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Proof of Bijection

  1. Feb 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the derivate of [tex]y = \frac {1} {2} ax^{2} + bx[/tex] is a bijection, when [tex] a, b, x \in \Re [/tex]

    3. The attempt at a solution
    y' = 2ax + b is a linear mapping, where [tex] a, b, x \in \Re [/tex].
    The mapping is [tex]\Re \rightarrow \Re [/tex].

    The mapping is an injection as each element in the domain maps to codomain.
    The mappning is a surjection as elements in the domain maps all elements in the
    codomain.
    (I am not sure about the proofs for the injection and surjection)

    Thus, the derivate is bijection, since it is an injection and surjection.
     
  2. jcsd
  3. Feb 20, 2009 #2

    HallsofIvy

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    No, that's the definition of "function". In order to be an injection, each distinct member of the domain must map to a distinct member of the codomain: if [itex]x\ne y[/itex] then [itex]f(x)\ne f(y)[/itex]. That is the same as saying, "if f(x)= f(y) then x= y".

    Prove:
    injection: if f(x)= f(y) (that is 2ax+ b= 2ay+ b) then x= y. That's simple algebraic manipulation.
    surjection: Given any real number y, there exist x such that f(x)= y. If f(x)=2ax+ b= y. If you can solve for x, it certainly exists!
     
  4. Feb 21, 2009 #3
    Thank you!
    So the point is for injection that each element in the domain equals the one in codomain. For intstance, f(x) = f(y), then x = y.
    In contrast to surjection, there is any real number y such that f(x) = y. This gives f(x) = 2ax + b = y. The bottom line is that if you can solve for all x, a surjection exists.
     
  5. Feb 21, 2009 #4

    HallsofIvy

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    The "bottom line" is to prove for all y there is such and x.
    Here, of course, if y is any number and y= ax+ b, then ax= y- b so x= (y-b)/a.

    Notice, by the way, that your original problem:
    is imposible (because it's not true) unless you put an additional condition on a!
     
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