Math 326  Exam 2 Solutions  12 Oct 2010
1. (a) (2 points) State precisely but concisely what it means when
a

b
.
(
∃
k
∈
Z
)(
b
=
ka
)
(b) (3 points) Does 12

0? why or why not?
Yes, because 0 = 12
·
0
(c) (2 points) State precisely but concisely what it means for a number to be rational.
(
x
∈
Q
)
⇐⇒
(
∃
a,b
∈
Z
)(
b
6
= 0)(
x
=
b/a
)
(d) (3 points) Is 4.613 rational? why or why not?
Yes because 4
.
613 =
4613
1000
2. Prove by contradiction that for any integer
n
, it is impossible for
n
to equal both
3
q
1
+
r
1
and 3
q
2
+
r
2
, where
q
1
,q
2
,r
1
,r
2
∈
Z
and 0
≤
r
1
,r
2
<
3, and
r
1
6
=
r
2
.
Proof 1.
Let
q
1
,q
2
∈
Z
and suppose that for some
r
1
and
r
2
,
3
q
1
+
r
1
= 3
q
2
+
r
2
3(
q
1

q
2
) =
r
2

r
1
If
r
1
= 0
then
r
2
= 1 or
r
2
= 2.
r
2
= 1 =
⇒
r
2

r
1
= 1 =
⇒
q
1

q
2
= 1
/
3
6∈
Z
r
2
= 2 =
⇒
r
2

r
1
= 2 =
⇒
q
1

q
2
= 2
/
3
6∈
Z
If
r
1
= 1
then
r
2
= 0 or
r
2
= 2.
r
2
= 0 =
⇒
r
2

r
1
= 0

1 =

1 =
⇒
q
1

q
2
=

1
/
3
6∈
Z
r
2
= 2 =
⇒
r
2

r
1
= 2

1 = 1 =
⇒
q
1

q
2
= 1
/
3
6∈
Z
If
r
1
= 2
then
r
2
= 0 or
r
2
= 1.
r
2
= 0 =
⇒
r
2

r
1
= 0

2 =

2 =
⇒
q
1

q
2
=

2
/
3
6∈
Z
r
2
= 1 =
⇒
r
2

r
1
= 1

2 =

1 =
⇒
q
1

q
2
=

1
/
3
6∈
Z
Hence all possible cases lead to contradiction so the conjecture is proven.
Proof 2.
Let
q
1
,q
2
∈
Z
and suppose that for some
r
1
and
r
2
,
3
q
1
+
r
1
= 3
q
2
+
r
2
3(
q
1

q
2
) =
r
2

r
1
3 =
r
2

r
1
q
1

q
2
1
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View Full DocumentBut 0
≤
r
1
,r
2
<
3, i.e.,
r
1
= 0
,
1
,
or 2 and
r
2
= 0
,
1
,
or 2, where
r
1
6
=
r
2
; hence

r
1

r
2
 ≤
2 and therefore
3 =
r
2

r
1
q
1

q
2
≤
2
q
1

q
2
Since
r
1
6
=
r
2
, from the equation 3(
q
1

q
2
) =
r
2

r
1
we conclude
q
1
6
=
q
2
. Hence

q
1

q
2
 ≥
1 =
⇒
1

q
1

q
2

≤
1
Hence by substitution 3
≤
2 which is a contradiction.
.
Proof 3.
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 Spring '08
 WOUTERS
 Math, Algebra, Prime number, Integer factorization, Fundamental theorem of arithmetic, contraposition

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