Discrete Mathematics- An Open Introduction - 3rd Edition, 2016a

Selected Solutions 341
(d) Such an element is 2 (in fact, that is the only element in the codomain
that is not in the range). In other words, 2 is not the image of any
element under f ; nothing is sent to 2.
0.4.2.
(a) This is neither injective nor surjective. It is not injective because
more than one element from the domain has 3 as its image. It is not
surjective because there are elements of the codomain (1, 2, 4, and 5)
that are not images of anything from the domain.
(b) This is a bijection. Every element in the codomain is the image of
exactly one element of the domain.
(c) This is a bijection. Note that we can write this function in two line
( )
1 2 3 4 5
notation as f
.
5 4 3 2 1
( )
1 2 3 4 5
(d) In two line notation, this function is f
. From this
1 1 2 2 3
we can quickly see it is neither injective (for example, 1 is the image
of both 1 and 2) nor surjective (for example, 4 is not the image of
anything).
0.4.5. There are 8 functions, including 6 surjective and zero injective
functions.
0.4.7.
(a) f is not injective, since f (2) f (5); two different inputs have the
same output.
(b) f is surjective, since every element of the codomain is an element of
the range.
( )
1 2 3 4 5
(c) f
.
3 2 4 1 2
0.4.9. f (10) 1024. To find f (10), we need to know f (9), for which we
need f (8), and so on. So build up from f (0) 1. Then f (1) 2, f (2) 4,
f (3) 8, .... In fact, it looks like a closed formula for f is f (n) 2 n . Later
we will see how to prove this is correct.
0.4.10. For each case, you must use the recurrence to find f (1), f (2) ...
f (5). But notice each time you just add three to the previous. We do this 5
times.
(a) f (5) 15.
(b) f (5) 16.
(c) f (5) 17.