Division Without Division

Question

Ok. I'm having a bit of a problem with a mathematical task my friend challenged me with, find the answer to two divided numbers without any division and the method used has to work for all whole numbers, so let's say I have the numbers $13 / 3$, by using repeated subtraction that gives me the number $4$ with the remainder of 1 but how can I find the decimal number with without using division, normally I could just do $1 / 3$ to find it but as I explained earlier I can't use division, is there a way to do this? Thanks in advance. (Sorry for bad grammar.)

Answer 1

Since multiplication is allowed,

$$1/3=(10/3)/10\tag1$$

$$10/3=3+1/3\tag2$$

substitute $(2)$ into $(1)$

$$1/3=(3+1/3)/10$$

If we rewrite $1/3=x$,

$$x=(3+x)/10$$

or, $x_n=(3+x_{n-1})/10$

We don't need to do division by $10$, just shift the decimals over $1$.

I recommend starting with $x_0=0$.

For example,

$$x_0=0$$

$$x_1=3/10=0.3$$

$$x_2=3.3/10=0.33$$

And so on... you can see we will get $x=0.333\dots$

Works for dividing two positive numbers.

Another example could be

$$x_n=(2+2x_{n-1})/10$$

where $x=1/4$.

$$x_0=0$$

$$x_1=2/10=0.2$$

$$x_2=2.4/10=0.24$$

$$x_3=2.48/10=0.248$$

And again, it gets closer and closer to $1/4$, or $0.25$.

(Not so sure about a general formula, you might have to make one based on remainders and stuff)

Once the number gets larger than $10$, you have to use an altered version of my formula using $100$, $1000$, or more, shifting the decimal as needed.

To get the algorithm used here to calculate $1/n$, rewrite it like $(1)$ above, $(10/n)/10$, or different numbers as needed.

$(2)$ Rewrite $10/n$ by subtracting out, so that you may get $a+r(1/n)$, where $r$ is the remainder.

Then substitute $(2)$ into $(1)$ to get the formula desired.

And to prove that dividing by $10$ is a shift in decimal places without the use of division

$$\frac{\sum_{-\infty<n<\infty}d_n10^{-n}}{10}=\sum_{-\infty<n<\infty}d_n10^{-(n+1)}$$

Basically the same statement written in math, where $d_n$ is the $n$th digit to the right of the decimal point.

Now, if we multiply both sides by $10$ and distribute the $10$ into the sum, we get a true equality.