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Math Help - Quick confirmation

  1. #1
    Junior Member
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    Quick confirmation

    Hi All, this shouldn't take anyone worth their salt too long.... i'm just wondering if this is correct:

    If the task is to prove that:

    (u,v)\rightarrow (u+v,u-v) is a bijection from \mathbb{R}^2onto \mathbb{R}^2...

    Do we go about it this way?

    To show injectivity:

    Assume there exists (u_1,v_1)\neq(u_2,v_2) such that: u_1+v_1 = u_2 + v_2, and also that u_1 - v_1 = u_2 - v_2.



    Isolating u_1 in the first equation, we can sub in to the second to get u_1=u_2 and then we can repeat the similar process, just isolating v_1 to arrive at v_1=v_2?

    Then to show surjectivity:

    Do we just pick any (x,y) in \mathbb{R}^2. We set (x,y)=(u+v,u-v) and then solve for u and v in terms of x and y. I get u=\frac{x+y}{2} and v=\frac{x-y}{2}??
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  2. #2
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    Re: Quick confirmation

    Quote Originally Posted by mathswannabe View Post
    if this is correct:
    If the task is to prove that:
    (u,v)\rightarrow (u+v,u-v) is a bijection from \mathbb{R}^2onto \mathbb{R}^2
    Yes, that works.
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