The Strong Principle of Mathematical Induction
Han Duong
April 16, 2013
Han Duong
The Strong Principle of Mathematical Induction
Problem 6.44
Consider the sequence F1 , F2 , F3 , · · · , where
F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5 and F6 = 8.
The terms of this sequence are called Fibonacci numbers.
Han Duong
The Strong Principle of Mathematical Induction
Problem 6.44
Consider the sequence F1 , F2 , F3 , · · · , where
F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5 and F6 = 8.
The terms of this sequence are called Fibonacci numbers.
(a) Define the sequence of Fibonacci numbers by means of a
recurrence relation.
Han Duong
The Strong Principle of Mathematical Induction
Problem 6.44
Consider the sequence F1 , F2 , F3 , · · · , where
F1 = 1, F2 = 1, F3 = 2, F4 = 3, F5 = 5 and F6 = 8.
The terms of this sequence are called Fibonacci numbers.
(a) Define the sequence of Fibonacci numbers by means of a
recurrence relation.
(b) Prove that 2 | Fn if and only if 3 | n.
Han Duong
The Strong Principle of Mathematical Induction
Part (a): Fn as a recurrence relation
We can observe that
F3 = F2 + F1 ,
Han Duong
The Strong Principle of Mathematical Induction
Part (a): Fn as a recurrence relation
We can observe that
F3 = F2 + F1 ,
F4 = F3 + F2 ,
Han Duong
The Strong Principle of Mathematical Induction
Part (a): Fn as a recurrence relation
We can observe that
F3 = F2 + F1 ,
F4 = F3 + F2 ,
F5 = F4 + F3 ,
Han Duong
The Strong Principle of Mathematical Induction
Part (a): Fn as a recurrence relation
We can observe that
F3 = F2 + F1 ,
F4 = F3 + F2 ,
F5 = F4 + F3 ,
and
F6 = F5 + F4 .
Han Duong
The Strong Principle of Mathematical Induction
Part (a): Fn as a recurrence relation
We can observe that
F3 = F2 + F1 ,
F4 = F3 + F2 ,
F5 = F4 + F3 ,
and
F6 = F5 + F4 .
These observations suggest that the recurrence relation is
Fn = Fn−1 + Fn−2
where n ≥ 3 and n ∈ Z.
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Proof.
Suppose 3 | n.
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Proof.
Suppose 3 | n. Then n = 3k for some k ∈ Z.
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Proof.
Suppose 3 | n. Then n = 3k for some k ∈ Z. Note that
n − 3 = 3k − 3 = 3(k − 1).
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Proof.
Suppose 3 | n. Then n = 3k for some k ∈ Z. Note that
n − 3 = 3k − 3 = 3(k − 1).
Since k − 1 ∈ Z, it follows that n − 3 is also divisible by 3.
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Proof.
Suppose 3 | n. Then n = 3k for some k ∈ Z. Note that
n − 3 = 3k − 3 = 3(k − 1).
Since k − 1 ∈ Z, it follows that n − 3 is also divisible by 3.
Now suppose 3 | (n − 3).
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Proof.
Suppose 3 | n. Then n = 3k for some k ∈ Z. Note that
n − 3 = 3k − 3 = 3(k − 1).
Since k − 1 ∈ Z, it follows that n − 3 is also divisible by 3.
Now suppose 3 | (n − 3). Then n − 3 = 3m for some integer m.
Han Duong
The Strong Principle of Mathematical Induction
Lemma: 3 | n ⇔ 3 | n − 3.
Proof.
Suppose 3 | n. Then n = 3k for some k ∈ Z. Note that
n − 3 = 3k − 3 = 3(k − 1).
Since k − 1 ∈ Z, it follows that n − 3 is also divisible by 3.
Now suppose 3 | (n − 3). Then n − 3 = 3m for some integer m.
Since
n = n − 3 + 3 = 3m + 3 = 3(m + 1)
and m + 1 ∈ Z, it follows that 3 | n.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Let P(n) be the open sentence
P(n) :
2 | Fn ⇔ 3 | n.
We induct on n.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Let P(n) be the open sentence
P(n) :
2 | Fn ⇔ 3 | n.
We induct on n.
Base case
For n = 3, we have F3 = 2.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Let P(n) be the open sentence
P(n) :
2 | Fn ⇔ 3 | n.
We induct on n.
Base case
For n = 3, we have F3 = 2. Clearly 2 | F3 and 3 | 3. It follows that
the implication 2 | F3 ⇒ 3 | n is a true statement since both
hypothesis and conclusion are true.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Let P(n) be the open sentence
P(n) :
2 | Fn ⇔ 3 | n.
We induct on n.
Base case
For n = 3, we have F3 = 2. Clearly 2 | F3 and 3 | 3. It follows that
the implication 2 | F3 ⇒ 3 | n is a true statement since both
hypothesis and conclusion are true. Similarly, 3 | 3 ⇒ 2 | F3 is also
true.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Let P(n) be the open sentence
P(n) :
2 | Fn ⇔ 3 | n.
We induct on n.
Base case
For n = 3, we have F3 = 2. Clearly 2 | F3 and 3 | 3. It follows that
the implication 2 | F3 ⇒ 3 | n is a true statement since both
hypothesis and conclusion are true. Similarly, 3 | 3 ⇒ 2 | F3 is also
true. Hence P(3) is true.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Now suppose that P(3), P(4), P(5), · · · , P(k) are true statements,
and consider Fk+1 = Fk + Fk−1 .
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Now suppose that P(3), P(4), P(5), · · · , P(k) are true statements,
and consider Fk+1 = Fk + Fk−1 .
We need to show that 2 | Fk+1 ⇔ 3 | (k + 1).
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Now suppose that P(3), P(4), P(5), · · · , P(k) are true statements,
and consider Fk+1 = Fk + Fk−1 .
We need to show that 2 | Fk+1 ⇔ 3 | (k + 1).
Since Fk = Fk−1 + Fk−2 , we may write
Fk+1 = (Fk−1 + Fk−2 ) + Fk−1 = 2Fk−1 + Fk−2 .
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Now suppose that P(3), P(4), P(5), · · · , P(k) are true statements,
and consider Fk+1 = Fk + Fk−1 .
We need to show that 2 | Fk+1 ⇔ 3 | (k + 1).
Since Fk = Fk−1 + Fk−2 , we may write
Fk+1 = (Fk−1 + Fk−2 ) + Fk−1 = 2Fk−1 + Fk−2 .
Suppose 3 | (k + 1).
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Now suppose that P(3), P(4), P(5), · · · , P(k) are true statements,
and consider Fk+1 = Fk + Fk−1 .
We need to show that 2 | Fk+1 ⇔ 3 | (k + 1).
Since Fk = Fk−1 + Fk−2 , we may write
Fk+1 = (Fk−1 + Fk−2 ) + Fk−1 = 2Fk−1 + Fk−2 .
Suppose 3 | (k + 1). Then by the previous lemma, 3 | (k + 1) − 3.
That is, 3 | (k − 2).
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Now suppose that P(3), P(4), P(5), · · · , P(k) are true statements,
and consider Fk+1 = Fk + Fk−1 .
We need to show that 2 | Fk+1 ⇔ 3 | (k + 1).
Since Fk = Fk−1 + Fk−2 , we may write
Fk+1 = (Fk−1 + Fk−2 ) + Fk−1 = 2Fk−1 + Fk−2 .
Suppose 3 | (k + 1). Then by the previous lemma, 3 | (k + 1) − 3.
That is, 3 | (k − 2). The inductive hypothesis implies
3 | (k − 2) ⇔ 2 | Fk−2 .
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Now suppose 2 | Fk+1 , and rewrite
Fk+1 = 2Fk−1 + Fk−2
as
Fk+1 − 2Fk−1 = Fk−2
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Now suppose 2 | Fk+1 , and rewrite
Fk+1 = 2Fk−1 + Fk−2
as
Fk+1 − 2Fk−1 = Fk−2
Clearly the left hand side is divisible by 2. Therefore Fk−2 is
divisible by 2.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Now suppose 2 | Fk+1 , and rewrite
Fk+1 = 2Fk−1 + Fk−2
as
Fk+1 − 2Fk−1 = Fk−2
Clearly the left hand side is divisible by 2. Therefore Fk−2 is
divisible by 2. However, 2 | Fk−2 ⇔ 3 | (k − 2).
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Now suppose 2 | Fk+1 , and rewrite
Fk+1 = 2Fk−1 + Fk−2
as
Fk+1 − 2Fk−1 = Fk−2
Clearly the left hand side is divisible by 2. Therefore Fk−2 is
divisible by 2. However, 2 | Fk−2 ⇔ 3 | (k − 2). Therefore
3 | (k − 2).
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Now suppose 2 | Fk+1 , and rewrite
Fk+1 = 2Fk−1 + Fk−2
as
Fk+1 − 2Fk−1 = Fk−2
Clearly the left hand side is divisible by 2. Therefore Fk−2 is
divisible by 2. However, 2 | Fk−2 ⇔ 3 | (k − 2). Therefore
3 | (k − 2). And by the previous lemma, 3 | (k + 1).
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Now suppose 2 | Fk+1 , and rewrite
Fk+1 = 2Fk−1 + Fk−2
as
Fk+1 − 2Fk−1 = Fk−2
Clearly the left hand side is divisible by 2. Therefore Fk−2 is
divisible by 2. However, 2 | Fk−2 ⇔ 3 | (k − 2). Therefore
3 | (k − 2). And by the previous lemma, 3 | (k + 1). Hence
P(k + 1) is also true.
Han Duong
The Strong Principle of Mathematical Induction
Part (b): Proving 2 | Fn ⇔ 3 | n
Showing P(3), P(4), P(5), · · · , P(k) imply P(k + 1)
Since 2 | Fk−2 and 2 | (2Fk−1 ), it follows that 2 | Fk+1 .
Now suppose 2 | Fk+1 , and rewrite
Fk+1 = 2Fk−1 + Fk−2
as
Fk+1 − 2Fk−1 = Fk−2
Clearly the left hand side is divisible by 2. Therefore Fk−2 is
divisible by 2. However, 2 | Fk−2 ⇔ 3 | (k − 2). Therefore
3 | (k − 2). And by the previous lemma, 3 | (k + 1). Hence
P(k + 1) is also true. So by the strong principle of mathematical
induction, P(n) is true for all n ≥ 3.
Han Duong
The Strong Principle of Mathematical Induction
Final comments
Han Duong
The Strong Principle of Mathematical Induction
Final comments
The proof, as is, is an example of the general strong principle
of mathematical induction since we only proved P(n) to be
true for n ≥ 3.
Han Duong
The Strong Principle of Mathematical Induction
Final comments
The proof, as is, is an example of the general strong principle
of mathematical induction since we only proved P(n) to be
true for n ≥ 3.
Since our proof relied on the recurrence relation, we could
only start as low as n = 3.
Han Duong
The Strong Principle of Mathematical Induction
Final comments
The proof, as is, is an example of the general strong principle
of mathematical induction since we only proved P(n) to be
true for n ≥ 3.
Since our proof relied on the recurrence relation, we could
only start as low as n = 3.
We can prove P(1) and P(2) directly, without reliance on the
recursive formula. The result would be an example of the
strong principle of mathematical induction (i.e. no “general”
qualifier).
Han Duong
The Strong Principle of Mathematical Induction
Final comments
The proof, as is, is an example of the general strong principle
of mathematical induction since we only proved P(n) to be
true for n ≥ 3.
Since our proof relied on the recurrence relation, we could
only start as low as n = 3.
We can prove P(1) and P(2) directly, without reliance on the
recursive formula. The result would be an example of the
strong principle of mathematical induction (i.e. no “general”
qualifier).
There is one minor detail about the validity of this proof. The
entire proof relies on a recurrence relation—which we merely
observed, and have not proved. That is, the entire proof is
only valid provided Fn = Fn−1 + Fn−2 is indeed the correct
recursion. (It is, but how would we prove this?)
Han Duong
The Strong Principle of Mathematical Induction