Faculty of Mathematics
University of Waterloo
MATH 135
MIDTERM EXAM #1
Fall 2007
Monday 15 October 2007
19:00 – 20:15
Solutions
1. Let
a, b, c
∈
Z
. Suppose that
P
is the statement: “If
a

b
or
a

c
, then
a

bc
”.
(a) Write down the converse of
P
.
[2]
Solution
:
The converse is “If
a

bc
, then
a

b
or
a

c
.”
(b) Disprove the converse of
P
.
[2]
Solution
:
Suppose
a
= 4,
b
= 2 and
c
= 2.
Then
a

bc
, since 4

4, but
a
6 
b
and
a
6 
c
since 4
6 
2.
Thus,
a
= 4,
b
= 2 and
c
= 2 is a counterexample that disproves the converse of
P
.
(c) Write down the contrapositive of
P
.
[2]
Solution
:
The contrapositive is “If
a
6 
bc
, then
a
6 
b
and
a
6 
c
.”
2. Suppose that
x
1
= 10,
x
2
= 16 and
x
n
= 4
x
n

1

4
x
n

2
for
n
≥
3.
[8]
Prove that
x
n
= 3(2
n
+1
)

n
2
n
for all
n
∈
P
.
Solution
:
We prove the result by strong induction on
n
.
Base Cases
n
= 1:
x
1
= 10 and 3(2
2
)

1
·
2
1
= 3(4)

2 = 10
n
= 2:
x
2
= 16 and 3(2
3
)

2
·
2
2
= 3(8)

2(4) = 16
Therefore, the formula holds for
n
= 1 and
n
= 2.
Induction Hypothesis
Assume that the result is true for
n
= 1
,
2
, . . . , k
for some
k
∈
P
,
k
≥
2.
That is assume, that
x
n
= 3(2
n
+1
)

n
2
n
for
n
= 1
,
2
, . . . , k
.
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Induction Conclusion
Consider
n
=
k
+ 1. Then
x
k
+1
= 4
x
k

4
x
k

1
= 4(3(2
k
+1
)

k
2
k
)

4(3(2
k
)

(
k

1)2
k

1
)
(by Induction Hypothesis)
= 12(2
k
+1
)

12(2
k
)

4
k
2
k
+ 4(
k

1)2
k

1
= 2
k
(24

12)

2
k

1
(4
k
(2)

4(
k

1))
= 2
k
(12)

2
k

1
(4
k
+ 4)
= 2
k
(2
2
)(3)

2
k

1
(2
2
)(
k
+ 1)
= 3(2
k
+2
)

(
k
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 Spring '09
 MARSHMAN
 Math, Logic, Algebra, Mathematical Induction, Greatest common divisor, Diophantine equation, Structural induction, Linear Diophantine equation

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