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Homework 3, Math 245, Fall 2010
Due Friday, October 1 in class.
The solution of each exercise should be at most one page long. If you can, try to write
your solutions in LaTex. Each question is worth 2 points.
1. The Fibonacci numbers (
F
n
)
n
≥
1
are deﬁned as follows:
F
1
=
F
2
= 1 and
F
n
=
F
n

1
+
F
n

2
for every
n
≥
2.
Prove that
F
n
=
±
1+
√
5
2
²
n

±
1

√
5
2
²
n
√
5
for every
n
≥
1.
Proof.
We will use strong induction on
n
to prove that
F
n
=
±
1+
√
5
2
²
n

±
1

√
5
2
²
n
√
5
.
Base Case:
n
= 1. In this case,
F
1
= 1 and
±
1+
√
5
2
²
1

±
1

√
5
2
²
1
√
5
=
2
√
5
2
√
5
= 1.
n
= 2. In this case,
F
2
= 1 and
F
2
=
±
1+
√
5
2
²
2

±
1

√
5
2
²
2
√
5
=
6+2
√
5
4

6

2
√
5
4
√
5
=
4
√
5
4
√
5
= 1
.
Thus, the base case is true.
Induction Step: Assume that
F
i
=
±
1+
√
5
2
²
i

±
1

√
5
2
²
i
√
5
for every
i
, 1
≤
i
≤
k
. We will prove
that
F
k
+1
=
±
1+
√
5
2
²
k
+1

±
1

√
5
2
²
k
+1
√
5
(1)
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View Full Document Using the recurrence relation of the Fibonacci numbers and the induction hypothesis,
we get
F
k
+1
=
F
k
+
F
k

1
=
±
1+
√
5
2
²
k

±
1

√
5
2
²
k
√
5
+
±
1+
√
5
2
²
k

1

±
1

√
5
2
²
k

1
√
5
=
±
1+
√
5
2
²
k
+
±
1+
√
5
2
²
k

1
√
5

±
1

√
5
2
²
k
+
±
1

√
5
2
²
k

1
√
5
=
±
1+
√
5
2
²
k

1
±
1+
√
5
2
+ 1
²
√
5

±
1

√
5
2
²
k

1
±
1

√
5
2
+ 1
²
√
5
=
±
1+
√
5
2
²
k

1
·
3+
√
5
2
√
5

±
1

√
5
2
²
k

1
·
3

√
5
2
√
5
=
±
1+
√
5
2
²
k

1
·
6+2
√
5
4
√
5

±
1

√
5
2
²
k

1
·
6

2
√
5
4
√
5
=
±
1+
√
5
2
²
k

1
·
±
1+
√
5
2
²
2
√
5

±
1

√
5
2
²
k

1
·
±
1

√
5
2
²
2
√
5
=
±
1+
√
5
2
²
k
+1
√
5

±
1

√
5
2
²
k
+1
√
5
.
This ﬁnishes our proof.
2. Let
f
:
A
→
B
and
g
:
B
→
C
be two functions. Show that if
g
◦
f
is injective, then
f
is injective. Show that if
g
◦
f
is surjective, then
g
is surjective.
Proof.
First, we show that if
g
◦
f
is injective, then
f
injective.
Let
a,b
∈
A
such that
f
(
a
) =
f
(
b
). Applying
g
to both sides we obtain
g
(
f
(
a
)) =
g
(
f
(
b
))
which means (
g
◦
f
)(
a
) = (
g
◦
f
)(
b
). As
g
◦
f
is injective, this implies
a
=
b
and shows
f
is injective.
Secondly, we show that if
g
◦
f
is surjective, then
g
is surjective.
Let
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This note was uploaded on 12/07/2011 for the course MATH 245 taught by Professor Cioaba during the Fall '10 term at University of Delaware.
 Fall '10
 CIOABA
 Math

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