Before attempting to divide species or even series, we should be sure we know how to divide numbers. But now that we know that the digits of a number can be expressed as a species in which each term is a power of the base, we will review dividing with this in mind.
We will use the Anglo notation for division instead of the Eurasian one. Remember that the Anglo notation places Dividend D, divisor d, quotient q and remainder r as
_____q_____
d) D
r
For example, 6358 / 17
______________
17) 6358
We start by selecting the first two digits of 6358, 63, since taking only the 6 would be smaller than the divisor d=17. We search for a number that multiplied by 17 gives less than 63 but as close (from below) as 63 as we can. This is 3. Notice how get 12 as a partial remainder.
___3__________
17) 6358
51
-____
12
Now it is time to lower the next digit of D, which is 5. We use a top down arrow to indicate this motion:
___3__________
17) 6358
51|
-___v
125
Now we repeat the process for 125/17, which gives 7, and we lower the next digit, 8.
___37_________
17) 6358
51||
-___v|
125|
119|
-____v
0068
We repeat the process one more time and we find a zero remainder, which indicates the end of the division.
___374________
17) 6358
51||
-___v|
125|
119|
-____v
0068
68
-_____
00
Let's try now 127 / 4.
____3_1.7_5__
4) 1 2 7
1 2 |
-___ v
0 0 7
4
- |
___v
30
28|
-__v
020
20
-__
00
It is of paramount importance to be able to reconstruct a division in these terms:
Dividend = divisor · quotient + remainder D = d·q + r .
or, by dividing everything by the divisor,
D/d = q + r/d = q + s .
Here, s=r/d is the supplement of the division. Why the name supplement, as in series? Because it is the same concept!
We are proposed to divide D by d. Then, we give a tentative result q, which is not the perfect solution. Of course, if we can give the exact solution at once, give it! But for difficult divisions, many times it is quite easy to give a tentative solution q. But we also recognise this solution to be approximate, so that we say that there is a correction term, the supplement, which, if added, would complete the solution.
So, when dividing 10/3, instead to jump to the exact solution, we can say "well, I know that 3 is quite an approximation, so I will say that the quotient is q=3". But then, we know this is not the end of the story, so we write 10 = 3·3 + 1, where 1 is the remainder, or 10/3 = 3 + 1/3, where 1/3 is the supplement. We know that 1/3 is what is missing in order to get a perfect solution. Also notice that 1/3 is not yet a number, but a division that we need to carry out.
We can now do the same process for the supplement. We need to calculate 1/3, so we can say, OK, we can give a tentative solution. Doing 10/3 instead of 1/3 (and later we divide by 10 again), we get 10/3 = 3 + 1/3, or dividing by 10, we get 1/3 = 0.3 + 1/30. So now we know that our result is 3.3 + 1/30, where 1/30 is the new supplement. And so on.
Another thing that is good to take into account is that, when we have a division D/d and we multiply up and down by a number, say a, we usually say "well, it does not matter, since the a cancels up and down and nothing changes". Well, be careful! It is true that q remains the same, but if D/d = q + r/d, then (a·D)/(a·d) = q + (a·r)/(a·d), which means that *the remainder becomes multiplied by a*. This is many times overlooked. If we divide 4/3, we get q=1 and r=1. But if we multiply up and down by 10, the division 40/30 has the same q=1 but with a new r=10.
Let's divide the following species.
(x³ - 2·x² - 4 ) / (x - 3) = D / d .
i) Divide the 1st term of D by highest term of d. That is, divide x³ by x and place the result above the x² term, since it is a x² term:
_________x²___________ .
x - 3) x³ - 2·x² + 0·x - 4
ii) Now take the result (q) of i) and multiply it by d. Place the result under D to do a subtraction and subtract. Notice that D - q·d = r, so the obtained x² is the remainder.
_________x²___________
x - 3) x³ - 2·x² + 0·x - 4
x³ - 3·x²
-__________
0 + x²
iii) Bring the next term of D to the result of subtraction, which constitutes the *interim dividend*.
_________x²___________
x - 3) x³ - 2·x² + 0·x - 4
x³ - 3·x² |
-__________ v
0 + x² + 0·x <--- interim dividend
iv) Repeat the process using the interim dividend. Here it consists in x² / x = x (the greatest term in the interim dividend by the x in (x-3)), so we add x to the q. We then multiply this x by d and place under interim to subtract and get the new remainder.
_________x²_+___x_____
x - 3) x³ - 2·x² + 0·x - 4
x³ - 3·x² | |
-__________ v |
0 + x² + 0·x |
x² - 3·x |
-_________ v
0 + 3·x - 4
We repeat the process again:
_________x²_+___x_+_3_
x - 3) x³ - 2·x² + 0·x - 4
x³ - 3·x² | |
-__________ v |
0 + x² + 0·x |
x² - 3·x |
-_________ v
0 + 3·x - 4
3·x - 9
-_______
0 + 5
Conclusion, q = x² + x + 3 and r = 5, or just
x³ - 2·x² - 4 = (x - 3)·(x² + x + 3) + 5 .
Let's divide 345/12, but this time written as
(3X² + 4X¹ + 5X⁰) / (X¹ + 2X⁰) or (3X² + 4X + 5) / (X + 2) ,
where X = 10. Let's keep the usual division in parallel for the sake of comparison.
_______________ _________
X + 2) 3X² + 4X + 5 12) 345
Now we must take a crude approximation, which is to divide 3X² by X, giving 3X. With numbers, this is equivalent to do 300 divided by 10, which gives 30.
________3X_____ ___30____
X + 2) 3X² + 4X + 5 12) 345
Next step is to do q·d and subtract it from D to get the remainder:
________3X_____ ___30____
X + 2) 3X² + 4X + 5 12) 345
3X² + 6X 360
-_____________ -_____
0 - 2X + 5 -15
How interesting this is! With regular number division we would have chosen 29 as quotient because (it seems) we are afraid to exceed the dividend. But there is nothing to be afraid of exceeding it. The only result is that we get a negative remainder. So what? Not a problem at all!
Let's continue the process by dividing the interim dividend (-2X+5) by the X in (X+2), which gives -2.
________3X - 2_ ___30_-_2____
X + 2) 3X² + 4X + 5 12) 345
3X² + 6X 360
-_____________ -_____
0 - 2X + 5 -15
- 2X - 4 -24
-__________ -_____
0 + 9 9
Conclusion:
(3X² + 4X + 5) = (X + 2)·(3X - 2) + 9 345 = 12·28 + 9 .
Not only it is interesting to compare the *specious* division with its number analogue. We also liberate ourselves from the tight constraint of choosing the greatest quotient that leaves a positive or null remainder. Now we know that if go for crude estimations of q, the next steps will compensate for that. Maybe we will end up having negative remainders or even more steps than necessary, but the process leads to the correct result no matter what.
For example, divide again 345 / 12, but this time in a crazy way:
_________
12) 345
Instead of choosing a 3, let's choose a 2 (or a 20, which is equivalent).
__20_____
12) 345
240
-____
105
The only problem here is that the remainder is still greater than the divisor. So what? No problem! We just continue dividing it. Now we fancy 105/12 as 10,and we don't care if the result comes greater or smaller than 105. It is just crude and easy.
__20_+_10___
12) 345
240
-____
105
120
-____
-15
Oh, a negative remainder. Not sorry! Let's continue with a clear -15/12 ~ -1 this time.
__20_+_10_-_1___
12) 345
240
-____
105
120
-____
-15
-12
-___
-3
Again a negative remainder. And now we should be done, but let's go for some decimals. -3/12=-1/4 give a -0.25, because suddenly we say to ourselves "why, instead of being crude, we cannot choose two figures at once?". Let's go for it!
__20_+_10_-_1_-_0.25___
12) 345
240
-____
105
120
-____
-15
-12
-___
-3
-3
-____
0
So our result is 20 + 10 - 1 - 0.25 and, since we got zero remainder, it must be exact! We do 20+10-1-0.25 and get 28.75, which is the result!
Please take some time to appreciate the incredible flexibility and robustness of this relaxed method. We are usually taught with fear of not choosing the proper quotients, thinking that a mistake would lead us into a fatal chain of further errors. Why aren't we told that choosing a quotient too big or too small only takes a different path to the same result? There is an elasticity in the method that allows you to choose. If you need to consider 20/11, it is so unnatural to choose 1 as quotient and having 9 as remainder! Isn't a more elegant path to choose 2 as quotient and -2 as remainder? In fact, this would lead us to the result much faster! Also, if you see two digits of a quotient, take them both without fear.