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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?

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    $\begingroup$ Do you admit logarithms? If so, we can very easily define division using subtraction: $$a/b = \exp\left(\log \frac{a}{b}\right) = \exp(\log a - \log b).$$ $\endgroup$ – Emily Aug 24 '12 at 18:14
  • $\begingroup$ doesn't exponents and logarithms come we define multiplication and division $\endgroup$ – Monkey D. Luffy Aug 24 '12 at 18:19
  • $\begingroup$ 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. $\endgroup$ – Emily Aug 24 '12 at 18:43
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    $\begingroup$ Reminded me this good old question. $\endgroup$ – user2468 Aug 25 '12 at 2:41
  • $\begingroup$ @Arkamis, what is this set whose supremum is $b^x$? Sounds very interesting. $\endgroup$ – goblin Dec 25 '14 at 7:06
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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 $4$.

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  • $\begingroup$ I remember re-implementing the built-in integer division and modulo functions of (insert language here) being a common programming exercise... :D $\endgroup$ – J. M. is a poor mathematician Aug 24 '12 at 23:45
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    $\begingroup$ @J.M. I had it on an exam; language was assembly for Zilog Z80 processors! $\endgroup$ – user2468 Aug 25 '12 at 3:37
  • $\begingroup$ Note this is horribly inefficient, computationally speaking. $\endgroup$ – Thomas Aug 25 '12 at 4:44
  • $\begingroup$ Seems a bit arbitrary that we always want to stop at zero. Nothing more than definition right? $\endgroup$ – Ovi Jun 3 '16 at 22:58
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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 $5$.

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

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

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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 $6.5$.

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  • $\begingroup$ $\frac{n}a\neq n-a-a-a\ldots$. Certainly it is not true that $$\frac{72}{9}=72-9-9-9-9-9-9-9-9$$ $\endgroup$ – Milo Brandt Jan 24 '15 at 22:27
  • $\begingroup$ Okay, I edited it so it would make more sense. $\endgroup$ – ReliableMathBoy Jan 24 '15 at 22:33

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