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We can divide a number by subtraction and stop at the remainder as shown here - How to divide using addition or subtraction.

But how do we continue to divide the remainder by subtraction ? I looked on google and could not find such answers. They don't go beyond the remainder.

For example, lets say we have $7/3$. $7-3 = 4$ $4-3 = 1$ So, we have $2$ & $(1/3)$. How do we do the $1/3$ division using only subtraction ?

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math mongol
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If you want to divide beyond that point, that is, to divide the remainder, then presumably you want the full decimal expression of a non-integer. In order to find the decimal expansion for $1/3$ via subtraction, we would have to say the following: $$ 1/3 = \frac1{10}\times (10/3) $$ That is, find $10/3$, and shift the answer over by a decimal place. So, we have $$ 1/3 = \frac1{10}\times 10/3\\ 10-3-3-3 = 10-3\times 3 = 1\implies\\ 1/3 = \frac1{10}\times(3R1) = \frac1{10}\left(3+\frac13\right) $$ To say this another way, we know that if you take the decimal expansion for $\frac13$ and multiply it by $10$ (or shift the decimal place to the right), you get $3$ more than the decimal expansion for $\frac13$. What does this recursive definition tell you about that decimal expansion?

By a similar process, we can find any terminating or repeating decimal. In other words, we have long division.

Ben Grossmann
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  • Please note that I dont want to use multiplication or division operators to do this. – math mongol Aug 10 '13 at 07:07
  • If you want to get a decimal representation of a non-integer, you have to acknowledge that numbers after the decimal point are divided by $10$. What I have done is as close as you can get to division-free division, since the only division carried out is division by $10$, corresponding to a shift of the decimal point. The only multiplication I've done here is integer multiplication, which can alternately be done by repeated subtraction as shown. – Ben Grossmann Aug 10 '13 at 13:59