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ArsDigita University
Month 2: Discrete Mathematics  Professor Shai Simonson
Problem Set 4 – Induction and Recurrence Equations
Thanks to Jeﬀrey Radcliﬀe and Joe Rizzo for many of the solutions.
Pasted together and modiﬁed by Dimitri Kountourogiannis
1. What’s wrong with the following proofs by induction?
(a) All binary strings are identical. The proof is by induction on the size of the string. For
n=
0 all
binary strings are empty and therefore identical. Let
X = b
n
b
n

1
...b
1
b
0
be an arbitrary binary
string of length
n+
1. Let
Y = b
n
b
n

1
...b
1
and
Z = b
n

1
...b
1
b
0
. Since both
Y
and
Z
are
strings of length less than
n+
1, by induction they are identical. Since the two strings overlap,
X
must also be identical to each of them.
ANSWER
This is not a subtly ﬂawed proof. It is a terribly ﬂawed proof. The problem is with the induction
step. It is never true that the pasting together of two identical strings is identical to the strings.
Furthermore, the
n
= 1 case fails for another reason: A string of length 1 is not the pasting
together of two strings of length 0.
(b) Any amount of change greater than or equal to twenty can be gotten with a combination of ﬁve
cent and seven cent coins. The proof is by induction on the amount of change. For twenty cents
use four ﬁvecent coins. Let
n
>
20 be the amount of change. Assume that
n
= 7
x +
5
y
for some
nonnegative integers
x
and
y.
For any
n
>
20, either
x
>
1, or
y
>
3. If
x
>
1, then since 3(5)
– 2(7) = 1,
n+
1 = 5(
y
+3)+7(
x
–2). If
y
>
3, then since 3(7) – 4(5) = 1,
n+
1 = 7(
x
+3)+5(
y
–4).
In either case, we showed that
n+
1 = 7
u +
5
v
where
u
and
v
are nonnegative integers.
ANSWER
For any
n >
20, either
x >
1 or
y >
3 is false. For example, if
n
= 21, then 21 = 0
·
5 + 3
·
7.
2. Prove by induction that:
(a) The
n
th Fibonacci number,
F
n
equals
G
n
=
1
√
5
"
(
1 +
√
5
2
)
n

(
1 +
√
5
2
)
n
#
where
F
0
= 0 and
F
1
= 1.
ANSWER
Base Case: These hold since
G
0
=
1
√
5
"
(
1 +
√
5
2
)
0

(
1 +
√
5
2
)
0
#
=
1
√
5
(1

1) = 0
and
G
1
=
1
√
5
"
(
1 +
√
5
2
)
1

(
1 +
√
5
2
)
1
#
=
1
√
5
"
1 +
√
5
2

1

√
5
2
#
= 1
Induction Step: Suppose that
G
n
=
F
n
and
G
n

1
=
F
n

1
We would like to show
G
n
+1
=
F
n
+1
Let
a
= (
1
2
+
√
5
2
)
Let
b
= (
1
2

√
5
2
)
We will use the fact that
a
,
b
are roots of
r
2

r

1 = 0 and hence
a
2
=
a
+ 1
b
2
=
b
+ 1
1
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View Full DocumentStarting with
G
n
+1
,
G
n
+1
=
1
√
5
(
a
n
+1

b
n
+1
) =
1
√
5
(
a
2
a
n

1

b
2
b
n

1
)
=
1
√
5
((
a
+ 1)
a
n

1

(
b
+ 1)
b
n

1
)
=
1
√
5
(
a
n

b
n
) +
1
√
5
(
a
n

1

b
n

1
))
=
F
n
+
F
n

1
=
F
n
+1
So the induction step is proven. QED.
(b) The sum of the geometric series 1 +
a
+
a
2
+ ...+
a
n
equals (1–
a
n
+1
)/(1–
a
), where
a
does not
equal one.
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 Spring '09

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