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.