How to divide using addition or subtraction

We can multiply $a$ and $n$ by adding $a$ a total of $n$ times.

$$n \times a = a + a + a + \cdots +a$$

Can we define division similarly using only addition or subtraction?

• Do you admit logarithms? If so, we can very easily define division using subtraction: $$a/b = \exp\left(\log \frac{a}{b}\right) = \exp(\log a - \log b).$$ Aug 24, 2012 at 18:14
• doesn't exponents and logarithms come we define multiplication and division Aug 24, 2012 at 18:19
• Nope. We can define exponents and logarithms without requiring multiplication or division -- in a manner of speaking. We define $b^x$ as the supremum of a very specific subset of real numbers. This definition does not require that we define $b^x = b\cdot b \cdot b \cdots b$ some $x$ times. In fact, this definition works for any possible value of $x$. Logarithms can be defined in a similar manner. To justify this definition, we require that multiplication is an assumed property of the field of real numbers. We don't need to define exponentiation as repeated multiplication. Aug 24, 2012 at 18:43
• Reminded me this good old question.
– user2468
Aug 25, 2012 at 2:41
• @Arkamis, what is this set whose supremum is $b^x$? Sounds very interesting. Dec 25, 2014 at 7:06

To divide $60$ by $12$ using subtraction:

\begin{align*} &60-12=48\qquad\text{count }1\\ &48-12=36\qquad\text{count }2\\ &36-12=24\qquad\text{count }3\\ &24-12=12\qquad\text{count }4\\ &12-12=0\qquad\;\text{ count }5\;. \end{align*}

Thus, $60\div 12=5$.

You can even handle remainders:

\begin{align*} &64-12=52\qquad\text{count }1\\ &52-12=40\qquad\text{count }2\\ &40-12=28\qquad\text{count }3\\ &28-12=16\qquad\text{count }4\\ &16-12=4\qquad\;\text{ count }5\;. \end{align*}

$4<12$, so $64\div 12$ is $5$ with a remainder of $4$.

• I remember re-implementing the built-in integer division and modulo functions of (insert language here) being a common programming exercise... :D Aug 24, 2012 at 23:45
• @J.M. I had it on an exam; language was assembly for Zilog Z80 processors!
– user2468
Aug 25, 2012 at 3:37
• Note this is horribly inefficient, computationally speaking. Aug 25, 2012 at 4:44
• Seems a bit arbitrary that we always want to stop at zero. Nothing more than definition right?
– Ovi
Jun 3, 2016 at 22:58
• @Ovi: Not arbitrary at all. Nov 4, 2021 at 19:34

If $n$ is divisible by $b$ ($\frac{n}{b}$ is a whole number), then keep doing $n - b - b - b - b - b - \cdots - b$ until the value of that is $0$. The number of times you subtract $b$ is the answer. For example, $\frac{20}{4} \rightarrow 20 - 4 - 4 - 4 - 4 - 4$. We subtracted '$4$' five times, so the answer is $5$.

You can also use additions. One should use results from intermediate calculations to speed up.

Let us divide 63 by 12. $$\begin{split} 12+12=24,&\qquad\textrm{count }1+1=2\\ 24+24=48,&\qquad\textrm{count }2+2=4\\ 48+24=72,&\qquad\textrm{count }4+2=6\textrm{ (exceeded 63)}\\ 48+12=60,&\qquad\textrm{count }4+1=5\textrm{ (so we try adding less)}\\ 63-60=3,&\qquad\textrm{(calculation of the remainder)}\\ \end{split}$$

You can define division as repeated subtraction:$${72\over 9}=72-9-9-9-9-9-9-9-9$$Subtracting by $9$ eight times is the same as subtracting by $72$ since $9\cdot8=72$. So, the answer is $8$. Also, this is why ${n\over a}=n-a-a-a-a\cdots$ for whatever whole number $a$ is other than zero.

If you have a remainder, then you just do this:$${13\over 2}=13-2-2-2-2-2-2-1$$as you just saw, subtracting by $2$ six times is the same as subtracting by $12$ since $2\cdot6=12$, but there's a remainder of $1$ being sutracted, so it's the same as subtracting by $13$ since $2\cdot6+1=13$, so the answer is $6$ R$1$ or $6.5$.

• $\frac{n}a\neq n-a-a-a\ldots$. Certainly it is not true that $$\frac{72}{9}=72-9-9-9-9-9-9-9-9$$ Jan 24, 2015 at 22:27
• Okay, I edited it so it would make more sense. Jan 24, 2015 at 22:33