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
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
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).
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
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?
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