SC/MATH 1019B - HOMEWORK 2
DUE OCTOBER 23, 2024
Solutions to the problems below must be brought to class on October,23 2018. Solu-
tions may by typed or neatly hand written. You must clearly indicate which problem
you are solving. All solutions must be fully justified.
# 1.
(a) Given an example of a function
f
: N → N which is injective but not
surjective.
(a) Given an example of a function g : N → N which is surjective but not
injective.
An example of an injective but not surjective function is f : N → N by f (n) = n+5.
If
f
(n
1
)
= f
(n
2
), then n
1
+
5
=
n
2
+
5 and so n
1
=
n
2
. Therefore
f
is injective.
To see that
f
is not surjective consider 1 ∈ N (the codomain). If
f
(n)
=
1,
then n
+
5
=
1 and n
=
−4. However, −4
/∈
N (the domain) so
f
is not
surjective.
An example of a surjective but not injective function is g : N → N by g(n)
=
[n/2♩. Indeed for any n ∈ N (the codomain) take 2n ∈ N (the domain) and
g(2n)
=
n. The function g is not injective since g(2)
=
1
=
g(3) but 2 /= 3.
# 2. Let A, B, and C be nonempty sets. Let
f
: A → B and g : B → C be
functions. Show that if g
◦
f
is one-to-one, then
f
must be one-to-one. Is it
true that g must also be one-to-one?
Let us use proof by contrapositive. Assume that
f
is not one-to-one. Then there
exists a
1
, a
2
∈ A with a
1
/= a
2
such that
f
(a
1
)
= f
(a
2
). This means
(g
◦
f
)(a
1
)
=
g(f (a
1
))
=
g(f (a
2
))
=
(g
◦
f
)(a
2
)
and g
◦
f
is not one-to-one.
To see an example where g
◦
f
is one-to-one, but that g is not one-to-one tkae
A = {0}, B =
Z, and
C = {0}
with
f
(0)
=
0 and g(n)
=
0 for n ∈ Z. Then g is
not one-to-one since g(0)
=
0
=
g(1), but g
◦
f
:
{0}
→
{0}
is one-to-one.
#
3. Solve the recurrence relation given by a
1
=
2 and a
n
=
2na
n−1
for n
>
1.
We claim that a
n
=
2
n
n!. Indeed a
1
=
2
=
2
1
1!. Assume that a
n
=
2
n
n! for some
n ≥ 1, then
a
n+1
= 2(n + 1)a
n
= 2n(2
n
n!) = 2
n+1
(n + 1)!.
1
