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Math Help - Injection and Surjection

  1. #1
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    Injection and Surjection

    Could you see if what i have done is right for the following questions:

    Ive got to find whether they are injection surjection or bijection for any integer to any integer so z to z (sorry i cant use latex)

    1. f(x) = x^3

    bijection

    2. g(x) = x-3

    bijection

    3. h(x) = 3x+1

    bijection

    4. i(x) = (x^2) -1

    Surjection
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  2. #2
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    First three are bijections, but the last one it's not, can you see why?
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    First three are bijections, but the last one it's not, can you see why?
    is it because i(n1) = i(n2)

    one of them can be negative and one can be positive so n1 does not equal n2?
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  4. #4
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    Meaning that for different points you got the same images, then yes.
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  5. #5
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    Specfically, i(2)= 4+1= 5= i(-2).
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Krizalid View Post
    First three are bijections, but the last one it's not, can you see why?
    No...1 and 3 are not surjective.

    In 1, what maps to, say, 2? The image is only powers of 3, \{ \ldots -9, -3, -1, 0, 1, 3, 9, 27, \ldots \}.

    In 3, what maps to, say, 2? Not every integer is of the form 3x+1 for x \in \mathbb{Z}...
    Last edited by Swlabr; January 1st 2010 at 02:52 AM.
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  7. #7
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    Consider x=\sqrt[3]2.
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  8. #8
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    Quote Originally Posted by Swlabr View Post
    No...1 and 3 are not surjective.

    In 1, what maps to, say, 2? The image is only powers of 3, \{ \ldots -9, -3, -1, 0, 1, 3, 9, 27, \ldots \}.

    In 3, what maps to, say, 2? Not every integer is of the form 3x+1 for x \in \mathbb{Z}...

    Editing a typo.
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  9. #9
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    Quote Originally Posted by Krizalid View Post
    Consider x=\sqrt[3]2.
    But \sqrt[3]2 \notin \mathbb{Z}...
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  10. #10
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    The original post is hard to read. But it does say that the functions are \mathbb{Z}\mapsto\mathbb{Z}.
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  11. #11
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    My bad, didn't read that!
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  12. #12
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    Aaahh right i see.

    My next question it says determine the inverse function wherever this exists.

    How do we find out if an inverse function exists and then is it just 1 over the equation?
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  13. #13
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    Quote Originally Posted by adam_leeds View Post
    Aaahh right i see.

    My next question it says determine the inverse function wherever this exists.

    How do we find out if an inverse function exists and then is it just 1 over the equation?
    No, the inverse function is NOT the reciprocal. I assume that by the time you are asked to find inverse functions, you have seen the definition and should know that. There is an unfortunate notation: we use f^{-1} which, with numbers, implies the reciprocal (the multiplicative inverse) but the definition for functions is " f^{-1} is the inverse function to f if and only if f^{-1}(f(x))= x and f(f^{-1}(x))= x for all x". That is, f^{-1} "undoes" whatever f does to x and vice versa.

    A standard way of finding an inverse for a function, f:
    1) Write the equation y= f(x).
    2) Swap x and y: x= f(y).
    3) Solve for y.

    For example, your first function if f(x)= x^3 so you would write that as y= x^3 and then swap x and y to get x= y^3. To solve for y, you would use the cube root, the "opposite" of cubing, to get y= \sqrt[3]{x} or, equivalently, y= x^{1/3}.

    However, these functions, we were told, were "from Z to Z". If that is still true, then this function does not have an inverse function because the cuberoot of an integer is not necessarily an integer.
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  14. #14
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    So the inverse of y = x - 3
    is
    x = y - 3

    y = x + 3

    this is an inverse as all integers go to an integer
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  15. #15
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    for 3

    y = 3x + 1
    x = 3y + 1
    (x-1)/3 = y

    y = (x-1)/3

    not an inverse as it becomes a fraction not an integer when you put an integer in for x.
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