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Simple question. Why is the following true?

$$\frac{ab}{c}=\frac{a}{c}b$$

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    $\begingroup$ $b=\frac{b}{1}$ $\endgroup$
    – randomgirl
    Jun 14, 2020 at 0:09
  • $\begingroup$ What are $a,b,c$? Please clarify your question with an edit. $\endgroup$
    – mrtaurho
    Jun 14, 2020 at 0:11

5 Answers 5

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Dividing by $1$ does not change a number, so $b=\frac b1$

When multiplying two fractions, simply multiply the numerators and multiply the denominators.

So $$\frac acb=\frac ac\frac b1=\frac{a\times b}{c\times1}=\frac{ab}c$$

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Using the basic properties of fractions: $$\frac{ab}{c} = \frac{a\cdot b}{c\cdot1} = \frac{a}c\cdot\frac{b}1=\frac{a}{c}\cdot b$$

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If you rearrange the equality to this:

${\frac{ab}{c}} = {\frac{a}{c}} * {\frac{b}{1}}$

You can see why this equality works. You multiply the a and b getting you ab, and multiply c and 1 to get c. Thus, the equation works.

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It's just a property of fractions $$\frac{ab}{c}=\frac{a\cdot b}{c}=\frac{a}{c}\cdot b=\frac ac b$$

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I am adding another only because it would be nice to see a proof using basic axioms. Let us assume $a, b, c$ belong to a field. In a field multiplication is associative and commutative and each non-zero element $y$ has a multiplicative inverse $y^{-1}$' Division, $x/y$, is then defined to be $x y^{-1} $.

In this case, then, assuming $c \neq 0$, \begin{align} \frac{ab}{c} &\equiv (ab)c^{-1} & \text{definition of division}\\ &= a(bc^{-1}) & \text{multiplication is associative} \\ &= a(c^{-1}b) & \text{multiplication is commutative} \\ &= (ac^{-1}) b & \text{multiplication is associative, again} \\ &=\frac{a}{c} b & \text{definition of division} \end{align}

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