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Thread: explain the second property of fractions

  1. #1
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    explain the second property of fractions

    property: a/b / c/d = a/b * d/c
    description: when dividing fractions invert the divisor and multiply.

    can someone explain this to me with proof? i don't understand why it works.
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  2. #2
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    Re: explain the second property of fractions

    $\begin {align*}
    &\dfrac{\dfrac a b}{\dfrac c d} = \\ \\

    &\dfrac{\dfrac a b}{\dfrac c d} \cdot \dfrac d d = \\ \\

    &\dfrac{\dfrac{a d}{b}}{\dfrac {c d}{d}} = \\ \\

    &\dfrac{\dfrac{ad}{b}}{c} = \\ \\

    &\dfrac{ad}{bc}

    \end{align*}$
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  3. #3
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    Re: explain the second property of fractions

    I would demonstrate this similarly, but slightly differently.

    First, it is important to know that anything divided by itself is 1.

    So, $\dfrac{a}{a} = 1$ for all $a\neq 0$. We can make this more complicated.

    $\dfrac{blah}{blah} = 1$

    $\dfrac{\tfrac{a}{b}}{\tfrac{a}{b}} = 1$

    So long as the numerator and denominator are equal, you get 1.

    Next, any number times 1 gives the original number.

    $a = a\cdot 1$

    Any number divided by 1 gives the original number:

    $\dfrac{a}{1} = a$

    And finally, multiplication is commutative:

    $ab = ba$

    Commutativity is true as part of a fraction, as well:

    $\dfrac{ab}{cd} = \dfrac{ba}{cd} = \dfrac{ab}{dc} = \dfrac{ba}{dc}$

    $\begin{align*}\left( \dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}\right) & = \left( \dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}\cdot 1 \right) \\ & = \left( \dfrac{\tfrac{a}{b}}{\tfrac{c}{d}}\cdot \dfrac{\tfrac{d}{c}}{\tfrac{d}{c}} \right) \\ & = \left( \dfrac{\tfrac{a}{b}\cdot \tfrac{d}{c}}{\tfrac{c}{d}\cdot \tfrac{d}{c}} \right) \\ & = \left( \dfrac{\tfrac{ad}{bc}}{\tfrac{cd}{dc}} \right) \\ & = \left( \dfrac{\tfrac{ad}{bc}}{\tfrac{cd}{cd}} \right) \\ & = \left( \dfrac{\tfrac{ad}{bc}}{1} \right) \\ & = \left( \dfrac{ad}{bc} \right) \\ & = \left( \dfrac{a}{b}\cdot \dfrac{d}{c} \right) \end{align*}$

    I went through a lot more steps to clarify in a bit more detail what is happening at each step, but romsek's approach is correct, as well.
    Thanks from topsquark
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