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Math Help - subspace

  1. #1
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    Smile subspace

    hello.
    (a);\left \{ (x,y)\in \mathbb{R}^{2},\exp x=\exp y\right \}
    i have to prove that a is a subspace.
    i took two elements from (a), (x,y) and (x',y') and i must have (x+x',y+y')\in a.
    so i wrote \exp (x+x')-\exp (y+y')=\exp x \exp x'-\exp y\exp y'
    where i went wrong ?
    i know that a can be written as (a);\left \{ (x,y)\in \mathbb{R}^{2}, x= y\right \},but i want to know where was my error.
    thanks.
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  2. #2
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    Quote Originally Posted by Raoh View Post
    hello.
    (a);\left \{ (x,y)\in \mathbb{R}^{2},\exp x=\exp y\right \}
    i have to prove that a is a subspace.
    i took two elements from (a), (x,y) and (x',y') and i must have (x+x',y+y')\in a.
    so i wrote \exp (x+x')-\exp (y+y')=\exp x \exp x'-\exp y\exp y'
    where i went wrong ?
    i know that a can be written as (a);\left \{ (x,y)\in \mathbb{R}^{2}, x= y\right \},but i want to know where was my error.
    thanks.
    No where went you wrong: e^xe^{x'}=e^{y}e^{y'}...why you tried to make a substraction there I can't say.

    Tonio
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  3. #3
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    Smile

    Quote Originally Posted by tonio View Post
    No where went you wrong: e^xe^{x'}=e^{y}e^{y'}...why you tried to make a substraction there I can't say.

    Tonio
    i tried to show a zero.
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  4. #4
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    Since exp is an invertible function, isn't \{(x,y)|exp(x)= exp(y)\} exactly the same as \{(x,y)| x= y\}?
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  5. #5
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    Smile

    God ! i forgot the fact that \exp(x+x')=\exp(x)\exp(x')=\exp(y)\exp(y')=\exp(y+  y') since
    \exp(x)=\exp(y) and \exp(x')=\exp(y')
    thanks everyone.
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  6. #6
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    You don't have to call me "God"- you can use my webname!
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    You don't have to call me "God"- you can use my webname!
    i meant "oh God !"
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