3. Method of long division in numbers and species. Example with zero remainder (no decimals) 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 we 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 Example with decimals 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 Remainder and supplement 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 of jumping 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. Long division of species/polynomials. Let's divide the following species. (x^3 - 2·x^2 - 4 ) / (x - 3) = D / d i) Divide the 1st term of D by highest term of d. That is, divide x^3 by x and place the result above the x^2 term, since it is a x^2 term: __________x^2___________ x - 3) x^3 - 2·x^2 + 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 x^2 is the remainder. __________x^2___________ x - 3) x^3 - 2·x^2 + 0·x - 4 x^3 - 3·x^2 -___________ 0 + x^2 iii) Bring the next term of D to the result of subtraction, which constitutes the *interim dividend*. __________x^2___________ x - 3) x^3 - 2·x^2 + 0·x - 4 x^3 - 3·x^2 | -____________ v 0 + x^2 + 0·x <--- interim dividend iv) Repeat the process using the interim dividend. Here it consists in x^2 / 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. __________x^2_+___x_____ x - 3) x^3 - 2·x^2 + 0·x - 4 x^3 - 3·x^2 | | -____________ v | 0 + x^2 + 0·x | x^2 - 3·x | -___________ v 0 + 3·x - 4 We repeat the process again: __________x^2_+___x_+_3_ x - 3) x^3 - 2·x^2 + 0·x - 4 x^3 - 3·x^2 | | -__________ v | 0 + x^2 + 0·x | x^2 - 3·x | -_____________ v 0 + 3·x - 4 3·x - 9 -_______ 0 + 5 Conclusion, q = x^2 + x + 3 and r = 5, or just x^3 - 2·x^2 - 4 = (x - 3)·(x^2 + x + 3) + 5 . Long division with numbers considered as species. Let's divide 345/12, but this time written as (3X^2 + 4X^1 + 5X^0) / (X^1 + 2X^0) or (3X^2 + 4X + 5) / (X + 2) , where X = 10. Let's keep the usual division in parallel for the sake of comparison. _______________ _________ X + 2) 3X^2 + 4X + 5 12) 345 Now we must take a crude approximation, which is to divide 3X^2 by X, giving 3X. With numbers, this is equivalent to do 300 divided by 10, which gives 30. _________3X_____ ___30____ X + 2) 3X^2 + 4X + 5 12) 345 Next step is to do q·d and subtract it from D to get the remainder: _________3X_____ ___30____ X + 2) 3X^2 + 4X + 5 12) 345 3X^2 + 6X 360 -_____________ -_____ 0 - 2X + 5 -15 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) 3X^2 + 4X + 5 12) 345 3X^2 + 6X 360 -_____________ -_____ 0 - 2X + 5 -15 - 2X - 4 -24 -__________ -_____ 0 + 9 9 Conclusion: (3X^2 + 4X + 5) = (X + 2)·(3X - 2) + 9 345 = 12·28 + 9 . 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.