 In this video, I wanna prove some important properties, very elementary properties of rings. So let R be a ring and let R, little R and little S be just arbitrary elements of that ring. So the first property says that if you multiply any element of the ring by zero, you get back zero. That zero has this dominance property. Sometimes it's called an absorption property that zero is dominant that anything times zero gives you back zero. This is a fact in any ring. Likewise, if you take R times negative S, now remember negative S does not mean negative one times S. What this means is that negative S is the additive inverse of R. So be very careful with that in this situation. If you take R times negative S, that is if you take R times the inverse of S, then that's the same thing as the inverse of R times S which is equal to the inverse of RS. So these additive inverses can kind of be moved around. Likewise, if you take negative R times negative S, this gives you RS, this gives us the principle that a double negative is actually a positive. In which case what we mean is if you take the additive inverse of R times by the additive inverse of S, this is the same thing as the product of R and S themselves. So let's see the proof of these arguments right here, the proof of these statements. So the first one, why is multiplication by zero always give you zero? Well, the significance here is that zero is the additive identity. So note, if I take any element A, don't know why I switched my mind to A, could have been R but whatever, it's just an arbitrary element. If you take zero times A, well since zero is the additive identity, that means zero plus zero is equal to zero, right? The additive identity is idempotent as an element. You can replace zero with zero plus zero. But since we're in a ring, A will distribute over the ring addition and give us zero A plus zero A. Now, this is an equation now living in an additive group. We could subtract zero A from both sides. We could subtract zero A because whatever zero A is at the element of the ring, it has an additive inverse. In which case those then cancel out, they cancel out over here. In which case as we add this additive inverse, it cancels because we're in a Boolean group and we're gonna get the important observation that zero is equal to zero times A. And therefore we get zero equals zero A. That then proves this statement. And we used in this argument right distribution. Using left distribution, a similar argument can be constructed to show that A zero is equal to zero. And this proves then property A. So zero is a dominant element inside of our ring. The next one here, let's, we wanna prove that R times the inverse of S is equal to the inverse of R times S, okay? The way we're gonna argue about this is that as these are additive inverses in an A Boolean group, we can use the uniqueness of additive inverses to prove this statement. So if I take AB, again, I don't know why I switched from RS to AB, but whatever, that's the name of the symbols here. If you take AB plus negative A times B, we can use right distribution to basically factor out the B. In which case you're gonna get A plus negative A times B, which has A and negative A are additive inverses. A plus its inverse will give you zero and zero B like we saw a moment ago is equal to zero. So we get that AB plus negative AB equals zero. In which case then negative AB is acting like an additive inverse of A times B. And if it walks like an inverse and quacks like an inverse, then it is the inverse, because inverses are unique. So we then can infer that negative AB is equal to the inverse of negative A times B. So this would be like negative A times B like that. I can drop the parentheses because this property B makes it unambiguous. Negative A times B is the same thing as negative the product A times B. And we accomplish this using right distribution because we factored here and we use uniqueness of inverses. We use that zero times B is equal to zero. Okay, we can move everything to the left. We have left distribution. We have B zero is equal to zero. We have uniqueness of inverses. So by a similar argument, we can show that A times negative B is equal to negative AB. So we can basically move the inverse around and that's perfectly fine inside of a ring, okay? And so then negative A times negative B, the last property, we wanna show that double negative is actually a positive. The inverse of A times the inverse, the inverse of A times the inverse of B, well, by property B, you can basically take the inverse of A out in which case we get the inverse of A times B, negative B there, right? So that is you're taking the inverse of the product of A times negative B. But then we can use property B twice and we pull it out and we end up with the inverse of the inverse of the product of A times B, all right? So we use property B twice. We can pull out the negative twice essentially. Now in a group, because inverses are unique, the double inverse is actually the original element. So if you take the inverse of the inverse, you get back, it's just the identity function. So negative negative AB is equal to AB and thus proving property C. So this shows us that in a ring, you get this double negative is a positive property. This thing we're quite used to. And that's because the rings we're most used to are like the ring of integers, rational numbers, real numbers, complex numbers. This is a property for every ring. So that's where we get this double negative business. Now I mentioned also in this video that when you see something like negative C, I don't mean negative one times C. What I mean is the additive inverse of, why am I saying C here? I'm talking about the negative inverse of S right here. But it turns out the reason why we call additive inverses negative S or negative R is because it actually is multiplication by negative one. Cause after all, if you're in a ring of unity, if you don't have a one, then it doesn't apply. But if you are in a ring with unity, that is you do have a number one, then we can actually argue that negative S is equal to negative one times S. And it's very simple to see this because you start off with negative one times S. Negative one times S means you take the additive inverse of one, the unity, and you times it by S. But by property B, this is equal to negative one times S. That is we're taking the additive inverse of one times S as one is the unity, you just get S. And so we see that negative one times S is equal to negative one if you're in a ring with unity.