# Finding 1-1 function

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• Sep 25th 2013, 06:07 PM
stribor40
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
• Sep 25th 2013, 07:04 PM
chiro
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).
• Sep 26th 2013, 02:12 AM
stribor40
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
• Sep 26th 2013, 02:56 AM
chiro
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?
• Sep 26th 2013, 05:54 AM
stribor40
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|
• Sep 26th 2013, 06:05 AM
Plato
Re: Finding 1-1 function
Quote:

Originally Posted by stribor40
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
$\displaystyle f(x) = \left\{ {\begin{array}{rl}{2x,}&{x \ge 0}\\{2|x| - 1,}&{x < 0}\end{array}} \right.$
• Sep 27th 2013, 08:57 AM
stribor40
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
• Sep 27th 2013, 09:09 AM
Plato
Re: Finding 1-1 function
Quote:

Originally Posted by stribor40
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 $\displaystyle f(a)=f(b)$ then

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

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

$\displaystyle \\a\ge 0~\&~b<0\\2a=-2b-1\\\text{what is the contradiction there ?}$
• Sep 28th 2013, 08:47 AM
stribor40
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
• Sep 28th 2013, 09:08 AM
Plato
Re: Finding 1-1 function
Quote:

Originally Posted by stribor40
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 $\displaystyle 0\in\mathbb{N}$.
• Sep 28th 2013, 02:39 PM
stribor40
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
• Sep 28th 2013, 02:58 PM
stribor40
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
• Sep 28th 2013, 03:31 PM
Plato
Re: Finding 1-1 function
Quote:

Originally Posted by stribor40
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 $\displaystyle 0\notin\mathbb{N}$ define
$\displaystyle f(x) = \left\{ {\begin{array}{rl}{2x+2,}&{x \ge 0}\\{2|x| - 1,}&{x < 0}\end{array}} \right.$.

Now $\displaystyle f$ is a bijection $\displaystyle \mathbb{Z} \leftrightarrow \mathbb{N}$
• Sep 28th 2013, 05:04 PM
chiro
Re: Finding 1-1 function
If you can't use zero then shift everything one unit to the right.
• Sep 28th 2013, 05:27 PM
votan
Re: Finding 1-1 function
Quote:

Originally Posted by Plato
There is no general agreement on that point.
Most who work in the area of foundations do say that $\displaystyle 0\in\mathbb{N}$.