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Math Help - simplification?

  1. #1
    Newbie
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    Oct 2009
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    4

    simplification?

    a-a
    _
    a - a
    _
    b

    not sure if that makes sense like that but im not entirely sure how to write it, basically a take away a over a take away a over b, simplified.

    Would I be right to make the last part

    a a
    _-_
    1 b

    so it turns into 1-b because the top parts cancel each other out, so I'm left with a-a
    _
    1-b?

    Or is that completely on the wrong track?
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  2. #2
    Super Member
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    Did you mean \frac{a-a}{\frac{a-a}{b}}?
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  3. #3
    Newbie
    Joined
    Oct 2009
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    Nope. Hmm, how to explain it.

    a-z
    _
    q - y
    _
    b

    a=z=q=y, but hopefully it gives more clarification, its a minus z over q minus y over b.

    edit: nope, cant get the underscores to stay spaced out how id like ><
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  4. #4
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    I think you mean:

    \frac{(\frac{a - z}{q - y})}{b}, well if this is the case let's simplify it:

    \frac{a-z}{q-y} \cdot \frac{1}{b},

    \frac{a - z}{(q - y) \cdot b}

    There isn't any possible way to simplify this, if I apply this to your question:

    \frac{\frac{a-a}{a-a}}{b},

    \frac{a-a}{a-a} \cdot \frac{1}{b}

    1 \cdot \frac{1}{b} = \frac{1}{b}

    It can't be simplified any further.

    In case you have the thing that Defunkt wrote:

    \frac{a-a}{\frac{a-a}{b}}
    (a - a) \cdot \frac{1}{\frac{a-a}{b}}
    (a - a) \cdot \frac{b}{a-a} = 0 \cdot \frac{b}{0}

    not possible because we divide by a-a = 0

    We could've also seen this by:

    \frac{0}{\frac{0}{b}} = \frac{0}{0} (this is undefined)
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