 Alright, so we're ready to introduce the standard algorithm for multiplication. So let's go back to our earlier example. We found 38 times 62, well literally this is the sum of 38, 62s, and so what I did is I found 36s, 30, 62s, 8, 62s, all together 38, 62s, and here's what I was adding together. Arithmetic is bookkeeping and more bookkeeping, and then finally here is our product, the sum of 38, 62s. And what I'm going to do is I'm going to move around the pieces to see where the standard algorithm comes from. And so if I rearrange these pieces a little bit, I can eventually get to the standard algorithm. So before we go any further, it's worth writing out the full values of all of our partial values, partial products here, and our partial sums. So here, I didn't write anything in this column because I didn't have any ones here, but I should have a zero here. This is not 180, this is 1800, 010, 01s, likewise 400s, 810, 01s, and so on. So I'll fill in the zeros there. And you know how the standard algorithm is set up. We put down one factor, 62, we put down the other factor, 38, we write that multiplication sign, and then what do I do? Well, I multiply this factor by the ones digits. So that's 8 times 2 gives me 16, that's 8 times 6, well, that's actually 8 times 60 gives me 480. And then I multiply the tens digit by the first factor here. So that's 3 times 2 is 6, wait a minute, that's really 30 times 2. So I really don't want to write this 6 here, it should actually be a 60, so I'll write that in the correct place. And 3 times 6 is 18, except that's really 30 times 60 is going to be 1800. And then I add the partial products, and again using my totals below method. Now the thing to notice here is that every one of these products here corresponds to one of the products we found here. So we did 8 times 2, well that was really finding 8 twos, and there's our 16. We did 8 times 6, that gave us 48, except that was really 480. And again that reflects the fact that it's really 8 times 6 tens. So that's 8 times 6 tens, 480 right there, and so on down the line. So every one of these partial products here corresponds to a partial product over here. And then we did the sum in exactly the same way we used our totals below method. And there's our sum of partial products, and again our actual product 2356. Now you might notice this actually isn't the standard algorithm, there's a lot more here than we usually write for the standard algorithm. What do we do? Well we generally go 62 times 8 and figure out what that is, and that's 496. So again that's combining the 8 times 2 with the 8 times 60 into a single product. And then we go 3 times 62, that is going to be 186, oh wait we have to slide that over one place because it's really 62 times 30, it's 1860. And then we add the partial products together. So we add 496, 186, let me get our answer 2356. And if I remove all the scaffolding, the process becomes an incomprehensible jumble of digit pushing. I got this 496 by doing something. I got this 186 by multiplying 3 times 62, but I had to write it over here because I can't write it down here because the 3 is over here. And well, why would you want to do that? Well the standard algorithm is worth learning because it saves space. Maybe it's not really worth learning as an independent algorithm. The amount of space saving that I get here comes at a cost of incomprehensibility. The standard algorithm, digit pushing, and yeah it saves space, but you lose the meaning of multiplication as repeated addition and arithmetic in general as bookkeeping. So maybe we should ask is it really worth saving space? Well it is reasonable to want something more efficient, so the question we might ask is can we modify the standard algorithm? So it is more compact, but it retains a connection with what we're doing when we multiply. So let's take a look at that. So if I apply the standard algorithm to the product 38 times 62, I get this. And to make the connection more clear, again the thing to remember is that we're multiplying 62 by 38, or 62 by 38. And what this suggests is maybe I want to break those factors into these components either literally writing them out, or again in the interest of efficiency at the very least mentally breaking these apart into 60 into 30 and 8. And what I did when I did the other method is I multiplied 30 by 6, 3060s, that's 1800, 32s, 120, and then 860s and 82s. And then I added everything together and I found my overall product. And as before what happened is I had 30, 62s, and 8, 62s. So all together I had 38, 62s, which is what I'm looking for as this product. And now if I want to make this more efficient, I'll remove the scaffolding. I really don't need all these little notes here. They're explaining what I'm doing, but I don't really need them once I understand what's going on. The important thing is that what's left over still retains this connection between the process of multiplication as repeated addition. The other thing it does is that there's none of these arbitrary rules like you have to write this 186 under the 3. It's 1800, you're writing it as 1800. And so our algorithm here, well, it takes up a little bit more space than the standard algorithm, but is it really worth saving two lines of space to compact everything into an incomprehensible jumble? And I would say I'd suggest that the answer is no, it's not actually worth it. And so what we have here is a modified standard algorithm that retains the connection with multiplication as a repeated addition. And in fact, this particular method of multiplying is known as the method of partial products, which is kind of a misnomer, because really this is also a method of partial products. It's just that this method makes it a little bit easier to see what those partial products are. Now, there's actually a couple of advantages to this form. One is that if I compare two multiplications both done incorrectly, one by the standard algorithm and the other using partial products, it's a lot easier to identify the error in a partial product problem, which also means it's easier to check your answer. So let's take a look at my partial products problem first. So what am I supposed to be getting? Well, this is supposed to be 20 times 30, 600, then what do I need to do? Well, I need to multiply 20 times 8, 160. I need to multiply 7 times 30. And oops, there's my mistake. That 7 times 7 times 30 is not 21. It's actually 210. So let's fix that. So I'll fix that. That also changes the sum, so I'll obliterate the sum briefly. And never assume that we only made one mistake. So the last thing we need to multiply 7 times 8 is going to be 56. And so we fixed the error there. And now we'll add things together to find our products. So this is going to be 1026. Now the important thing to notice here is how easy it is to check our work here. We just verify 20 times 30, 20 times 8, 7 times 30, 7 times 8. And we were able to check our work fairly quickly and re-add when we found a mistake. The easier it is to check our work, the more likely we are to actually do the check. And consequently, the more likely we are to write down a correct answer at the end. On the other hand, let's try to check our work for the standard algorithm. And so what do I have to do? Well, what did I do with the standard algorithm? Well, I got 7 times 38. And so this first line here, I had to figure out what 7 times 38 was. Well, that takes a bit more effort to figure that out. But I can run the computation, 7 times 38, 7 times 8 is 56. Write down 6, carry the 5, 7 times 3 is 21 plus 5 is 26. Write that down in some order there. And so there's my product, okay, there we go. And then let's see, 2 times 38, so let's see. So that's going to be 2 times 38 is 76, and that all seems correct. Maybe I mis-added about 6 and 6 is 12, 13. The products are correct here. But the thing we forgot here is that when we applied the standard algorithm, the second product has to be shifted over one place. So the actual product is going to be over here someplace. And so now I can fix that and then add to find our sum, 6 to 1000. And the thing again to notice here is that two problems arise. First of all, the actual process of checking is a lot harder. I have to redo a multiplication that is not an easy one. The other thing I have to do is I have to make sure that I've also put my partial products in the correct places. Which means that unless I understand and can apply the standard algorithm 100% correctly every time, I can make a mistake here that I will have a very hard time catching. It's a lot harder to make that same sort of mistake over here, because I'm writing down the full values of the partial product. And there's none of this. Well, you got to write this number over in a different column, stuff that we have to remember with the standard algorithm. It's important to keep in mind there is a difference between what is easy and what is familiar. And that's probably the most important lesson to learn.