Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
Do you admit logarithms? If so, we can very easily define division using subtraction: $$a/b = \exp\left(\log \frac{a}{b}\right) = \exp(\log a - \log b).$$ –  Arkamis Aug 24 '12 at 18:14
    
doesn't exponents and logarithms come we define multiplication and division –  Monkey D. Luffy Aug 24 '12 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. –  Arkamis Aug 24 '12 at 18:43
1  
Reminded me this good old question. –  user2468 Aug 25 '12 at 2:41

3 Answers 3

up vote 15 down vote accepted

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 $4$.

share|improve this answer
    
I remember re-implementing the built-in integer division and modulo functions of (insert language here) being a common programming exercise... :D –  J. M. Aug 24 '12 at 23:45
    
@J.M. I had it on an exam; language was assembly for Zilog Z80 processors! –  user2468 Aug 25 '12 at 3:37
    
Note this is horribly inefficient, computationally speaking. –  Thomas Aug 25 '12 at 4:44

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 $5$.

share|improve this answer

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.