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Math Help - Finding 1-1 function

  1. #1
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    Finding 1-1 function

    I am trying to find function f:Z->N that is 1-1.
    Does that mean that every elements is Z (.....-3,-2,-1,0,1,2,3....) has to map to every element of (0,1,2,3.....)
    of just every element of Z has to map to some distinktive element of N
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    Re: Finding 1-1 function

    Hey stribor40.

    You have it correct with regard to your definition of the mapping. Basically for every element in Z you map to a single unique element in N (also N is usually from 1 onwards instead of 0 onwards).

    A hint if you are stuck: try bunching numbers of the same magnitude in lots of 2's (-1,1) (-2,2) and so on and map them to pairs in N (1,2) (3,4).
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    Re: Finding 1-1 function

    Since it says f:Z->N how do i know what actual elements are in Z and
    N? I know that Z are all integers and N are all naturals but to solve this
    problem how do i know what actual numbers are in each set i am trying to
    find function that is 1-1
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  4. #4
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    Re: Finding 1-1 function

    You should take the hint I said above. You have an input z (an element of Z) and you mapping to n (an element of N).

    Consider the hint above where you map each pair of numbers of the same magnitude to a joined pair in the naturals. Let f(0) = 0.

    f(1) = 2, f(2) = 4, f(3) = 6 : what is the pattern here?
    f(-1) = 1, f(-2) = 3, f(-3) = 5 : what is the pattern here?

    Now can you unify those functions into one single function that produces the right answer for the positive and negative integer inputs?
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    Re: Finding 1-1 function

    All i see is something like this...

    f(x) = if x>=0 then 2x
    else if x<0 then |x|
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    Re: Finding 1-1 function

    Quote Originally Posted by stribor40 View Post
    All i see is something like this...

    f(x) = if x>=0 then 2x
    else if x<0 then |x|
    That function is not one-to-one because f(1)=f(-2).

    Try
    f(x) = \left\{ {\begin{array}{rl}{2x,}&{x \ge 0}\\{2|x| - 1,}&{x < 0}\end{array}} \right.
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    Re: Finding 1-1 function

    how would that prove above function to be 1-1 when we have 2 cases. for example by following definition i can easily prove that this function is 1-1
    f(x)=5x-6

    i have to show that f(a) = f(b) implies a=b

    5a - 6 = 5b-6
    5a-5b = 6-6
    5a-5b=0
    5a=5b
    a=b yes

    how to prove above function in similar fashion
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  8. #8
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    Re: Finding 1-1 function

    Quote Originally Posted by stribor40 View Post
    how would that prove above function to be 1-1 when we have 2 cases. for example by following definition i can easily prove that this function is 1-1
    f(x)=5x-6

    i have to show that f(a) = f(b) implies a=b
    If f(a)=f(b) then

    a \ge 0\;\& \;b \ge 0 \Rightarrow \;2a = 2b \Rightarrow \;a = b

    a < 0\;\& \;b < 0 \Rightarrow \; - 2a - 1 =  - 2b - 1 \Rightarrow \;a = b


    \\a\ge 0~\&~b<0\\2a=-2b-1\\\text{what is the contradiction there ?}
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    Re: Finding 1-1 function

    You cant map 2x because when input is zero.
    f(0) = 2*0 = 0 and there is no zero in N or am i wrong here
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    Re: Finding 1-1 function

    Quote Originally Posted by stribor40 View Post
    You cant map 2x because when input is zero.
    f(0) = 2*0 = 0 and there is no zero in N or am i wrong here
    There is no general agreement on that point.
    Most who work in the area of foundations do say that 0\in\mathbb{N}.
    But follow whatever your textbook says.
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    Re: Finding 1-1 function

    I looks like this function is not 1-1 because in cases where a >=0 and b<0 so functions is not 1-1
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    Re: Finding 1-1 function

    Would this function be onto f:Z->N

    f(x) = { if x<= 0 then 1
    If x > 0. Then x

    It hits every element in N
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    Re: Finding 1-1 function

    Quote Originally Posted by stribor40 View Post
    Would this function be onto f:Z->N
    f(x) = { if x<= 0 then 1
    If x > 0. Then x
    It hits every element in N
    If we say that 0\notin\mathbb{N} define
    f(x) = \left\{ {\begin{array}{rl}{2x+2,}&{x \ge 0}\\{2|x| - 1,}&{x < 0}\end{array}} \right..

    Now f is a bijection \mathbb{Z} \leftrightarrow \mathbb{N}
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  14. #14
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    Re: Finding 1-1 function

    If you can't use zero then shift everything one unit to the right.
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    Re: Finding 1-1 function

    Quote Originally Posted by Plato View Post
    There is no general agreement on that point.
    Most who work in the area of foundations do say that 0\in\mathbb{N}.
    But follow whatever your textbook says.
    That what's my knowledge. I noticed that some publications use WHOLE numbers set, and Natural numbers. the 0 is included in the whole numbers set and excluded from the natural numbers set. I don't know what is the rationale for this classification.
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