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Math Help - abstract algebra!

  1. #1
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    abstract algebra!

    I really need help with this question..

    Which of the following are true for all real numbers a,b>0? For those that are not always true, give specific values of a,b>0 where they fail

    (a) 1/a +1/b = 1/(a+b)

    (b) 1/(a) + b = (1+b)/a

    (c) 1/a +1/b = (a+b)/ab
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  2. #2
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    I like c)
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    If you can't do this question, you will have some trouble in an algebra course. Do you know how to add fractions?
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  4. #4
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    I have very little maths back ground. I need an example to be able to understand
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  5. #5
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    \frac{1}{a}+\frac{1}{b}

    You need to make a common denominator, try a \times b = ab

     = \frac{1}{a} {\color{red}\times 1}+\frac{1}{b} {\color{red}\times 1}<br />

     = \frac{1}{a} {\color{red}\times \frac{b}{b}}+\frac{1}{b} {\color{red}\times \frac{a}{a}}<br />

     = \frac{b}{ab} +\frac{a}{ab}<br />

     = \frac{b+a}{ab}

     = \frac{a+b}{ab}
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  6. #6
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    thanks, but i do understand that, What im confused is how does that relate wiht this question.

    would u please be able to demonstrate on (a), (b) and/or (c)
    thank you
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  7. #7
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    Quote Originally Posted by tim_mannire View Post
    thanks, but i do understand that, What im confused is how does that relate wiht this question.

    would u please be able to demonstrate on (a), (b) and/or (c)
    thank you

    I have demonstrated c) as it is able to be proved, the others can't be.

    Job done.
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  8. #8
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    Great, I finally understand.

    But for where it says "For those that are not always true, give specific values of a,b>0 where they fail." how can i do that ?
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  9. #9
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    also, what about fractions like these?

    a/(a+b) = 1/(1+b)


    a/(a+b) = 1/b


    a/(a+b) = 1/(1+(b/a))
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  10. #10
    MHF Contributor Bruno J.'s Avatar
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    To show that a statement doesn't hold, all you need to do is find a counterexample to what the statement asserts. For instance if you want to disprove

    \frac{1}{a}+\frac{1}{b} = \frac{1}{a+b},

    you can just notice that

    \frac{1}{1}+\frac{1}{1} \not = \frac{1}{1+1}.

    Just try a couple of values and you'll soon find a pair which violates the assertion.
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