Why do we add a zero to dividend during long division? 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.
 A: 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.
A: 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 $0.0$.
A: 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!
A: 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.
A: 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.
A: One appends (not adds) the zero to keep the columns straight during subtraction. (Paraphrasing other answers.)
A: 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 $3/2=1.5$.
