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Long addition can be done "digit by digit" when you stack the numbers, being sure to carry and align them at their decimal places.

Long subtraction can be done digit by digit when you stack the numbers, again aligning them by decimal point and borrowing as necessary.

Long multiplication can be done digit by digit too.

What about long division? So for example, 67 divided by 4, the 4 goes under the 67 and they are aligned against the right side. What process do you follow? It would have to work for long numbers, like 746 divided by 105.

Getting the answer as a quotient + remainder is okay, as the remainder can be taken over the divisor for the final fraction and left that way.

(Yes I know how to do traditional long division, where the dividend goes inside the division box and you start writing the quotient above that, multiplying each digit of the quotient by the entire divisor, subtracting, bringing down the next digits, repeat. I want to know why addition, subtraction, and multiplication allow you to perform that operator one digit at a time, yet apparently division does not.)

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    https://en.wikipedia.org/wiki/Fourier_division – 2017-02-26
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    @Bye_World a - (b+c) =/= a-b + a-c , but subtraction can work longhand digit by digit. I'm not sure if I copied your thing precisely. Too bad there are not variables for binary operators themselves. – 2017-02-26
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    The problem is $654*k=600*k+50k+4*k $ is an easy separation. But $k/654 = w/600+y/50+z/4$ is not so straightforward. I think it can be done but it's not straightforward. – 2017-02-26

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$\dfrac{67}{4}=\dfrac{40 + 27}{4}=\dfrac{40 + 20 +7}{4}=\dfrac{40 + 20 + 4 +3}{4}$

$=\dfrac{40}{4} + \dfrac{20}{4} +\dfrac{4}{4} + \dfrac{3}{4}$

$=10 + 5 + 1 + \dfrac{3}{4}$

$=16 +\left( \dfrac{30}{4}\right). \dfrac{1}{10}$

$=16 + \left(\dfrac{28 + 2 }{4}\right). \dfrac{1}{10}$

$=16+\left(\dfrac{28}{4}+\dfrac{2}{4}\right). \dfrac{1}{10}$

$= 16 + \left(7+0.5\right)\dfrac{1}{10}$

$=16 + 0.7 + 0.05 =16.75$

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    Ask me in comments if you didn't understand anything. – 2017-02-26
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    I understand 10+5+1+3/4 = 16+3/4 = 16.75. I actually don't understand all the next lines, lol, but that's okay. However, can you do the same thing with a multi-digit divisor? E.g., 67/12. How would you break up the fractions? – 2017-02-26
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    Do you know multiples of 12? Then take highest multiple of 12 number which is less than 67 and break it into small numbers. – 2017-02-26
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    @DrZ214 I multiplied and divided with $10$ to $\frac 34$ – 2017-02-26
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    Yes I know multiples of 12. I was proposing an easy example of a multi-digit divisor to see how you would break up the fractions. I want an algorithm that can do any long division, and I mean long, like 6842411/9645. – 2017-02-26
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Take $\frac{45.00}{20}$ We then say $20$ goes into $4$ zero times, so we put a $0$ above the 4 and carry the $4$ above the $5$, and then we put a $2$ above the $5$, since $20$ goes into $45$ $2$ times remainder $5$. Then we carry over the remainder into the decimal point and see how many times $20$ goes into $50$. This would give us $2$ remainder $4$. So above the $.0$, we would put $2$, etc etc...

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    Technically you are dividing by 20, instead of dividing by 2 and then 0 (digit by digit). You cannot divide by zero, so presumably the process, if it exists, has a special rule or maybe even skips right over it. Maybe, can you redo your example for 45/21 and explain it a bit more? – 2017-02-26
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So obviously addition and subtraction can be written it terms of digits for numbers $10^{x_n} +/- ... +/- 10^{x_0} $, so it's pretty simple to see where the digit by digit idea lines up. Multiplication is a little more complicated, but still relatively simple. if we were to multiply this number: $10^{x_n} +/- ... +/- 10^{x_0} $ by a random number, let's say 451, we can use the distributive property of multiplication. However, coming to the division, the distributive property doesn't exist. I suppose one could rewrite $\frac{a}{b}$ as $\frac{1}{b}*a$ and then distribute the terms of $a$ onto $b$, but remember that $\frac{1}{b} + \frac{1}{c}$ != $\frac{1}{b+c} $

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The Trachtenburg fast division technique is what you want to look at. It will require some different mental operations to become familiar with, but it will allow you to write down the answer with 2 extra lines of working figures (that are used in subtracting from.) Not for the faint of heart. From the youtube example at here the work looks like this:

8 1 5 1 / 5 7 = 1 4 3

31 25 01

8 24 17 0

The 2 lower lines help in subtracting parts of the answer. Problems with longer divisors build on add more subtracting pieces but don't get substantially more difficult (as long as you can follow the steps in your head.)