MATH 135
Fall 2007
Assignment #2
Due: Wednesday 26 September 2007, 8:20 a.m.
N.B. Assignments 3 to 9 will not be distributed in class. You must download them from the course
Web site.
HandIn Problems
1. A sequence of integers is deﬁned by
x
1
= 2,
x
2
= 86,
x
m
+2
=

2
x
m
+1
+ 15
x
m
for
m
≥
1. Prove
that
x
n
= 4
·
3
n
+ 2(

5)
n
for all
n
∈
P
.
2. A sequence of integers is deﬁned by
y
1
= 2,
y
2
= 11,
y
3
= 2,
y
m
+3
= 3
y
m
+1

2
y
m
for
m
≥
1.
Prove that
y
n
= (

2)
n
+ 3
n
+ 1 for all
n
∈
P
.
3. Suppose that
x, y
∈
R
. Prove that
x
2

4
xy
+ 5
y
2

4
y
+ 4 = 0 if and only if (
x, y
) = (4
,
2).
4. Prove by contradiction that log
10
3 cannot be written in the form
m
n
where
m
and
n
are positive
integers.
5. Prove that if
x
6
=
y
and
a
6
= 0, then
x
y
6
=
x
+
a
y
+
a
.
6. To prove a statement of the form “If
H
then
C
1
or
C
2
” (where
H, C
1
, C
2
are mathematical
statements), we can assume that
H
is TRUE and that
C
1
is FALSE, and prove that
C
2
is
TRUE. (If
C
1
happened to be TRUE, we would be done, so we assume that
C
1
is FALSE and
prove that
C
2
must be TRUE.)
Suppose that
m
and
n
are integers.
(a) Prove that if 3
m
2
+ 6
m
+ 5
n
2
+ 3 is odd, then
m
is even or
n
is odd.
(b) Is the converse TRUE or FALSE? Either prove or give a counterexample.
7. An expression of the form
a
n
x
n
+
a
n

1
x
n

1
+
···
+
a
1
x
+
a
0
with
n
≥
0 and
a
n
, a
n

1
, . . . , a
1
, a
0
∈
Q
(or
R
) is called a polynomial in
x
with coeﬃcients from
Q
(or
R
).
If
a
n
6
= 0, the polynomial is said to have degree
n
. (The degree of a polynomial is the highest
power of
x
that has a nonzero coeﬃcient.) If all of the coeﬃcients are 0, the polynomial is
called the zero polynomial and its degree is not deﬁned.
(a) In each part, state whether the expression is a polynomial. If it is a polynomial, state its
degree.
i.
√
2
x
3

πx
+ 3
ii. 2(
√
x
)
3

πx
+ 3
iii. 4
(b) In each part, perform the calculation, simplifying as much as possible.
i. (
x
2

√
2
x
+ 1)(
x
2
+
√
2
x
+ 1)
ii. (
x
2
+ 3
x
+ 1)(
x
3
+
x
2

2) + 3
x
4
+ 4
x
2
+ 5
...continued
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View Full Document8. A pile starts with
n
sticks, with
n
≥
2. The pile is split into two piles, and the product of
the sizes of the piles is calculated. One of these new piles is then split into two piles, with the
product of the sizes of these two new piles calculated and added to (the sum of) the previous
product(s). This process continues until we have
n
piles with 1 stick each.
For example, starting with 5 sticks, we could split 2, 3 (giving 6), then 2, 2, 1 (giving 6 + 2),
then 1, 1, 2, 1 (giving 6 + 2 + 1), then 1, 1, 1, 1, 1 (giving 6 + 2 + 1 + 1 = 10).
Prove using strong induction that, no matter how the pile is split, the total at the end is always
1
2
(
n
2

n
).
9. In the country of Alwynnia, there are 10 cent, 20 cent and 50 cent coins. In her purse, Maddie
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 Fall '08
 ANDREWCHILDS
 Math, Logic, Integers, Mathematical Induction, Natural number, Maddie

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