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Math Help - Identity proof

  1. #1
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    Identity proof

    I have to prove that if |x| <> 1, y <> 0, y<>-x then identity is valid:

    \frac{x^2-1}{y^2+xy}\cdot(\frac{1}{1-\frac{1}{y}}-1)\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}


    I have try to prove it.This what I have done:

    \frac{x^2-1}{y^2+xy}\cdot(\frac{1}{y-1})\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}

    \frac{(1-x^2)(-1+xy^3+y^4-y)}{(y^2+xy)(y-1)(1-x^2)}=\frac{y^2+y+1}{y}

    \frac{(-1+xy^3+y^4-y)}{(y^2+xy)(y-1)}=\frac{y^2+y+1}{y}

    \frac{-1+xy^3+y^4-y}{y^3-y^2+xy^2-xy}=\frac{y^2+y+1}{y}

    \frac{y(-1+xy^3+y^4-y)}{y(y^2-y+xy-x)}=y^2+y+1

    \frac{-1+xy^3+y^4-y}{y^2-y+xy-x}=y^2+y+1

    -1+xy^3+y^4-y=(y^2-y+xy-x)(y^2+y+1)

    -1+xy^3+y^4-y=-x+xy^3+y^4-y

    x=1


    So identity is not valid.
    Am I right?
    Last edited by DenMac21; January 14th 2006 at 03:04 PM.
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  2. #2
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    Did you make a mistake when you multiplied the 3 terms by the 4 terms?
    And I also do not understand what all that stuff in the beginning is?
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  3. #3
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    Quote Originally Posted by ThePerfectHacker
    Did you make a mistake when you multiplied the 3 terms by the 4 terms?
    I don't think so, I checked.

    Quote Originally Posted by ThePerfectHacker
    And I also do not understand what all that stuff in the beginning is?
    What do you mean exactly?
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  4. #4
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    I also checked, confirmed.

    That means that if the identity is true then x=1 which means that if x is anything else from 1 then it is false. Which means it cannot be an identity.
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  5. #5
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    Quote Originally Posted by DenMac21
    I have to prove that if |x| <> 1, y <> 0, y<>-x then identity is valid:

    \frac{x^2-1}{y^2+xy}\cdot(\frac{1}{1-\frac{1}{y}}-1)\cdot\frac{1-xy^3-y^4+y}{1-x^2}=\frac{y^2+y+1}{y}


    I have try to prove it.This what I have done:

    [...skip...]

    x=1


    So identity is not valid.
    Am I right?
    As in proving trig identities, here it is better to check first if the probvlem identity is true for a numerical substitution.
    Try x=0 and y=2.
    You get the LHS to be 13/4, while the RHS to be 7/2.
    LHS <> RHS, hence, not an identity.

    So no need to crack cranium for those x's and y's.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    I also checked, confirmed.

    That means that if the identity is true then x=1 which means that if x is anything else from 1 then it is false. Which means it cannot be an identity.
    The " |x| <> 1, y <> 0, y<>-x" is meant to be the set of conditions:

    x^2 \ne 1, y \ne 0, y \ne -x

    RonL
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