Long Division

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The other day someone asked:

Good question. The answer is yes, and I solved it using a technique the interlocutor didn't know:

OK. If you're interested in the actual algorithm then simply skip to the end. However, I'm going to derive the algorithm, starting with long division.

So let's start with the division problem, computing 999999 divided by 7.

Modern teaching doesn't use long division any more, apparently. Instead they use a thing called "chunking". In fact, "chunking" is nothing more that long division done with more guesses and less formality.


        1
      +-------------
    7 | 9 9 9 9 9 9
        7
       ---
        2 9 9 9 9 9
So we start by ignoring everything except the most significant part. We ask how many 7s we can take out from the first digit. The answer is just one. So we have


    999999 = 700000 + 299999
We can see that laid out formally at right.

That table then says that

 9999999 / 7 = 100000 + 299999/7 

We now iterated. We can remove 4 chunks of 7 from 29, so that's removing 40000 lots of 7 from 299999. Again, we lay that out in the tabular fashion:


        1 4
      +-------------
    7 | 9 9 9 9 9 9
        7
       ---
        2 9 9 9 9 9
        2 8
       -----
          1 9 9 9 9

So we are, at each stage, deciding what the next digit should be, multiply by 7, subtract from the remaining amount, and then repeat. This is laid out in a formal, tabular fashion. At each stage we have an intermediate sum that makes sense.


        1 4 2 8 . .
      +-------------
    7 | 9 9 9 9 9 9
        7
       ---
        2 9 9 9 9 9
        2 8
       -----
          1 9 9 9 9
          1 4
         -----
            5 9 9 9
            5 6
           -----
              3 9 9
Here we can see that 999999 = 7 * 142800 + 399. At each stage we reduce the remainder, subtracting a chunk of 7s from it.

This can be used for polynomials too. So, for example:


                           x^2 - 2x + 2
            +---------------------------------
x^2 + x - 3 | x^4 - x^3 - 3x^2 + 8x - 5
              x^4 + x^3 - 3x^2
             ------------------
                  -2x^3        + 8x
                  -2x^3 - 2x^2 + 6x
                 -------------------
                        + 2x^2 + 2x - 5
                          2x^2 + 2x - 6
                     -------------------
                                      1
As you can see, we can divide $x^4-x^3-3x^2+8x-5$ by $x^2+x+3$ and we get $x^2+2x+2$ with a remainder of 1.

So we can perform long division with polynomials in exactly the same way as we can perform long division with numbers, and in effect, it's just a formalised way of doing "chunking."

So, on to Square Roots By Long Division ...


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