Long division taken apart, continued

Now that we have taken long division apart, in the last posting, and made a multiplication table for the divisor ${\displaystyle 21}$, in the posting before that, we are ready to look at the standard long division algorithm, with Jakow Trachtenberg's trick.

We are dividing ${\displaystyle 1543}$ by ${\displaystyle 21}$, using the multiplication table

${\displaystyle \left[\begin{array}{cc}0& 0\\ 1& 21\\ 2& 42\\ 3& 63\\ 4& 84\\ 5& 105\\ 6& 126\\ 7& 147\\ 8& 168\\ 9& 189\end{array}\right]}$.

Start by writing


${\displaystyle 21)\overline{1543}.}$

From the table, see that no whole ${\displaystyle 21}$s go into ${\displaystyle 1}$ or into ${\displaystyle 15}$, but that ${\displaystyle 7}$ whole ${\displaystyle 21}$s, but not ${\displaystyle 8}$, go into ${\displaystyle 154}$ :


${\displaystyle 21)\overline{1543}.}$
${\displaystyle 147.}$

Subtract ${\displaystyle 1470}$ from ${\displaystyle 1543}$ leaving ${\displaystyle 73}$.


${\displaystyle 21)\overline{1543}.}$
${\displaystyle \underline{1470}.}$
${\displaystyle 73.}$

From the table, again, ${\displaystyle 3}$ whole ${\displaystyle 21}$s, i.e. ${\displaystyle 63}$, go into ${\displaystyle 73}$, leaving ${\displaystyle 10}$.

${\displaystyle 73.}$
${\displaystyle 21)\overline{1543}.}$
${\displaystyle \underline{1470}}$
${\displaystyle 73.}$
${\displaystyle \underline{63}.}$
${\displaystyle 10.}$

This is the stopping point for divmod division, so that

For decimal division, one continues ${\displaystyle 21}$s into ${\displaystyle 100}$ go ${\displaystyle 4}$ times, i.e. ${\displaystyle 84}$ leaving ${\displaystyle 16}$.

${\displaystyle 73.}$
${\displaystyle 21)\overline{1543}.}$
${\displaystyle \underline{1470}}$
${\displaystyle 73.}$
${\displaystyle \underline{63}.}$
${\displaystyle 10.0}$
${\displaystyle \underline{8.4}}$
${\displaystyle 1.6}$

This continues until one has reached the desired degree of accuracy.

Long division taken apart

When we divide, we are trying to find out how many times the divisor (e.g. 21) goes into the dividend (e.g. 1543).

We could just repeatedly subtract, keeping a tally of the number of times we have subtracted, stopping when further subtraction would give a negative result.  At least, that is how we could do divmod division.

Keeping such a tally would be tedious, however.  It is better to subtract large but convenient multiples of the divisor, than smaller multiples, keeping tally of these separately as we go.

In our usual decimal system of numeration, powers of ten are particularly convenient multipliers.

In 1543, the highest nonzero column is the thousands column.  We can therefore try subtracting thousands of ${\displaystyle 21}$s.  We could not subtract ${\displaystyle 21}$ thousand even once from ${\displaystyle 1}$ thousand and anything without the result being negative.  So we need ${\displaystyle 0}$ thousands of ${\displaystyle 21}$s, and we still have ${\displaystyle 1}$ thousand and something.

Next, we try hundreds of ${\displaystyle 21}$s.  Again, we cannot subtract even ${\displaystyle 21}$ hundreds even once from ${\displaystyle 15}$ hundred and something, without the result being negative.  So we need ${\displaystyle 0}$ hundreds of ${\displaystyle 21}$s, and we still have ${\displaystyle 15}$ hundred and something.

Next, we try tens of ${\displaystyle 21}$s.  We can subtract ${\displaystyle 21}$ tens from ${\displaystyle 154}$ tens and something.  Indeed, we can subtract ${\displaystyle 7}$ times, a total of ${\displaystyle 147}$ tens, without getting negative result.  So now we have that ${\displaystyle 154}$ tens and ${\displaystyle 3}$ is the same as ${\displaystyle 7}$ lots of ${\displaystyle 21}$ tens, and then ${\displaystyle 7}$ tens and ${\displaystyle 3}$.

Next we try whole ${\displaystyle 21}$s.  We can subtract ${\displaystyle 21}$ from ${\displaystyle 73}$ just ${\displaystyle 3}$ times without going negative.  This subtracts a total of ${\displaystyle 63}$ from ${\displaystyle 73}$, leaving ${\displaystyle 10}$.

So now we have that ${\displaystyle 1543}$ is the same as ${\displaystyle 7}$ lots of ${\displaystyle 21}$ tens, and ${\displaystyle 3}$ lots of ${\displaystyle 21}$ ones, and a further ${\displaystyle 10}$.

This can instead be understood as ${\displaystyle 70}$ lots of ${\displaystyle 21}$, and ${\displaystyle 3}$ lots of ${\displaystyle 21}$, and ${\displaystyle 10}$ more.  This in turn is ${\displaystyle 73\times 21}$ and ${\displaystyle 10}$.  i.e.

${\displaystyle 1543=(73\times 21)+10.}$

If we are doing divmod division, we can stop :



But if, instead of a remainder, we want the quotient to have a fractional part expressed as a decimal, then we keep going.

We have ${\displaystyle 10}$ that still needs to be divided by ${\displaystyle 21}$.  So now we try subtracting tenths of ${\displaystyle 21}$.  ${\displaystyle 10}$ is ${\displaystyle 100}$ tenths.  We can subtract ${\displaystyle 4}$ lots of tenths of ${\displaystyle 21}$, i.e. ${\displaystyle 84}$ tenths, from ${\displaystyle 100}$ tenths, leaving ${\displaystyle 16}$ tenths.

Next, we try subtracting hundredths of ${\displaystyle 21}$ from ${\displaystyle 160}$ hundredths.  ${\displaystyle 7}$ hundredths of ${\displaystyle 21}$ is ${\displaystyle 147}$ hundredths, which when subtracted from ${\displaystyle 160}$ hundredths leaves ${\displaystyle 13}$ hundredths.

Stopping at this point, we find that

${\displaystyle 1543=(73.47\times 21)+0.13}$,

which we can write

.

The preceding discussion has been rather lengthy--such work is usually set out in a much more compressed format.  Unfortunately, long division seems to have been taught procedurally without adequate preparation, so that the elided computational form entrains elided thinking.  Even among the few in these calculator-infested days who can still actually do long division, a substantial fraction can give no convincing account of why the standard long division algorithm works.

Building ad hoc multiplication tables

Proportion tables are sometimes useful for doing long division. 

For many people, the most difficult part of that standard long division algorithm is estimating which multiple of the divisor to subtract.  Jakow Trachtenberg taught a simple trick for avoiding this difficulty.

In the next posting, we are going to divide 1543 by 21, using the standard long division algorithm and Trachtenberg's trick.

The first step is to invest some time making a table of multiples of ${\displaystyle 21}$, up to the ${\displaystyle 9\times 21}$.  The ${\displaystyle 0}$ and ${\displaystyle 1}$ rows are obvious :

${\displaystyle \left[\begin{array}{cc}0& 0\\ 1& 21\end{array}\right]}$.

The ${\displaystyle 2}$ row can be found by doubling the second row :

${\displaystyle \left[\begin{array}{cc}0& 0\\ 1& 21\\ 2& 42\end{array}\right]}$.

The ${\displaystyle 3}$ row can be found by adding the ${\displaystyle 1}$ row and the ${\displaystyle 2}$ row :

${\displaystyle \left[\begin{array}{cc}0& 0\\ 1& 21\\ 2& 42\\ 3& 63\end{array}\right]}$.

The ${\displaystyle 4}$ row can be found either by adding the ${\displaystyle 1}$ row to the ${\displaystyle 3}$ row, or else by doubling the ${\displaystyle 2}$ row.  One picks whichever is more convenient :

${\displaystyle \left[\begin{array}{cc}0& 0\\ 1& 21\\ 2& 42\\ 3& 63\\ 4& 84\end{array}\right]}$.

The ${\displaystyle 5}$ row can be found either by adding the ${\displaystyle 2}$ row and the ${\displaystyle 3}$ row, or else by adding the ${\displaystyle 1}$ row and the ${\displaystyle 5}$ row.  Again, one picks whichever is more convenient :

${\displaystyle \left[\begin{array}{cc}0& 0\\ 1& 21\\ 2& 42\\ 3& 63\\ 4& 84\\ 5& 105\end{array}\right]}$.

One continues in this way, constructing the next row opportunistically, until at last one has :

${\displaystyle \left[\begin{array}{cc}0& 0\\ 1& 21\\ 2& 42\\ 3& 63\\ 4& 84\\ 5& 105\\ 6& 126\\ 7& 147\\ 8& 168\\ 9& 189\end{array}\right]}$.

This can be checked by casting out nines, if one knows how to do that.  (If not, it needs to be the subject of yet another post.)

Now we are ready divide anything by ${\displaystyle 21}$.

${\displaystyle \left[\begin{array}{ccc}0& 0& \left(0\right)\\ 1& 21& \left(3\right)\\ 2& 42& \left(6\right)\\ 3& 63& \left(0\right)\\ 4& 84& \left(3\right)\\ 5& 105& \left(6\right)\\ 6& 126& \left(0\right)\\ 7& 147& \left(3\right)\\ 8& 168& \left(6\right)\\ 9& 189& \left(0\right)\end{array}\right]}$.

This checks out, so we can rely on the table we have made.