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MATH 290 – EXAM 2 – SOLUTIONS
1. (10 pts.) Prove that for any integers
a
,
b
, and
c
, if
a

b
and
a

c
then
a

(
b

c
).
Solution:
Let
a
,
b
, and
c
be integers. Assume that
a

b
and
a

c
. We must show that
a

(
b

c
).
Since
a

b
there is an integer
n
such that
b
=
na
, and since
a

c
there is an integer
m
such that
c
=
ma
. Hence
b

c
=
na

ma
= (
n

m
)
a
. Since
n

m
is an
integer, there is an integer
k
such that
b

c
=
ka
. Therefore
a

(
b

c
).
2. (10 pts.) Prove by contrapositive that for all natural numbers
n
, if
n
2
is odd then
n
is
odd.
Solution:
Let
n
be a natural number. Assume that
n
is not odd. We must show that
n
2
is not odd. Since
n
is not odd,
n
is even. Hence there exists a natural number
k
such that
n
= 2
k
. Hence
n
2
= 4
k
2
= 2(2
k
2
). Since 2
k
2
is an integer,
n
2
is even,
and therefore is not odd.
3. (14 pts.) Prove that for all integers
a
and
b
,
a
+
b
is even if and only if
a
and
b
are even
or
a
and
b
are odd.
Solution:
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This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.
 Fall '08
 Alligood,K
 Advanced Math, Integers

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