I think it's time to talk a bit more about division. In my last math diary, Fundamental Understanding of Mathematics XXVI, I talked about prime factorization -- the peculiar fact that all integers are either primes, or can be produced by multiplying primes. I outlined a way to prove this, and got into a bit of theoretical, rather than applied mathematics. Since this series is also about teaching math, I think I need to go back and do some more practical stuff.
Bear with me on this one. Rather than try to do a long division problem in one fell swoop, we're going to chip away at it, bit by bit, until it's done. If you know how to divide and learned in American schools, this treatment may seem odd, even incomprehensible. However, if you, or someone you know, doesn't know how to divide reliably, and is a bit shaky with multiplication, this method might just break the logjam.
We will do an example problem: 27348 divided by 362. A multiple digit integer divided by another multiple digit integer.
We have 27,348 items, and we are putting them into 362 boxes. The division will tell us how many will be in each box when all the boxes have the same number of items in them, and we no longer have enough items to put at least one more into each box. Or, we could look at it as if we had 27,348 cards, and were dealing them to 362 players, and kept dealing until we didn't have enough cards to deal another round to everyone. Or, we have jurisdiction over 27,348 inches of riverfront, and the law says we must allow each fisherman 362 inches of riverfront to fish, and we want to know how many fishing licences we can issue each day.
If we had a lot of time, we could simply subtract 362 over and over again (as if we were dealing rounds of a single card to each player) and keep track of the number of rounds we've dealt, but that would get incredibly tedious. Still, just as multiplication is just repeated addition of the same number, over and over, division can be done by repeated subtraction. But we can be clever about it. We don't have to deal the cards one at a time.
Suppose we dealt 10 cards at a time, or 100 cards at a time. Now, we do have to do a bit of multiplication, but it's fairly easy to do: multiplying by ten we just stick a zero on the end of the number, and multiplying by one hundred we paste on two zeros. It's fairly easy to remember, too: 10 is 1 with one zero pasted on the end, and 100 is 1 with two zeros after the one.
Well, 362 players would get 3,620 cards with a ten at a time round, and would get 36,200 cards with a 100 at a time round. That second number is too high, we don't have that many cards. And the first number is way too low. But this tells us that the answer is going to be between 10 and 100, and will probably be closer to 100 than to 10. We can use this later to check our work. If we get, say, 437, then we've made a mistake somewhere.
Now, we assume we can't multiply well, but we can add. If we deal two cards to everyone, then we need 362+362 cards, or 724 cards to deal a round of two cards. If we deal 20 cards at a time, that would be 7,240 (multiply by 10: stick a zero on the end), and if we dealt 200 cards at a time we'd be way over 100, which we already know is too high, so we don't have to bother thinking about it.
We can also add two plus two, and find out that dealing four cards would use 724+724= 1448 cards, and dealing 40 cards at a time uses 14,480 cards per round.
That's a lot of words, lets make a little chart:
Let's do some repeated subtraction with these numbers
We start with 27,348: the number we are dividing up. We deal 40 cards to each player and we have 27,348 - 14,480 = 12,868 cards left, and each player has 40 cards. Can we do that again? Well, no, we don't have 14,480 cards left. If we did, of course we'd deal out another 40 cards to each player, because we are trying to deal the cards as quickly as we can, but we just don't have enough.
So, let's give everyone 20 more cards, which uses up 7,240 cards. Now we are left with 12,868 - 7,240 = 5628 cards, and we've dealt a total of 60 cards to each of our 362 players. We don't have enough cards left to give everyone another 20 cards, but we do have enough to give everyone 10 cards, so we do that next.
5628 - 3620 = 2008 cards left, 70 total dealt to each player.
Now we are getting down to the smaller numbers. We can't give everyone 10 more, but we have enough to deal 4 to each player, so that's our next step.
2008 - 1448 = 560 cards left, 74 cards in each player's hand. We don't have enough cards to give everyone 2 cards each, but we can give them one. We do so:
560 - 362 = 198 cards left, and each player has 75 cards. We don't have enough to give everyone another card, so we can either call those 198 cards a remainder, or we can cut each one into 362 pieces, and give everyone 198 of those tiny pieces, or 198/362. So, there's our answer
27348 divided by 362 is 75 remainder 198, or 75 + 198/362.
We can put all these numbers into a chart, too.
The chart would be a bit easier to work with, if we put the subtractions in the same column as the number of cards left to deal, since we are actually doing a series of subtractions.
Here's what it looks like without the words, with the number of groups -- 362 -- written at the top (above a line so it doesn't get added in to the answer by mistake) and our specially built "multiplication table" off to the side.
The answer, 75 remainder 198, is found at the bottom.
This is division using partial products. Instead of playing "the price is right" with an imperfectly memorized multiplication table and only dealing with some of the digits of the dividend, we start by building a custom made "multiplication table" (by adding the divisor to itself.) That means we don't have to guess which number to use, we simply compare the amount we'd need with the amount we've got left, and there is either enough or there isn't. There's no weird "bringing down" nonsense, just simple subtractions. We write the problem at the top in the same order we say it "27,348 divided by 362" is written with the 27,348 coming first, and the 362 coming after. Finally, as with addition, subtraction and multiplication problems, the answer is found at the bottom, not at the top.
This is long division, without the confusing shortcuts of the traditionally taught algorithm.
As a student's confidence in calculating increases, the shortcuts can be added.
Questions? I wonder what needs to be added to this to make it clearer.
Have fun in the comments.