Download Some notes for Week #2

These are supplementary notes for Week 2’s lectures. It does not cover all of the material discussed
in lectures, but rather is intended to provided extra explanations of material related to the lectures.
The required text is Peter J. Eccles’ book An Introduction to Mathematical Reasoning. The section
and page numbers refer to his book.
Sigma notation (also called summation notion) is for example
n
X
ak = a1 + a2 + · · · + an .
k=1
I.5. The induction principle
5.1. Proof by induction (p. 39)
Formulas which hold for all positive integers are good candidates for proof by the method of induction.
We illustrate this method by an example.
Theorem 1 For any positive integer n,
1 + 3 + 5 + · · · + (2n − 1) =
n
X
(2k − 1) = n2 .
k=1
In other words, the sum of the first (i.e., the smallest) n odd positive integers equals n2 .
Proof.
1. Base case: We verify the formula for n = 1. In this case the formula is true since when n = 1,
1 + 3 + 5 + · · · + (2n − 1) = 1 = n2 .
(1)
2. Inductive step: Suppose that the formula is true for some integer n = k. We shall prove that
then the formula is true for the next integer n = k + 1. In other words, if some case is true,
then the next case is true. Since the formula is assumed true for n = k, we have
1 + 3 + 5 + · · · + (2k − 1) = k 2 .
(2)
We then compute (to verify the n = k + 1 case) that1
1 + 3 + 5 + · · · + (2 (k + 1) − 1) = (1 + 3 + 5 + · · · + (2k − 1)) + (2k + 1)
= k 2 + (2k + 1)
= (k + 1)2 ,
where we used (2) to obtain the second equality (in essence, the key step). Hence formula (1) is
true for n = k + 1.
3. By induction we are done: That is, mathematical induction says that we have proved (1) for
all positive integers n. Commentary: Why does induction work? We may think of it in various, essentially equivalent,
ways:
1
In the first line we used that 2 (k + 1) − 1 = 2k + 1. Also, 2k − 1 is the odd integer immediately preceding 2k + 1.
1
1. General method: Let P (n) denote the nth formula you wish to proof. The proposition is: P (n)
is true for all positive integers n. Mathematical induction says that this proposition follows if
one can:
(a) (Base case) Prove that P (1) is true.
(b) (Inductive step) Prove that if P (k) is true for some positive integer k, then P (k + 1) is
true. I.e., prove P (k) ⇒ P (k + 1).
2. Domino effect: Think of P (n) as being the nth domino and imagine it topples if and only if it
is true.
The proposition says all dominos topple. Mathematical induction says to prove this, we just
need to prove:
(a) P (1) being true, which says the 1st domino topples.
(b) P (k) ⇒ P (k + 1), which says that if the k th domino topples, then the (k + 1)th domino
topples.
The idea is: (a) says P (1) is true. P (1) is true and (b) imply that P (2) is true. P (2) is true and
(b) imply that P (3) is true. P (3) is true and (b) imply that P (4) is true. And so on.
In other words, the 1st domino topples implies the 2nd domino topples, which implies the 3rd
domino topples, which implies the 4th domino topples, etc.
Proposition 2 We have Bernoulli’s inequality
(1 + x)n ≥ 1 + nx
for all non-negative integers n and real numbers x > −1.
Proof. Homework problem.
Proposition 3 We have
n
X
1
1
1
1
n
+
+ ··· +
=
=
1·2 2·3
n (n + 1)
i (i + 1)
n+1
i=1
for all positive integers n.
Proof. Base case: n = 1. This is true because
1
X
i=1
Inductive step: Suppose
k+1
X
i=1
Pk
1
i=1 i(i+1)
1
1
1
1
=
= =
.
i (i + 1)
1·2
2
1+1
=
k
k+1
for some positive integer k. Then
k
X
1
1
1
=
+
i (i + 1)
i (i + 1) (k + 1) (k + 2)
i=1
P
k
1
1
+
(using ki=1 i(i+1)
=
k + 1 (k + 1) (k + 2)
k (k + 2) + 1
k 2 + 2k + 1
=
=
(k + 1) (k + 2)
(k + 1) (k + 2)
=
(k + 1)2
(k + 1) (k + 2)
k+1
=
.
k+2
=
2
k
k+1 )
This is the formula for n = k + 1, so we are done by induction. Proposition 4 For all positive integers n,
12 + 22 + · · · + n2 =
Proof. Base case: Exercise.
Inductive step: Suppose 12 + 22 + · · · + k 2 =
n (n + 1) (2n + 1)
.
6
k(k+1)(2k+1)
6
for some positive integer k. Then
12 + 22 + · · · + (k + 1)2 = (12 + 22 + · · · + k 2 ) + (k + 1)2
k (k + 1) (2k + 1)
=
+ (k + 1)2
6
k (k + 1) (2k + 1) + 6 (k + 1)2
=
6
k (2k + 1) + 6 (k + 1)
= (k + 1)
6
2
2k + 7k + 6
= (k + 1)
6
(k + 2) (2k + 3)
= (k + 1)
6
(k + 1) ((k + 1) + 1) (2 (k + 1) + 1)
=
.
6
This is the formula for n = k + 1. So we are done by induction.
5.2. Changing the base case (p. 44)
Theorem. For any integer n ≥ 3 we have
3n ≥ n3 .
Proof. (1) Base case. We have 33 ≥ 33 .
(2) Inductive step. Suppose k ≥ 2 is an integer such that 3k ≥ k 3 . Then
3k+1 = 3 · 3k ≥ 3k 3 .
Assuming the Claim that 3k 3 ≥ (k + 1)3 , we obtain 3k+1 ≥ 3k 3 ≥ (k + 1)3 and we are done by
induction.
We prove the claim backwards:
3
3k ≥ (k + 1)
since
1
31/3 −1
3
(k + 1)3
3≥
=
k3
⇐=
⇐=
31/3 ≥ 1 +
⇐=
3≤k
1
k
⇐=
k+1
k
1
31/3 − 1 ≥
k
≤ 2.262 ≤ 3.
3
3
1 3
= 1+
k
1
⇐=
≤k
1/3
3 −1
5.3. Definition by induction (p. 45)
5.4. The strong induction principle (p. 48)
A fundamental sequence in nature, computer science, mathematics, etc. are the Fibonacci numbers.
The Fibonacci numbers are
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, . . .
You may have noticed the pattern: 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, etc. I.e., the
sum of two successive Fibonacci numbers is equal to the next one.
Denote by un the nth Fibonacci number. Then un is defined inductively by
u1 = 1,
u2 = 1,
uk+1 = uk−1 + uk
for k ≥ 2.
So u3 = u1 + u2 (k = 2), u4 = u2 + u3 (k = 3), etc.
Proposition 5 For any n ∈ Z+ , the 4nth Fibonacci number u4n is divisible by 3.
Proof. Homework problem.
Almost magically, the Fibonacci numbers are related to the quadratic equation:
x2 − x − 1 = 0.
By the quadratic formula, the roots to this equation are
√
√
1+ 5
1− 5
α=
, and β =
.
2
2
So
α2 − α − 1 = 0,
β 2 − β − 1 = 0.
(3)
Proposition 6 (The Binet formula) For each positive integer n,
un =
That is,
un =
αn − β n
√
.
5
(4a)
√ n
1+ 5
2
√ n
− 1−2 5
√
.
5
Before we prove this, we observe a simple formula which we shall use: For each k ≥ 2,
αk+1 = αk + αk−1 ,
β k+1 = β k + β k−1 .
Indeed, (5) follows from:
0 = α2 − α − 1 αk−1 = αk+1 − αk − αk−1
and the same for β.
Let P (n) denote the statement un =
αn√
−β n
,
5
which we wish to prove for all positive integers n.
Proof of Proposition 5.4.3. We use strong mathematical induction (see section 5.4).
4
(5)
(1) Base case: Starting from the right side of (4a) for n = 1
α1 − β 1
√
=
5
√
√
1+ 5
− 1−2 5
2 √
5
√
2 25
= √ = 1 = u1 ,
5
which is the left side of (4a) for n = 1. Thus P (1) is true. Similarly we may verify that P (2) is true.
(2) Strong inductive step: Suppose P (1) , P (2) , . . . , P (k) are all true for some positive integer
k. That is, suppose (4a) is true for n = 1, 2, . . . , k. We want to prove that P (k + 1) is true, i.e., that
(4a) is true for n = k + 1. To this end we compute
uk+1 = uk−1 + uk =
αk−1 − β k−1 αk − β k
√
+ √
5
5
since P (k − 1) and P (k) are true. Combining the α’s and the β’s and then using (5), we obtain
αk−1 + αk − β k−1 + β k
αk+1 − β k+1
√
√
=
,
uk+1 =
5
5
which says that P (k + 1) is true. This completes the strong inductive step.
(3) By strong induction, we have proved P (n) is true for all positive integers n. That is, we
have proved Proposition 6. Analysis. We are trying to prove a formula for the nth Fibonacci number by induction. The issue
is that the (k + 1)th Fibonacci number depends on the (k − 1)th and k th (it is their sum). So we need
the formulas for uk−1 and uk to derive the formula for uk+1 . In other words, we need to know that
both P (k − 1) and P (k) are true in order to prove that P (k + 1) is true. That this method works is
(a special case of) strong induction. Note that, before, in (regular) induction, we only needed to know
that P (k) is true in order to prove that P (k + 1) is true.
Exercise: Think about this in terms of dominos.
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