MI 4
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Mathematical Induction
Let’s investigate perhaps the world’s most famous sequence:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...
the Fibonacci sequence. Recall that a sequence is a function whose domain is the natural
numbers, and that some sequences are more naturally defined recursively. In the case of the
Fibonacci sequence, we have the definition
F (1) = 1,
F (2) = 1,
F (n) = F (n − 1) + F (n − 2),
n ≥ 2.
Also recall that we often use a subscript notation for sequences, so that the above definition
is more commonly written
F1 = 1,
F2 = 1,
Fn = Fn−1 + Fn−2 ,
n ≥ 2.
As an example, using n = 3 in the previous definition, we get
F3 = F2 + F1 ,
and then we can use the definitions of F2 and F1 to get F3 = 2. That this function is welldefined is closely related to a proof technique called mathematical induction. In other
words, although we do not give an explicit formula, we are pretty sure that F1000 can readily
be calculated using the recursive definition. Interestingly, this definition may be used to
derive the explicit formula
√ !n
√ !n !
1
1+ 5
1− 5
Fn = √
−
.
2
2
5
Mathematicians have studied the Fibonacci sequence and related sequences for centuries.
One well-known companion is the Lucas sequence, defined by
L1 = 1,
L2 = 3,
Ln = Ln−1 + Ln−2 ,
n ≥ 2.
Write the first ten terms in the Lucas sequence below:
IMSA
Induction 1.1
Spring 2012
MI 4
Name:
Note that the Lucas sequence has the same recursive definition as the Fibonacci sequence,
although different starting values. This, as we will see, means that the sequences are closely
related. As an example, use a spreadsheet to produce the first 100 values of each sequence.
Then, use this spreadsheet to find the ratios Fn+1 /Fn and Ln+1 /Ln . What do you observe?
Write your conclusion below:
Another pattern may be found in exploring Fibonacci series. Recall that a series is a sum
of terms in a sequence. Thus, we may define the series
Sn =
n
X
Fk .
k=1
Write down the first ten terms of the Fibonacci series by filling in the chart below (using
your spreadsheet!):
n
1
2
3
4
5
6
7
8
9
10
Fn
Sn
In calculating the terms of the series, a pattern should be evident. Write your observation
here, using the subscript notation:
Now the big question: how do you know that the pattern you observed continues? For
example, would the pattern hold even if you wrote 1000 terms of the series? If so, how
IMSA
Induction 1.2
Spring 2012
MI 4
Name:
would you justify it? Write a few sentences below giving a sound reason for the continuity
of the pattern. And don’t say, “It’s obvious.”
The technique of mathematical induction is the means whereby such “obvious” statements are proved. This technique has been foreshadowed with the Tower of Hanoi puzzle.
How did you solve it? Didn’t making progress depend on first solving a smaller, but related
problem? And once you solved this, you moved the next disk, and went on? A recursive
procedure was the most natural way to solve this puzzle, and mathematical induction may
be thought of as a recursive style of proof.
So let’s dive in! Let’s see an inductive proof of the conjecture you made about the Fibonacci
series.
First, we need to make sure the conjecture is formalized correctly. We hope to prove that
Sn = Fn+2 − 1
for all natural numbers n. Here we go!
First, is the statement true for n = 1? We found that S1 = 1. But
F1+2 − 1 = F3 − 1 = 2 − 1 = 1
as well, so our conjecture is valid for n = 1.
Now here is the inductive part: let’s assume that the statement is true for some value p.
Now let’s prove that the statement is true for the next value, p + 1. Can you see why this
would work? We know the statement is true for n = 1. But if it’s true for 1, then it’s true for
2. And if it’s true for 2, then it’s true for 3. And so on.... We conclude that the statement
is true for all natural numbers. This is the technique known as “mathematical induction.”
IMSA
Induction 1.3
Spring 2012
MI 4
Name:
So assume that
Sp = Fp+2 − 1.
Let’s look at Sp+1 . Now
Sp+1 =
p+1
X
Fk .
k=1
But recall the meaning of the summation notation; this allows us to say that
Sp+1 =
p
X
Fk + Fp+1 = Sp + Fp+1 .
k=1
But we assumed that Sp = Fp+2 − 1. So we can substitute this in, giving
Sp+1 = Fp+2 − 1 + Fp+1 = Fp+2 + Fp+1 − 1.
And now the important observation! We know that adding two consecutive Fibonacci numbers yields the next, so that
Fp+2 + Fp+1 = Fp+3 .
Thus,
Sp+1 = Fp+3 − 1.
But this is exactly what we want, since this may be rewritten
Sp+1 = F(p+1)+2 − 1.
In other words, the statement we wish to prove is the same statement as our assumption
Sp = Fp+2 − 1, except that “p” is replaced with “p + 1.” In other words, if our conjecture is
true for some value of p, it’s also true for the next.
Now this is our first induction proof. Typically, after the proof is begun, the detailed
argument above is replaced by a relatively shorter version, shown below. As a summary,
induction proofs follow a few important steps:
Basic Steps for an Induction Proof
1. Write down what you want to prove.
2. Make sure the base case is valid.
3. Write down what you need to prove for the induction step.
4. Then prove the induction step.
IMSA
Induction 1.4
Spring 2012
MI 4
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Proof:
(1) We want to prove that Sn = Fn+2 − 1 for all n ≥ 1.
(2) First, we consider the case n = 1. Then
S1 =
1
X
Fk = F1 = F3 − 1.
k=1
(3) For the induction step, we assume that Sp = Fp+2 − 1 for some p. Now we need to prove
this statement with “p” replaced by “p + 1”:
Sp+1 = Fp+3 − 1.
(4) Now we calculate Sp+1 .
Sp+1 =
p+1
X
Fk
k=1
=
p
X
Fk + Fp+1
by definition
k=1
= Fp+2 − 1 + Fp+1
= Fp+1 + Fp+2 − 1
= Fp+3 − 1
by assumption
by the definition of Fn .
This shorter version is often used because once you become familiar with proofs by induction,
you really don’t need all the details, just the essentials. So to keep it brief, short explanations
are often written to the right of important steps. Of course there are induction proofs which
do require much more explanation, but we won’t be concerned with those right now.
Let’s try another one before you get a chance! This time, Sn will be the series of the first n
Fibonacci numbers with odd subscripts. So for example,
S1
S2
S3
S4
IMSA
= F1 ,
= F1 + F3 ,
= F1 + F3 + F5 ,
= F1 + F3 + F5 + F7 .
Induction 1.5
Spring 2012
MI 4
Name:
Take a few minutes now to investigate the series Sn , and formulate a conjecture about it.
Your conjecture:
Sn =
Now let’s look at a proof of your conjecture. The first obstacle is how to write a conjecture
using summation notation. We have to remember that 2k − 1 is an explicit formula for the
sequence of odd numbers 1, 3, 5, 7, .... Thus, the sum of the first n Fibonacci numbers with
odd subscripts is
n
X
F2k−1 .
k=1
You should have noticed that this sum is just the next Fibonacci number after the last
number you added. Thus, our conjecture becomes
n
X
Sn =
F2k−1 = F2n .
k=1
Now for the proof!
Proof:
(1) We want to prove that Sn = F2n for all n ≥ 1.
(2) First, consider the case n = 1. Then
S1 =
1
X
F2k−1 = F1 = F2 .
k=1
(3) For the induction step, let’s assume that Sp = F2p for some value of p. Then we’ll show
that this statement is valid with “p” replaced with “p + 1”:
Sp+1 = F2p+2 .
(4) Now we explicitly calculate Sp+1 .
Sp+1 =
p+1
X
F2k−1
k=1
=
p
X
F2k−1 + F2p+1
by definition
k=1
= F2p + F2p+1
= F2p+2
by assumption
by the definition of Fn .
Make sure you understand every step here, as they are all important!
IMSA
Induction 1.6
Spring 2012
MI 4
Name:
Now, it’s your turn! Below you will be given some sums to investigate. Formulate your
conjectures – writing them in summation notation – and write induction proofs proving your
conjectures. You may use the briefer version of the proof, but be sure to write justifications
to the right as shown above.
1. What is the sum of the first n Lucas numbers?
2. What is the sum of the first n Lucas numbers with odd subscripts?
3. What is the sum of the first n Fibonacci numbers with even subscripts?
4. What is the sum of the first n squares of Fibonacci numbers?
5. Discover a different property of Fibonacci or Lucas numbers, and prove it!
IMSA
Induction 1.7
Spring 2012