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I know this might sound like an easy question, but I was thinking about it the other day and I couldn't think of any good reasons (aside from the fact that it works). So, why is it that when we multiply fractions, we multiply numerator to numerator, and denominator to denominator?


marked as duplicate by user147263, Micah, Joel Reyes Noche, Mark Bennet, Rebecca J. Stones Jun 24 '15 at 7:37

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By definition $\ x = a/b\ $ is the unique solution of $\ b\,x=a$

By definition$\ y = c/d\ $ is the unique solution of $\ d\,y = c$

Multiplying these equations $\,\Rightarrow\, bd\,xy = ac\ $ so $\ xy = (ac)/(bd),\ $ i.e. $\ \dfrac{a}b\dfrac{c}d = \dfrac{ac}{bd}$

Remark $\ $ Analogous "reductionist" arguments apply elsewhere, e.g.

By definition $\ x = \sqrt2 \ $ is the unique solution $>0\,$ of $\ x^2 = 2$

By definition $\ y = \sqrt 3\ $ is the unique solution $>0\,$ of $\ y^2 = 3$

Multiplying these equations $\,\Rightarrow\, (xy)^2= 6\ $ thus $\ xy = \sqrt 6,\ $ i.e. $\ \sqrt2 \sqrt 3 = \sqrt 6$


Going from first principles a fraction $\frac ab$ is equal to $a \times \frac 1b$. When you multiply this by another fraction $\frac cd = c \times \frac 1d$, you get: $a \times \frac 1b \times c \times \frac 1d$ which can be rearranged by commutativity and associativity into: $(a \times c) \times (\frac 1b \times \frac 1d) = \frac{ac}{bd}$

  • $\begingroup$ Your final result should (ac)/(bd). $\endgroup$ – The Puppet Master Jun 24 '15 at 2:19
  • $\begingroup$ @ThePuppetMaster Yes thanks. Doing this on a phone was harder than I thought, hence the typo. :) $\endgroup$ – Deepak Jun 24 '15 at 3:45
  • $\begingroup$ -1 I don't think that this answer includes a justification for $\frac{1}{b}\times\frac{1}{d}=\frac{1}{bd}$ $\endgroup$ – John Joy Jun 24 '15 at 13:56

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