See #2 of http://www.mathhelpforum.com/math-he...ts-proofs.html by kathrynmath.
A) For each part, find a function f: R -> R that has the desired properties: neither onto nor one-to-one
B)Under what conditions does A\(A\B) = B?
C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n element of N(natural numbers).
D) Given A = {1,2,3,4,5}, B = {2,3,4,5,6,7} and C = {a,b,c,d,e} state an example of f: A ->B, g: B-> C, such that g(f)(the composition of g onto f) is 1-1 but g is not 1-1.
See #2 of http://www.mathhelpforum.com/math-he...ts-proofs.html by kathrynmath.
C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n which is an element of N.
I need to use what was given(the line above is all that is given) to find: the image of f, if f is 1-1, if f is onto, and if f is onto I need to find its inverse, the domain of the inverse, and the range of the inverse.
here is what I have so far:
The Im(f) has to be all odd numbers because that is what you get when you plug numbers into f(n)=2n-1.
from there I get stumped.
Although the following remark is not applicable for this exercise, I would like to tell it.
Remark. Let be a function and be finite sets if f is onto, then it is one-to-one; vice versa, if f is one-to-one, then it is onto.
if , then can not be a function with an image which is a subset of (pick an element which is not in the set , and see that it is mapped into the set ).
Therefore, we must have .
If , then is one-to-one and onto.
If , then is only one-to-one.
Just, try to figure it out by yourself by letting , where and . Note that is strictly increasing, which indicates that it is one-to-one.
Then obtain the set by isolating ...
If is a function from to defined by , we see that for all .
This means that for every , there exists such that holds.
Note that for every , we may not find .
Therefore, we see that
holds.
Since , we this indicates that
Hence, the domain of can be picked to be any subset of , i.e. .
Now consider the following possible cases.
If , then is one-to-one and onto.
If , then is only one-to-one.
Hint. is strictly icreasing, and it is hence one-to-one.
I guess it is more clear now?