Math Is Fun Forum

Kryptonis

Member

Registered: 2011-03-03

Posts: 11

functions

a) Determine which of the followings are functions with domain X.
        i) (3 pts) X = {1, 3, 5, 7, 8} and R ={(1,7), (3,5), (5,3), (7, 7), (8,5)}
       
        ii) (3 pts) X = {-2, -1, 0, 1} and R = {(-2, 6), (0, 3), (1, -1)}
       
        iii) (3 pts) X is the set of real numbers and, for x ∈ X,
            g(x) = x^2 − 3x + 2, assume that the codomain is also X
       
        iv) (3 pts) X is the set of real numbers and, for x ∈ X,
            g(x) = sqrt(x^2 − 3x + 2) , assume that the codomain is also X
       
        v) (3 pts) X is the set of real numbers and, for x ∈ X, g(x) = log2 x , ,             assume that the codomain is also X
   
    b) Let Z = {...−2, −1, 0, 1, 2, ...} denote the set of integers. Suppose f :             Z→Z is a function, defined by:

        f (n) = {2 if is odd
               n/ 2 if is even

        i) (5pts) Prove or disprove that f is one-to-one (injective)

        ii) (5pts) Prove or disprove that f is onto (surjective).

Last edited by Kryptonis (2011-03-04 01:48:30)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: functions

hi kryptonis

To be a function it must be 'well defined' for all x in the domain.

(i)   Not sure about this notation ... haven't seen a function given this way

If this means 1 maps to 7, 3 maps to 5 ....etc, then yes it is a function because we know what f(x) is for all x.

(ii) This time we don't.  f(-1) is undefined so it is not a function.

(iii)  Did you mean:

If so, then yes, because squaring a real gives a real, so does times by -3 and so does adding some reals together so g(x) is real in the codomain.

(iv)  Same question ??  Do you want to post something else here?

(v)  logs are not defined for negative reals so, for example, g(-2) is undefined.

(b) 

an injective function is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain....Wiki

But f(-3)= f(-1) = f(1) = f(3) = 2 so it is not injective.

a function is said to be surjective or onto if its image is equal to its codomain. A function f: X → Y is surjective if and only if for every y in the codomain Y there is at least one x in the domain X such that f(x) = y ... Wiki

If n is an even integer the n/2 is an integer, because that's what even means.  Importantly, every integer, when doubled gives an integer so choose any y in the codomain, find 2y (= x) in the domain and we have f(x) = y for all y.

It is a surjection. 

Hope that helps,

Bob

Last edited by Bob (2011-03-04 01:54:06)

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Kryptonis

Member

Registered: 2011-03-03

Posts: 11

Re: functions

ty sir, much appreciated!

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: functions

You are welcome.

:-)

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob