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Math Help - Showing a piecewise function is one-to-one

  1. #1
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    Showing a piecewise function is one-to-one

    I'm trying to show that a piecewise function is one-to-one by showing that for all f(a) = f(b), a = b. The function is:
    (x-2)/4 if x/2 is odd
    -x/4 if x/2 is even.

    I can show f(a) = f(b) implies a = b if a/2 and b/2 are both odd, and i can if they are both even.

    Do i need to show f(a) = f(b) implies a = b for the case where a is even and b odd?
    I would think i need to, because it should be for all a and b, but when i tried to i got

    (a-2)/4 = -b/4 ==> a-2 = -b ==> a = -b+2.

    so here it would not be true that f(a) = f(b) implies a = b and thus the function is not one-to-one.

    I'm pretty sure that it is, though, and that i'm doing something wrong. The function is suppose to map positive even numbers to integers.
    (f(2) = 0----------f(4) = -1--------f(6) = 1--------f(8) = -2....
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  2. #2
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    Quote Originally Posted by Relmiw View Post
    I'm trying to show that a piecewise function is one-to-one by showing that for all f(a) = f(b), a = b. The function is:
    (x-2)/4 if x/2 is odd
    -x/4 if x/2 is even.
    What is the initial set for this mapping?
    It appears that you are mapping the even integers to the integers.
    Is that correct?
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  3. #3
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    Quote Originally Posted by Plato View Post
    What is the initial set for this mapping?
    It appears that you are mapping the even integers to the integers.
    Is that correct?
    I am suppose to come up with a function that maps positive even integers to integers and then show it is one to one and onto
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  4. #4
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    Quote Originally Posted by Relmiw View Post
    I am suppose to come up with a function that maps positive even integers to integers and then show it is one to one and onto
    Well then, in the future please post complete in details.
    Suppose \dfrac{a-2}{4}=\dfrac{-b}{4} .
    That means a=2n where n is a positive odd.
    And b=2m where m is positive even.
    But that leads to n=1-m. Is that possible?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Well then, in the future please post complete in details.
    Suppose \dfrac{a-2}{4}=\dfrac{-b}{4} .
    That means a=2n where n is a positive odd.
    And b=2m where m is positive even.
    But that leads to n=1-m. Is that possible?
    I'm not sure where you're getting that a = 2n where n is a positive odd and that b = 2m where m is a positive even.

    Wouldn't a = 2 - b where b is a positive odd and b = a - 2 where a is a positive even from simplifying (a-2)/4 = -b/4?
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  6. #6
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    Quote Originally Posted by Relmiw View Post
    I'm not sure where you're getting that a = 2n where n is a positive odd and that b = 2m where m is a positive even.
    I am getting it from the definition of your function.
    You wrote the function.
    I am showing the one case that you said you could not.
    In order for \dfrac{a}{2} to be odd, it is necessary that a=2n where n is a positive odd.
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  7. #7
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    Quote Originally Posted by Plato View Post
    I am getting it from the definition of your function.
    You wrote the function.
    I am showing the one case that you said you could not.
    In order for \dfrac{a}{2} to be odd, it is necessary that a=2n where n is a positive odd.
    Ok, I think I'm understanding your post about n = 1 - m now.
    This is not possible as n will be a negative integer for all m.

    But, this does not actually mean that the function is not one-to-one because we just showed f(a) /= f(b) if one is even and one is odd (a /= b).

    So basically, f(a) = f(b) implies a = b for both even and both odd. For one of each, neither are true so that also maintains it is one-to-one. I believe.
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  8. #8
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    You function is one-to-one.
    Can you prove that it is onto?
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  9. #9
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    Thank you for your help so far.

    I would try to show this is onto by showing that each individual piece is onto.
    First, (x-2)/4 if x/2 is odd

    so, f(e) = (e - 2) / 4
    y = (e - 2) / 4
    e = 4y - 2

    Now i must show that for all y in Z, e is a positive even initeger. I'm thinking 4y - 2 ==> 2(2k - 2) ==> 2(g) g in Z.
    The only problem is this only shows e is even, not positive.


    similarly, for -x/4 if x/2 is even.
    y = -e / 4
    e = -4y.

    Now i must show that for all y in Z, e is a positive even integer. I'm thinking -4y ==> 2(-2k) ==> 2(g), g in Z.
    But again, this doesn't mean e is positive.
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