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Suppose we want to divide 3 by 2.

By Long Division we first write 1 as the first digit as the quotient, then we subtract 2 from 3, then we add a zero to the remainder 1, and then add a decimal point after the quotient 1, then write 5 after the decimal point, getting the quotient as 1.5

I know it is indeed right as we can verify. But why do we add a zero and the decimal point in the quotient?

Thank you.

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    $\begingroup$ You are now filling in the tenths column. There was 1 one left over, which is 10 tenths. (Similarly of course 2 ones is 20 tenths, 3 ones is 30 tenths, etc) $\endgroup$ – immibis Jun 26 '18 at 0:37
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The principle behind long division is repeated subtraction. If we want to divide a number $p$ by a number $q$, what we are trying to do is subtract as large a multiple of $q$ from $p$ as we can, until the difference is $0$ (if possible).

Let's take a nastier example for illustration. Suppose we wish to divide $4101$ by $12$. We want to subtract multiples of $12$ from $4101$. We can start by subtracting multiples of $12 \times 10^2 = 1200$, since this is a very simple, large multiple of $12$. We have \begin{align*} 4101 - \color{red}{3} \times 12 \times 10^2 &= 4101 - 3600 = 501 \ge 0 \\ 4101 - \color{red}{4} \times 12 \times 10^2 &= 4101 - 4800 = -699 < 0. \end{align*} So, we can only take $\color{red}{3}$ lots of $12 \times 10^2$ from $4101$ before we've taken too much. When we do, we are left with $501$. We know we cannot take even $1$ lot of $12 \times 10^2$, so we try one order of magnitude less: $12 \times 10^1$. We have \begin{align*} 501 - \color{green}{4} \times 12 \times 10^1 &= 501 - 480 = 21 \ge 0 \\ 501 - \color{green}{5} \times 12 \times 10^1 &= 501 - 600 = -99 < 0. \end{align*} We can only take $\color{green}{4}$ lots of $12 \times 10^1$ from $501$. Note that, in total we have $$4101 - \color{red}{3} \times 12 \times 10^2 - \color{green}{4} \times 12 \times 10^1 = 4101 - \color{red}{3}\color{green}{4}0 \times 12 = 21 \ge 0.$$ From $21$, we can take a single multiple of $12 \times 10^0 = 12$, as \begin{align*} 21 - \color{purple}{1} \times 12 \times 10^0 &= 21 - 12 = 9 \ge 0 \\ 21 - \color{purple}{2} \times 12 \times 10^0 &= 21 - 24 = -3 < 0. \end{align*} In total, $$4101 - \color{red}{3}\color{green}{4}\color{purple}{1} \times 12 = 9 \ge 0.$$

From the remaining $9$, $12$ cannot be subtracted any positive integer number of times without becoming negative. But, this doesn't mean we have to stop. Maybe no more whole amounts of $12$ will go into $9$, but we can try fractional parts of $12$. Again, all we do is go down an order of magnitude: \begin{align*} 9 - \color{orange}{7} \times 12 \times 10^{-1} &= 9 - 8.4 = 0.6 \ge 0 \\ 9 - \color{orange}{8} \times 12 \times 10^{-1} &= 9 - 9.6 = -0.6 < 0. \end{align*} So, $$4101 - \color{red}{3}\color{green}{4}\color{purple}{1}.\color{orange}{7} \times 12 = 0.6 \ge 0.$$ Once more, \begin{align*} 0.6 - \color{blue}{5} \times 12 \times 10^{-2} &= 0.6 - 0.6 = 0 \ge 0 \\ 0.6 - \color{blue}{6} \times 12 \times 10^{-2} &= 0.6 - 0.72 = -0.12 < 0. \end{align*}

That is, $$4101 - \color{red}{3}\color{green}{4}\color{purple}{1}.\color{orange}{7}\color{blue}{5} \times 12 = 0 \implies \frac{4101}{12} = \color{red}{3}\color{green}{4}\color{purple}{1}.\color{orange}{7}\color{blue}{5}.$$

I'm hoping this answer conveys a better understanding of the logic behind here. The decimal point in question is being introduced around where we subtract two numbers with a different number of decimal places. For example, when compute $$9 - 8.4 = 0.6 \ge 0.$$

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  • $\begingroup$ In practice when doing this though you aren't really looking at how many times $12 \times 10^2$ goes into 4101. You are looking at how many times $12$ goes into $41$. you then get a remainder 5 and then the next step you add on the 0 from the original number and work out how many times you can divide 12 into 50. You then add the next digit in the original number to the remainder to get 12 into 20. Looked at this way it then becomes obvious that since 4101 is the same as 4101.0000000 then this is the source of the extra zeros. $\endgroup$ – Chris Jun 26 '18 at 9:29
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    $\begingroup$ I really like this answer but I think it would be improved by adding in the association with the division you actually do, the one where you actually add the zero in. As it stands this answer doesn't show the zero being added in anywhere. $\endgroup$ – Chris Jun 26 '18 at 9:31
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Whether you write the decimal point and zero from the beginning, or after you realize you need them, is immaterial. They're always "there."

Or, just to be cautious, you could have written $3.0000000000$, which was more zeroes than you "needed" to solve this problem.

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Decimals have infinitely many digits in both directions; when we only write finitely many of them, the intention is that every unwritten place has a zero.

When we do long division in the way you indicate, we have to actually use some of the places we had previously left unwritten. So we write in the zeroes.

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We do not add zero, because it already is there.

We just show it, just as in replacing $3$ with $3.0$,
or $0.01$ with $0.010000$ etc.

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The integer division of $3$ by $2$ yields the quotient $1$ and remainder $1$.

The integer division of $30$ by $2$ yields the quotient $15$ and remainder $0$.

Hence the division of $3.0$ by $2$ yields the quotient $1.5$ and remainder $0.0$.

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As others have pointed out, the zeros were already part of the decimal representation of the number; you just haven't written them yet in your example.

Here's what happens if we do write a few of those zeros to begin with. That is, instead of writing $3,$ we write $3.00000,$ instead of writing $2,$ we write $2.00000,$ we keep writing the decimal digits for other numbers as well, and we make sure always to write write the decimal point between the "ones" digit and the "tenths" digit.

We still multiply the divisor by each digit of the result in order to find the next number that must be subtracted. In particular,

$$ \begin{array}{rr} & 2.00000 \\[-3pt] \times & \underline{\phantom{0.0000}1} \\[-3pt] & 2.00000 \\ \\ & 2.00000 \\[-3pt] \times & \underline{\phantom{00000}0.5} \\[-3pt] & 1.000000 \\ \end{array} $$

following the usual rules of multiplication of decimal numbers.

Putting it all together, we get

$$ \require{enclose} \begin{array}{r} 1.50000 \phantom0 \\[-3pt] 2.00000 \enclose{longdiv}{3.00000} \phantom0\\[-3pt] \underline{2.00000} \phantom0\\[-3pt] 1.00000 \phantom0\\[-3pt] \underline{1.000000} \\[-3pt] 0.000000 \end{array} $$

It sure is less trouble to work this out if we wait until we actually need a zero digit before we write one!

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One appends (not adds) the zero to keep the columns straight during subtraction. (Paraphrasing other answers.)

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Let's say you want to get the “result” of $37/13$ with two decimal digits (truncated). As we know, the decimal expansion of $37/13$ is infinite and periodic, but there is a decimal number $a.bc$ ($a$, $b$ and $c$ denote digits) that differs from the actual quantity by less than $1/100$ (and is less than $37/13$).

The number $a.bc$ actually denotes $$ \frac{abc}{100} $$ and we are just pretending that $$ \frac{abc}{100}=\frac{37}{13} $$ that is, $$ abc=\frac{3700}{13} $$ Well, that's not really an equality, but it gives the idea. In better terms, we do the division of $3700$ by $13$ and discard the remainder $$ \begin{array}{r|l} 3700 & 13 \\ 110\hphantom{0} & \color{green}{284}\\ 60\\ 8 \end{array} $$ The quotient is $284$ and the remainder is $8$. We thus have $$ \frac{3700}{13}=284+\frac{8}{13} $$ Note that $\varepsilon=8/13<1$, so if we divide both sides by $100$, we obtain $$ \frac{37}{13}=2.84+\frac{\varepsilon}{100} $$ and indeed $\varepsilon/100<1/100$ as required.

Can you see the “added zeros”?

The same happens with your case, with the difference that the division leads to an “exact” decimal, because the divisor is $2$; actually we know that at most one decimal digit will appear (they would be at most two when dividing by $4$ and so on). Thus the idea is the same: we do $$ \frac{30}{2}=15 $$ and dividing by $10$ yields $3/2=1.5$.

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