CSE 20: Discrete Mathematics
Fall 2010
Problem Set 3
Instructor: Daniele Micciancio
Due on:
Wed. Oct. 20, 2010
Problem 1 (10 points)
(a)
Use the wellordering principle to prove that every positive integer number is either even or odd
,
where
“even(n)=
∃
x
∈
Z
.n
= 2
x
” and “odd(n)=
∃
x
∈
Z
.n
= 2
x
+ 1
”.
Proof:
Let
S
=
{
n
∈
Z
+
 ¬
even
(
n
)
∧
¬
odd
(
n
)
}
be the set of all positive integers that are neither even nor odd. We want to prove that
S
is
empty. Assume for contradiction that
S
is not empty. Since
S
is a nonempty set of positive integers,
by the wellordering principle it contains a minimal element
m
∈
S
. Since
m
is the smallest element
of
S
and
m

1
is smaller than
m
,
m

1
cannot be in
S
.
Notice that 1 is an odd integer because
1 = 2
·
0 + 1
. Therefore
1
/
∈
S
. In particular,
m
6
= 1
and
m

1
>
0
is also a positive integer. But
(by definition of
S
) a positive integer that is not in
S
is either even or odd. We consider the two cases
separately. If
m

1
is even, then
m

1 = 2
x
for some
x
∈
Z
, and
m
= 2
x
+ 1
is odd. If
m

1
is
odd, then
m

1 = 2
x
+ 1
for some
x
∈
Z
and
m
= 2
x
+ 2 = 2(
x
+ 1)
is even. In both cases, we got a
contradiction because
m
is neither even nor odd.
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 Fall '10
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