1
$\begingroup$

In mathematics we have the Addition operation from which subtraction and multiplication can be resulted. so it's not wrong to say they both have the concept of addition within them. $$ a*b = a+a+...+a \ \ \ \ \ (b\ times)$$ and $$ a-b = a+ (-b) .$$ division has the concept of multiplication within it: $$ \frac{a}{b} = a* \frac{1}{b}$$ but I'm not so sure if it is so, because there is still division in it in $ \frac{1}{b} $ and can't be decomposed any more as far as I know.

also, I assume division doesn't have the concept of addition in it: no way to say how $ \frac{a}{b} $ can be decomposed to addition, and also no way to say how it's possible to add "a", $ \frac{1}{b} $ times.

$ Q_1 $: so how is Division related to or can be derived from Addition and Multiplication?

$Q_2$ : what branch of math is this subject related to?

$\endgroup$

closed as unclear what you're asking by Andrés E. Caicedo, user21820, JonMark Perry, Claude Leibovici, mlc Nov 5 '17 at 9:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ In abstract algebra, there is a structure called a division ring that generalizes division from fields like $\mathbb{R}$ and $\mathbb{C}$ to a broader class of rings. $\endgroup$ – Michael Lee Aug 1 '17 at 10:46
  • $\begingroup$ Yes; inverse operation. $a - a=0$ as well as $a \times \dfrac 1 a =1$. $\endgroup$ – Mauro ALLEGRANZA Aug 1 '17 at 10:46
  • 1
    $\begingroup$ And the subject is arithmetic. $\endgroup$ – Mauro ALLEGRANZA Aug 1 '17 at 10:48
  • 1
    $\begingroup$ 1/b is the inverse of b, it is usually stated as an axiom in say analysis, where every element of a field, say, has a multiplicative inverse. I would also say fields is a part of abstract algebra, which goes much further in depth than analysis. $\endgroup$ – smokeypeat Aug 1 '17 at 10:50
  • 1
    $\begingroup$ You say that $\frac{a}{b}=a*\frac{1}{b}$ still has division in it, the exact same applies to $a-b = a + (-b)$, It's just that in this second notation we don't write the neutral element $0$ anymore, but you can: $a-b = a + (0-b)$. Both cases are examples of the exact same thing (note that in the first case, the $1$ is also the neutral element). yet in the case of substraction you are sure but in the case of division you are not? $\endgroup$ – Jens Renders Aug 1 '17 at 10:56
3
$\begingroup$

Division, as an operation, is the inverse of multiplication. So, the number $$\frac ab$$ is defined to be the unique solution to the equation $$b\cdot x = a.$$

This definition also explains why

  • $\frac a0$ is not defined for $a\neq 0$ (because there is no solution to the equation $0x=a$ in that case)
  • $\frac 00$ is not defined (because there is more than one solution to the equation $0\cdot x = 0$.
$\endgroup$
  • $\begingroup$ thank you, so there is no division it's basically multiplication for numbers greater than 0 and less than 1. is that it? $\endgroup$ – parvin Aug 1 '17 at 10:54
  • $\begingroup$ if that's true, how can you explain 0.5 * b? add b half times? $\endgroup$ – parvin Aug 1 '17 at 10:55
  • $\begingroup$ @parvin Well sure there is such a thing as division. It's just that yes, division is the same thing as multiplication by the inverse. And the same thing as solving an equation. $\endgroup$ – 5xum Aug 1 '17 at 10:55
  • 1
    $\begingroup$ @parvin You can have a real world example. I you have $b$ litres of milk, then $0.5\cdot b$ litres of milk is the amount of milk you must give each person if you want to split all the milk equally among $2$ people. $\endgroup$ – 5xum Aug 1 '17 at 11:00
  • 1
    $\begingroup$ @Arthur I was talking about the "there is no division it's basically multiplication for numbers greater than 0 and less then 1." analogy. $\endgroup$ – Matt Aug 1 '17 at 12:07

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