1. ## 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

2. ## 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).

3. ## 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

4. ## 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?

5. ## 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|

6. ## Re: Finding 1-1 function

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

7. ## 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

8. ## Re: Finding 1-1 function

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 $f(a)=f(b)$ then

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

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

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

9. ## 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

10. ## Re: Finding 1-1 function

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 $0\in\mathbb{N}$.

11. ## 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

12. ## 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

13. ## Re: Finding 1-1 function

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 $0\notin\mathbb{N}$ 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 $\mathbb{Z} \leftrightarrow \mathbb{N}$

14. ## Re: Finding 1-1 function

If you can't use zero then shift everything one unit to the right.

15. ## Re: Finding 1-1 function

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