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.
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
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 they are
identical.
Since the two strings overlap,
X
must also be identical to each of them.
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.
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 Spring '09
 Recursion, Recurrence relation, Fibonacci number, recurrence

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