CSE 20: Discrete Mathematics
Spring 2010
Problem Set 4 Solutions
Instructor: Daniele Micciancio
Due on:
Thu. May 6, 2010
In this problem set all variables range over the set of nonnegative integers. Problems 2, 3 and 4 are on
the well ordering principle which states that “Any non empty set of positive integers has a smallest element”,
and was covered in class on Thursday April 28. Alternatively, you can solve the problems using induction,
which you can ﬁnd in the textbook in unit IS, and will be covered in class on May 4.
Problem 1 (6 points)
Use the deﬁnition “a divides b” (namely,
∃
c.b
=
a
·
c
) to prove each of the following statements:
1. If
x

a
and
x

b
, then
x

(
a
+
b
)
Proof.
x

a
and
x

b
⇒ ∃
c,d.a
=
x
·
c,b
=
x
·
d.
⇒
(
a
+
b
) =
x
·
c
+
x
·
d
=
x
·
(
c
+
d
)
. Therefore
x

(
a
+
b
)
.
2. If
x

a
and
x

(
a
+
b
)
, then
x

b
Proof.
x

a
and
x

(
a
+
b
)
⇒ ∃
c,d.a
=
x
·
c,a
+
b
=
x
·
d
.
⇒
b
= (
a
+
b
)

a
=
x
·
d

x
·
c
=
x
·
(
d

c
)
.
Therefore
x

b
.
Problem 2 (10 points)
Remember the deﬁnition of even and odd:
Even
(
n
)
≡
(
∃
k.n
= 2
k
)
and
Odd
(
n
)
≡
(
∃
k.n
= 2
k
+ 1)
. For each
of the following statements, formulate the statement using predicate logic and the predicates Even and Odd,
and give a proof that the stamement is true:
•
For any even integer
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 Fall '08
 Foster
 smallest element, nonnegative integers

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