As far as I know for any rational numbers $a$, $b$ and $c$ the following is true: $(a\cdot b)/c = (a/c)\cdot b = a\cdot (b/c)$.

How is this property called?

  • $\begingroup$ I don't think this particular equality has a name. It's a combination of associativity, commutativity, and the definition of division by $c$ as multiplication by $c^{-1}$. $\endgroup$ – Najib Idrissi Jul 30 '13 at 9:54

This represents a mixture of properties associated with multiplication.

First we note that division by $c\neq 0$ is the same as multiplication by $c^{-1}=\cfrac 1c$, so we really are dealing with multiplication in disguise.

The fact that it doesn't matter where we put the brackets - so that $(ab)c^{-1}=a(bc^{-1})$ is the associative law for multiplication.

NB: if we use $/$ to represent the division operator, we need the brackets in, because the $/$ operation is not associative. That's why I converted to $c^{-1}$ form first, as this helps to make clear what is happening. If we use the $c^{-1}$ form we can omit the brackets.

The fact that the order does not matter, so that (omitting brackets) $abc^{-1}=ac^{-1}b$, depends on the commutative law for multiplication - that $ab=ba$. Again note carefully that the symbol "/" is not commutative $(a/b \neq b/a)$.

  • $\begingroup$ @Bentley4 What is not true that $a\cdot b^{-1} \neq a{-1} \cdot b$ - it is true that $a\cdot b^{-1}=b^{-1}\cdot a$. The $/$ symbol is sometimes useful in simple expressions, but can get confusing in lengthy ones - and of course gets used in spreadsheet formulae and things like that. $\endgroup$ – Mark Bennet Jul 30 '13 at 10:24
  • $\begingroup$ Isn't this the most succinct way of putting it: that $^{-1}$ notation of division is commutative [$ab^{-1}c^{-1}=ac^{-1}b^{-1}$] but non-associative [$(ab^{-1})c^{-1} \neq a(b^{-1}c^{-1})$]? $\endgroup$ – Bentley4 Jul 30 '13 at 11:17
  • $\begingroup$ Again you need to be a little careful. $(ab)^{-1}c^{-1}\neq ab^{-1}c^{-1}$, but $(ab^{-1})c^{-1}=a(b^{-1}c^{-1})=\frac a {bc}$. $\endgroup$ – Mark Bennet Jul 30 '13 at 11:26
  • $\begingroup$ I understand now why I made that last false statement. I wanted to write the equivalent of $(a(1/b/1/c))$ but I falsely inferred this was $a(b^{-1}c^{-1})$. I also understand your answer now, thanks! $\endgroup$ – Bentley4 Jul 30 '13 at 22:27

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