Functions: Injective, Surjective, or Bijective?

[Jamers328]

Here is the question:
Classify each function as injective, surjective, bijective, or none of these.

a) f: N -> N defined by f(n)=n+3
b) f: Z -> Z defined by f(n)=n-5
c) f: R -> R defined by f(x)=x^3 - x
d) f: [1, infinity) -> [0, infinity) defined by f(x)=x^3 - x
e) f: N -> Z defined by f(n)=n^2 - n
f) f: [3, infinity) -> [5, infinity) defined by f(x)= (x-3)^2 + 5
g) f: N -> Q defined by f(n)=1/n

where N are the natural numbers, Z are the integers, etc...

This is what I thought of so far:
a) Injective
b)
c) Surjective
d) Bijective
e) Injective
f)
g)

Please help... Thanks!

 

[daon]

b) Its a line within Z... What do you know about lines?
f) It is strictly increasing on its domain (hence injective). Suppose y=(x-3)^2+5, with y >= 5. Does that imply x >= 3? If so, its surjective.
g) Well, if 1/p=1/q does p=q (p and q positive integers)? Is every rational number able to be obtained this way?

 

[Jamers328]

Do you think I did a, c, d, and e correct?

Thanks for your help.
b) Injective
f) I think it is surjective then, so bijective
g) Yes, p would equal q. Bijective?

 

[daon]

Hmmm..

For b: Well, it is injective but its also surjective. If you take an integer k, is there some integer n such that f(n)=k? If so... what is it?
For f: Yes... but why you you think its surjective too? Theres a bit of minor algebra involved... make sure you show why.
For g: No... Is there a positive integer p such that f(p)=2/3?

The others look fine.

 

[Jamers328]

Wow I am terrible at these.

b) The integer k?
g) No, there is definitely not.

Thanks a lot for your help. I took this quiz last Thursday, probably did bad... but it is still good for me to understand these!

 

[daon]

No, the integer n such that f(n)=k, is k+5... since f(n)=f(k+5)=(k+5)-5=k. All that was really needed was to solve for n in the equation n-5=k. In general if there is a solution and it is unique (no \(\displaystyle \pm\)'s in which both solutions work for example) then your function is surjective.

The reason #1 wasn't surjective is because trying to solve n+3=k results in n=k-3. But this does not work for k=1,2,3. To be surjective, it must work for ALL elements in the set where f is being sent... in this case N=1,2,3,.... If we modify the domain to include (-1), (-2), and 0 then we would have a bijetcive function.

 

[Jamers328]

I really do understand now. With your help, and with my teacher explaining it the way you just did yesterday.

Thank you SO much. You're amazing. Sorry these are just confusing to me!