Question

1. Let f : Z to Z and g : Z to Z be functions defined by f (x) = 3x + 1 and g(x) = [x/2]
.
(a) Is fog one-to-one? Prove your answer.
(b) Is fog onto? Prove your answer.
(c) Is gof one-to-one? Prove your answer.
(d) Is gof onto? Prove your answer.

Answer 1

<b>one-to-one</b> (injection):
every element in the codomain is mapped by <i>at most</i> one element of the domain.

<b>onto</b> (surjection):
every element in the codomain is mapped by <i>at least</i> one element of the domain.

We can see that f(x) is both one-to-one but not onto, since
f(1)=4, f(2)=7, so codomain (Z) contains "holes" such as 5 and 6 which are not mapped.
On the other hand, g(x) is onto, but not one-to-one, since
g(2)=g(3)=1 (assuming [x/2] is the greatest integer function of x/2).

(a)
fog(x)=f(g(x)) is not one-to-one since
fog(2)=f(g(2))=f(1)=3(1)+1=4
fog(3)=f(g(3))=f(1)=3(1)+1=4
gives a counter-example.
(b)
fog(x)=f(g(x)) is not onto since the codomain (Z) still contain holes (such as 5 and 6 which are not mapped.

I will leave (c) and (d) for you as exercise.