Again, we'll work our way back a step. So ab times the vector v is going to be ab times each component of v. And in this case, my next step takes me directly to where I finished. And so I'm able to start at a times the quantity b times the vector v. And I'm able to go directly to the quantity ab times the vector. And so I have what can be called scalar associativity of scalar multiplication.