Is f: R->R one-to-one and onto?

[relyt]

Define f: R->R by the rule:

$$f(w) = \left\{ \begin{array}{lr} -5, if x < 0\\ 0, if x=0\\ +5, if x > 0 \end{array} \right.$$


if f onto and/or one-to-one?

These questions always get me. Can someone please help me walk through this problem? What do I look for first, etc?

those should be "if" (if x < 0, if x=0, if x > 0 etc)

Any help would be appreciated.$$\bar{}$$

 

[AUMathTutor]

I think I might be able to help you out.

A function is one-to-one if every element in the domain is mapped to a unique element in the range. Another way of saying that is that is that every element in the range is mapped to by at most one element in the domain. An easy way to check for this is to see whether the function passes the horizontal line test.

Your function is not one-to-one because 3 is in R and 10 is in R and f(3) = f(10) = +5. This means both 3 and 10 map to 5, or 5 is mapped to by both 3 and 10. You can draw a horizontal line y = 5 that passes through both points (3,5) and (10,5).


A function is onto if every element in the domain is mapped to by at least one element in the domain. Another way of saying this is that the range of the function is equal to the codomain. One way to show that something is one to one is to find a way to say which element in the domain maps to a given element in the range, and show that the method can find at least one element for any element in the codomain. An easy way of showing a function is not onto is to come up with one example of an element in the codomain which is not mapped to.

For instance, 10 is in R. But there are no x in R for which f(x) = 10. To prove this, note that any real number is positive, negative, or zero. If x is zero, f(x) = 0 =/= 10. If x is negative, x < 0 and f(x) = -5 =/= 10. If x is positive, f(x) = +5 =/= 10. The proof that f is not onto follows from considering these cases.


Does that answer your question?