one to one and onto functions

[mathsfun]

can somebody pls help me figure this problem out??

 

f : Z SUB 7 --> Z SUB 7, f([n] SUB 7) = [2n] SUB 7

 

is this one below correct??

 

f : R --> R, f(x) = 2x + 1

f'(x) = 2 > 0 and f(x) = 2x + 1 passes the horizontal line test so f is 1 to 1

but how do I figure out if it is onto??

 

 

is this one correct below??

 

f : Z -->Z, f(n) = 2n + 1

f'(n)= 2 > 0 and f(n) = 2n + 1 passes the horizontal line test so f is 1 to 1

but how do I figure out if it is onto??

 

 

 

thank you

[MandrakeRoot]

f is clearly injective since the image of any two distinct points is distinct (a simple property of the sum that you can without a doubt assume)

f is surjective since every element has an original in the case where you consider the function f from R into R ! (try finding which one that would be)

When taking the function from Z into Z this is not true since 2 does not have an original ! (In other words 2 is not in the range of your function !)

 

I've tried to explain it rather simple, i hope that worked out ?

 

Mandrake

[matt grime]

I am not sure that I would say this approach is remotely sound. It is philosophically dubious, and mathematically unnecessary, and since at no point do you use the fact you are reducing mod 7 you should worry. the map n to 2n mod 10 isn't bijective for instance.

 

The simple and better way is:

 

suppose f(n)=f(m) then 2n=2m mod 7

 

ie 7 divides 2(n-m), as 2 and 7 are coprime, it must be that 7 divides n-m, ie n=m mod 7. Thus proving f is injective.

 

 

f is surjective is now either trivial, by some larger result, or needs proving. Here is the larger result:

 

if f is a function between two finite sets and f is injective, then f is surjective.

 

proof: if f is not surjective, then there is some element not mapped to, hence, omitting this element, we can define a map that is injective from a set to one strictly smaller. by the pigeon hole principal this cannot be injective # so f was surjective.

 

Or, we could say:

 

Since 2 and 7 are coprime, there are integers a nd b such that

 

2a+7b=1.

 

Let [n] be in Z_7, then 2an+7bn=1, ie 2an=n mod 7. n ws arbitrary hence the map is surjective.

[mathsfun]

thank u, appreciate the input

 

also I am not worried bc all this math stuff for fun I ask questions bc I do really want to understand & it can difficult at times when u r learning on your own

[MandrakeRoot]

Asking questions is always good in order to understand things.

 

Mandrake