MATH 114
Final Exam  Solutions
1. Is the function
f
(
x
) =
x
2
from
N
to
N
(a) onetoone?
Yes. Suppose
f
(
x
1
) =
f
(
x
2
)
. For real numbers we have
x
2
1
=
x
2
2
→
x
1
=
±
x
2
. Since
x
1
and
x
2
are natural numbers,
x
1
=
x
2
.
(b) onto?
No. For example, 2 is not in the image because there is no
x
∈
N
such that
x
2
= 2
.
2. Use Mathematical Induction to prove that 2
n
< n
! for every positive integer
n
with
n
≥
4.
Basis step.
If
n
= 4
,
2
4
= 16
<
24 = 4!
.
Inductive step.
Assume
2
k
< k
!
. Then
2
k
+1
= 2
k
·
2
< k
!
·
2
< k
!(
k
+ 1) = (
k
+ 1)!
.
3. Let
P
(
x, y
) denote the proposition
y
=
x
+ 5 where
x
and
y
are positive integers. Determine
the truth value of the following propositions.
(a)
∀
x
∃
yP
(
x, y
)
True. For any positive integer
x
,
x
+5
is a positive integer, so we can choose
y
=
x
+5
.
(b)
∀
y
∃
xP
(
x, y
)
False. Counterexample: if
y
= 1
, there is no positive integer
x
that satisfies
1 =
x
+ 5
.
(c)
∃
y
∀
xP
(
x, y
)
False. We will show that
¬∃
y
∀
xP
(
x, y
)
is true.
¬∃
y
∀
xP
(
x, y
)
≡ ∀
y
¬∀
xP
(
x, y
)
≡ ∀
y
∃
x
¬
P
(
x, y
)
≡ ∀
y
∃
x y
6
=
x
+ 5
.
For any positive integer
y
we can choose
x
=
y
+ 4
. Then
y
=
x

4
, so
y
6
=
x
+ 5
.
4. Consider the following graph.
a
b
c
d
e
f
(a) How many vertices does this graph have?
6
(b) How many edges does this graph have?
8
(c) Is this graph bipartite?
Yes. Label the vertices as shown above. The set of vertices can be partitioned into the
following 2 sets:
V
1
=
{
a, b, e, f
}
and
V
2
=
{
c, d
}
. Then every edge connects a vertex in
V
1
and a vertex in
V
2
.
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 Spring '11
 rsharma
 2k, positive integer, two digits

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