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ArsDigita University
Month 2:
Discrete Mathematics  Professor Shai Simonson
Problem Set 4 – Induction and Recurrence Equations
1. What’s wrong with the following proofs by induction?
a. Every binary string contains identical symbols.
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
n1
…b
1
b
0
be an arbitrary binary string of length
n+
1.
Let
Y = b
n
b
n1
…b
1
and
Z = b
n1
…b
1
b
0
.
Since both
Y
and
Z
are strings of length less than
n+
1, by induction
each contains identical symbols.
Since the two strings overlap,
X
must also contain
identical symbols.
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.
2. Prove by induction that:
a. The
n
th Fibonacci number equals (1/
√
5)[(1/2 +
√
5/2)
n
– (1/2 –
√
5/2)
n
], where
F
0
=
0 and
F
1
= 1.
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.
c. 21 divides 4
n+
1
+ 5
2n
1
d. The determinant of the
n
by
n
square matrix below is equal to the determinant of B,
where B is a matrix of
m
by
m
(
m<n
), and I is the
m
–
n
identity matrix.
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 Spring '09

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