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This question already has an answer here:

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?

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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. $\ \sqrt2 \sqrt 3 = \sqrt 6$

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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: $(a \times c) \times (\frac 1b \times \frac 1d) = \frac{ac}{bd}$

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  • $\begingroup$ 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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