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1.5 hours wasted on this Complex Fraction; Darn these -1's for (1-b)'s

  1. Mar 6, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex] a - \frac {a}{1-\frac{a}{1-a}} [/tex]

    3. The attempt at a solution

    [tex] \frac{a}{1-a} [/tex] This might be the part where I am confused. I always get confused when I need to factor out a -1 from this fraction's denominator in order to put it into the form of [tex] \frac{a}{a-1} [/tex]

    So here is my attempt (3 other attempts were made, but that will take way too long to convert to latex (I'm newbie at it):

    [tex] a - \frac {a}{1-\frac{a}{1-a}} = a - \frac {a}{1+\frac{a}{a-1}} [/tex] This is the part where I factored out a -1 from the [tex] \frac{a}{1-a} [/tex]

    LCM of complex fraction = [tex]a-1[/tex]

    [tex]= a-\frac{a^2-a}{a-1+a} [/tex]

    [tex]= \frac {2a^2-a-a^2+a}{a-1+a}[/tex][tex] = \frac{a^2}{2a-1} [/tex]

    The book's answer: [tex]-\frac {a^2}{1-2a} [/tex]

    Arghh.. what the heck!!! :cry:
     
  2. jcsd
  3. Mar 6, 2009 #2

    gabbagabbahey

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    Your answer is mathematically equivalent to the books answer....just multiply the numerator and denominator by -1
     
  4. Mar 6, 2009 #3

    Mentallic

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    lol so you went through all this grief just because the book's answer was less simplified than yours. Hey, at least you're smarter than the book now :wink:

    What gabbagabbahey was saying is that:

    [tex]-\frac{a^2}{1-2a} =\frac{-a^2}{1-2a} =\frac{-(-a^2)}{-(1-2a)} =\frac{a^2}{2a-1}[/tex]
     
  5. Mar 7, 2009 #4
    gaaahhh... this is so ???????????????? High school math... you're always left wondering how these algorithms work, and you need to be 99.98 percentile to figure it out. Am I suffering the same feeling as most high school students? Do you think this is a bad sign that I will be miserable if I take up engineering, if I'm already clueless on this stuff?

    Anyways... time to start a new thread, this time with I think two complex fractions, one within another (in the denominator). :cry:
     
  6. Mar 7, 2009 #5

    Mentallic

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    This kind of stuff reminds me of how clueless our entire class was at proving trigonometric functions. So much manipulating and we were still getting nowhere fast. It just took practice, and now I rock at it! Just keep trying, you'll be fine.
    Besides, I'm sure you probably knew about multiplying the numerator and denominator by -1, but failed to notice that the answer in the book compared to your answer was just that, because of your frustration.
     
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