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  1. #1 inequality 
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    How do you solve this inequality (0<1/x<2) I'm in calculus, but I forgot how to do this. Thanks


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  3. #2  
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    How about inveting both side
    wgat you get is 0<x/1<1/2
    therefore x is
    0<x<1/2


    e = mc^2
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  4. #3  
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    How about inveting both side
    wgat you get is 0<x/1<1/2
    therefore x is
    0<x<1/2
    that is incorrect, in inequalities you cannot invert both sides

    3>2
    1/3 is not > 1/2

    Here is how to do it:-

    0<1/x<2
    Multiply by x:-
    0 < 1 < 2x

    1/2 < x

    therefore this inequality remains true for all values greater then 1/2.
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  5. #4  
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    thanks for the answers, but I found out how to do it veli had the right concept but you also have to flip the signs. so if the inequality was
    2<1/x<3 the answer will be 1/2>x>1/3. you have to flip the signs and take the reciprocal of every thing.
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  6. #5  
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    Hmm, I will try not to be offended but my answer is correct....

    If you flip the sign of Veli inequality you get x > 1/2

    I believe that is the same as my answer 1/2 < x.

    Both methods are fine...

    Still it is correct to flip the inequality when inverting. You also flip the inequality if you multiply by - 1.
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  7. #6  
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    ya you where right too .
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  8. #7  
    Forum Radioactive Isotope mitchellmckain's Avatar
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    Quote Originally Posted by veli
    How about inveting both side
    wgat you get is 0<x/1<1/2
    therefore x is
    0<x<1/2
    Hmmm... no one has mentioned that the left hand side is incorrect as well. If you include the limit called infinity in your number set then the correct operation would be....

    0<1/x<2
    to
    infinity > x > 1/2

    which brings up the intriguing question of how you would handle
    -2<1/x<2
    well you can seperate this into
    -2<1/x<-0 OR 0<1/x<2
    then our operation gives us
    -1/2 > x > -infinity OR infinity > x > 1/2
    Notice the use of the limit -0 to avoid the obviously wrong conclusion
    -1/2 > x > infinity OR infinity > x > 1/2 which makes no sense.
    I have also left out the possibility that 1/x = 0 under the assumption that a only real solutions were desired.

    Of course multiplying by x is more sensible using a more elementary level of mathematics but it is not as interesting.
    See my physics of spaceflight simulator at http://www.relspace.astahost.com

    I now have a blog too: http://astahost.blogspot.com/
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