Mathematical Induction - University Of Maryland...Jason Filippou (CMSC250 @ UMCP) Induction...

Mathematical Induction

Jason Filippou

CMSC250 @ UMCP

06-27-2016

Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 1 / 48

Outline

1 Sequences and seriesSequencesSeries and partial sums

2 Weak InductionIntro to InductionPractice

3 Strong Induction

4 Errors in proofs by mathematical induction


Sequences and series

Sequences and series


Sequences and series Sequences

Sequences



Definitions

Definition (Sequence)

A function a : N 7→ R is called a sequence.

Examples:

2, 4, 6, . . .10, 20, 301, 1, 2, 3, 5, 8, 13, 21, . . .

So, sequences can be either finite or infinite.

We will mostly care about infinite sequences.



Denoting sequences

A sequence can be enumerated...

a : a1, a2, . . . (or just a1, a2, . . . )c0, c1, c2, . . . (notice the indices)

Described through an explicit formula...

bk = 2k

rn = (n + 1)!

Or a recursive formula...

Fn+1 = Fn + Fn−1 ∀n ≥ 1



The arithmetic sequence

Definition

Let a : a0, a1, . . . be a sequence and ω ∈ R. If aj = aj−1 + ω ∀j ∈ N∗,a is an arithmetic sequence (or progression).

a0 and ω fully define the sequence.

So, how can I write ar?

a1 + r ∗ ω r ∗ a0 a0 + r ∗ ω r ∗ a0



The geometric sequence

Definition

Let a : a0, a1, . . . be a sequence and k ∈ R∗. If aj = c ∗ aj−1 ∀j ∈ N∗,a is a geometric sequence (or progression).

a0 and c fully define the sequence.

ar. How can I write it?

cr ∗ a0 ar0 arc0 a0 + cr


Sequences and series Series and partial sums

Series and partial sums



Definitions

Definition (Series)

Let a0, a1, . . . be any sequence. Then, the sum

+∞∑i=0

ai is called a series.

Definition (Partial sum)

Let n ∈ N. Then, the n-th partial sum of the series+∞∑i=0

ai, denoted

Sn, is the sum

n∑i=0

ai.

The partial sums themselves also form a sequence!



Definitions

Definition (Series)

Let a0, a1, . . . be any sequence. Then, the sum

+∞∑i=0

ai is called a series.

Definition (Partial sum)

Let n ∈ N. Then, the n-th partial sum of the series

+∞∑i=0

ai, denoted

Sn, is the sum

n∑i=0

ai.

The partial sums themselves also form a sequence!



Statements to prove!

To kickstart the discussion on induction, here are two theoremsconcerning partial sums:

Theorem (Closed form of the arithmetic progression partial sum)

If a is an arithmetic progression, Sn = n(a1+an)2 .

Theorem (Closed form of the geometric progression partial sum)

If a is a geometric progression and c 6= 1, Sn = a1(cn−1)c−1 .

Both of those theorems can be proven via (weak) mathematicalinduction!


Weak Induction

Weak Induction


Weak Induction Intro to Induction

Intro to Induction



Proof methods: The story so far...

S

Existential Stmt. ?

Existential Proof

Constructive Non-

constructive

Universal Proof

Direct Indirect

Contradiction ? ? ? Generic

Particular

+ + -

-

Universal Statement Contraposition Division into cases Exhaustion



Where induction fits

S

Existential Stmt. Universal Stmt

Existential Proof

Constructive Non-

constructive

Universal Proof

Direct Indirect

Contradiction Contraposition Exhaustion Cases

Generic Particular

+ + -

-

Induction



The penny proposition: Statement

Suppose I have at least 4¢ in my wallet.

Then, it turns out that all my money can be stacked as 2¢ and 5¢coins!



The penny proposition: Direct (non-inductive) proof

The penny proposition

Every dollar amount greater than 3 ¢ s can be paid with only 2¢ and5¢ coins.

Proof (Direct, by cases).

Suppose we have a total amount of C cents in our wallet. If C is aneven number, then the statement is trivial by the definition of evennumbers: We can just stack k 2¢ coins for some positive integer k. If Cis an odd number greater than 3, then it is 5 or greater. If it is 5, theproblem is trivial: We only need one 5¢ coin. But every odd dollaramount after 5¢ can be retrieved by adding any number of 2¢ coins,because of the definition of parity. We are therefore done in bothcases.

What do you think of this proof?



The principle



The principle

Principle of Weak Mathematical Induction

Assume P (n) is a predicate applied on any natural number n, anda ∈ N. If:

P (a) is true

P (k + 1) is true when P (k) is true ∀k ≥ a

then, ∀n ≥ a, P (n) is true.



The approach

Our task is to prove some proposition P (n), for all positiveintegers n ≥ n0.

Mathematical induction includes the following steps:

1 Inductive Base (IB): We prove P (n0). Most often, n0 will be0, 1, or 2.

2 Inductive hypothesis (IH): If k ∈ N is a generic particular suchthat k ≥ n0, we assume that P (k) is true.

3 Inductive Step (IS): We prove that P (k + 1) is true by makinguse of the Inductive Hypothesis where necessary.



The approach


Mathematical induction includes the following steps:1 Inductive Base (IB): We prove P (n0). Most often, n0 will be

0, 1, or 2.

2 Inductive hypothesis (IH): If k ∈ N is a generic particular suchthat k ≥ n0, we assume that P (k) is true.




The approach



0, 1, or 2.2 Inductive hypothesis (IH): If k ∈ N is a generic particular such

that k ≥ n0, we assume that P (k) is true.




The approach



0, 1, or 2.2 Inductive hypothesis (IH): If k ∈ N is a generic particular such

that k ≥ n0, we assume that P (k) is true.3 Inductive Step (IS): We prove that P (k + 1) is true by making

use of the Inductive Hypothesis where necessary.



The penny proposition, revisited

We want to prove that all amounts of pennies consisting of at least4 pennies can be built using only 2¢ and 4¢ coins.

The following proposition captures the essence of what we want toprove more generally:

The penny proposition, mathematical version

Let n be an integer equal to at least 4. Then, there exist integers m, ksuch that n = 2m + 5k.



Inductive proof of the penny proposition

Proof (By weak mathematical induction).

Let P (n) be the proposition that we want to prove, where n ≥ 4. Let r ∈ Nbe a generic particular for all integers equal to at least 4. Then:

Inductive base: For r = 4, P (4) holds, since for m = 2 and k = 0,4 = 2m + 5k.

Inductive hypothesis: For a specific r ≥ 4, assume P (r): there existintegers m, k such that r = 2m + 5k.

Inductive step: We want to prove P (r + 1): there exist m′, k′ ∈ N suchthat r + 1 = 2m′ + 5k′. By the inductive hypothesis, we have thatr = 2m + 5k ⇔ r + 1 = 2m + 5k + 1⇔ r + 1 = 2m + 5k + (6− 5)⇔r + 1 = 2 (m + 3)︸︷︷︸

m′

+5 (k − 1)︸︷︷︸k′

⇔ r + 1 = 2m′ + 5k′. So P (r + 1) holds.

Since r is a generic particular for the set of integers above 3, the result holdsfor every n ∈ {4, 5, . . . }. We are done.



Rules for authoring mathematical induction proofs

THOU SHALT NOT NEGLECT:

1 To state which variable you are inducing on (“Proof by induction onn, a, r, . . .

2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it

may seem.4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in

the inductive step.




THOU SHALT NOT NEGLECT:1 To state which variable you are inducing on (“Proof by induction on

n, a, r, . . .

2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it


the inductive step.





n, a, r, . . .2 Properly using your generic particulars.

3 Explicitly proving the inductive basis, no matter how obvious itmay seem.

4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in

the inductive step.





n, a, r, . . .2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it

may seem.

4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in

the inductive step.






may seem.4 Clearly assuming the inductive hypothesis

5 Explicitly mentioning where the inductive hypothesis is used inthe inductive step.







the inductive step.


Weak Induction Practice

Practice



Mini-quiz

Are the following partial sums (“Sn”s) of a series?

YES NO

1

100∑n=0

2 ∗ n

2

+∞∑n=0

2 ∗ n

3

+∞∑n=1

2 ∗ n

4

0∑n=0

2 ∗ i

5

100∑n=0

0



Three sums!

The following are some statements we will prove inductively inclass!

Note the different ways in which the theorems can be presented.

Theorem (Partial sum of a special-case arithmetic progression)

Let a be an arithmetic progression with a1 = ω = 1. Then, prove thatSn = n(n+1)

2 .

Theorem (Partial sum of a special-case geometric progression)

For r 6= 1 and n ∈ N, prove that∑n

i=0 ri = rn+1−1

r−1

Theorem (Some random theorem Jason found in an old book of his)

For n ≥ 1, 13 + 23 + · · ·+ n3 =[n(n+1)

2

]2.



Problems beyond sums

Mathematical induction is useful for proving other properties ofintegers as well!

Let’s prove the following theorems inductively:

Theorem

For all integers n ≥ 1, 22n − 1 is divisible by 3.

Theorem

For all integers n ≥ 3, 2n + 1 < 2n.

Theorem

For all integers n ≥ 2,√n < 1√

1+ 1√

2+ · · ·+ 1√

n.


Strong Induction

Strong Induction


Strong Induction

Nets instead of dominoes


Strong Induction

Strong induction principle

Principle of Strong Induction

Let n, a, b ∈ N : a ≤ b. If:

1 P (a), P (a + 1), . . . , P (b) are all true, and

2 ∀k > b, if P (i) is true ∀i : a ≤ i < k, then P (k) is true,

then, P (n) ∀n ≥ a.

Is there any relation to the principle of Weak Induction?

YES NO


Strong Induction

Strongly inductive proofs: Sketch

We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)

Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.

That’s our safety net!

Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)

Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.

Notice that the inductive hypothesis does not cover P (k)!

All 5 rules for cleanly authoring inductive proofs still apply.


Strong Induction

An example proof

Theorem (Divisibility of an integer by a prime)

Prove that any integer n > 1 is divisible by a prime number.

We already proved this one through a very complicated directproof by cases.

The strong inductive hypothesis allows for a much more consiseproof!

Notice that, in this case, r = 0.


Strong Induction

Reminder: recursively defined sequences

Strong induction is very popular for solving problems withrecursively defined sequences.

The Fibonacci sequence is perhaps the most famous such sequence.

Which one among those is a correct definition?1 F1 = 0, F2 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 32 F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 23 F0 = 1, F1 = 1, Fn+1 = Fn + Fn−1 ∀n ≥ 24 F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 2

1 2 3 4


Strong Induction

A proof on a recursive sequence

A statement on a recursive sequence

Let a : a1, a2, . . . be a sequence, recursively defined as follows:

a1 = 0

a2 = 0

ak =3abk/2c + 2 ∀k ≥ 3

Prove that an is even for each integer n ≥ 1.


Strong Induction

Proof

Proof via strong induction on k.

Let r ≥ 1 be an integer generic particular and P (n) be the proposition that we wantto prove. We proceed inductively:

Inductive Base: For r = 1 and r = 2, a1 and a2 are even, so P (1) and P (2)hold.

(Strong) Inductive Hypothesis: For some r > 2 and for all 1 ≤ i < r,assume that P (i) is true: ai is even.

Inductive step: We want to prove that P (r) is true, i.e that ar = 3abr/2c + 2is even. From a known theorem, we know that the product of an even and anodd integer is even, while the sum of two even numbers is also even. Since wecovered the cases r = 1 and r = 2 in the inductive basis, it is the case thatr ≥ 3, and for all such choices of r, br/2c < r. But this means that abr/2c iseven, by the inductive hypothesis. Therefore, ar is even, since it consists ofa sum of an even number (2) added to another even number (3 times an evennumber).

Since r was chosen arbitrarily from the set N∗, the result holds ∀n ∈ N∗.


Strong Induction

Another proof on a recursive sequence

The closed form for another recursive sequence

Let a be a sequence such that:

a1 = 1

a2 = 8

an = an−1 + 2an−2, ∀n ≥ 3

Prove that an = 3 · 2n−1 + 2(−1)n.


Strong Induction

Proof

Proof (by strong induction on n).

Let k be a generic particular for N∗, and P (n) be the proposition that we want to show. Weproceed inductively:

Inductive base: P (1) is the proposition: a1 = 3 · 20 + 2 · (−1)1 ⇔ a1 = 1. Since, bythe definition of the sequence a, a1 = 1, P (1) is true. P (2) is the proposition:a2 = 3 · 21 + 2(−1)2 ⇔ a2 = 8. Once again, this is true by the definition of a, so P (2)also holds.

(Strong) inductive hypothesis: For some k > 2, we assume P (i) is true∀i ∈ {1, 2, . . . , k − 1}, i.e ai = 3 · 2i−1 + 2 · (−1)i.

Inductive step: We want to prove P (k), i.e ak = 3 · 2k−1 + 2(−1)k. By thedefinition of an for n ≥ 3, we have:

ak = ak−1 + 2ak−2

= 3 · 2k−2 + 2(−1)k−1 + 2[3 · 2k−3 + 2(−1)k−2] (By Inductive Hypothesis)

= 3 · 2k−2 + 2(−1)k−1 + 3 · 2k−2 + 2 · 2(−1)k−2 (By distributing the factor 2)

= 3 · 2k−1 + 2[(−1)k−1 + 2(−1)k−2] (By grouping terms)

= 3 · 2k−1 + 2(−1)k (By equality of the red terms)

⇔ P (k) is true

Since our result was reached for an arbitrarily selected k,it holds ∀n ∈ N∗.


Strong Induction

Re-visiting an older theorem

Theorem (Divisibility by a prime)

Every integer n > 1 is divisible by a prime number.

Do you perhaps remember what the direct proof that we discussedpreviously looked like?

Strong induction will lead us to a much more consise proof!


Strong Induction

ProofProof (by strong induction on n).

Let q be a generic particular for the set {2, 3, . . . } and P (n) be the proposition we aretrying to prove. We proceed inductively:

Inductive base: For q = 2, we have that 2 is divisible by a prime number, namelyitself. Therefore, P (2) holds.

(Strong) Inductive hypothesis: For q > 2, we assume that P (i) holds∀i ∈ {2, 3, . . . , q − 1}, i.e i is divisible by a prime number.

Inductive step: We want to prove P (q), i.e that q > 2 is divisible by a prime

number. We distinguish between two cases:

1 q is prime. In this case, since q|q, we are done.2 q is not prime. Therefore, q is composite, which means that∃a, b ∈ N : q = a · b.a Since both a and b are above 1 (if one of them was 1, theother would have to be q, making q prime), we conclude that both of them haveto be smaller than q. By the inductive hypothesis, this means that thereexists some prime p that divides at least one of a or b, and by transitivity ofdivisibility we have that p|q. This is exactly P (q).

Since q was selected arbitrarily from {2, 3, . . . }, we have that the result holds for everyinteger in this set.

aWe only care about the positive divisors of q.


Strong Induction

Practice

Let’s split into teams and solve these problems!

Yet another recursive sequence

Suppose that the sequence s is such that: s0 = 12, s1 = 29 andsk = 5sk−1 − 6sk−2 ∀k ≥ 2. Prove that sk = 5 · 3k + 7 · 2k, ∀k ∈ N.

More Fibonacci? Sure, why not.

Show that the Fibonacci sequence follows an “odd-odd-even” pattern.More formally, if by F we denote this sequence, show that∀m ∈ N∗, F3m−2 and F3m−1 are odd, while F3m is even.

The “Tribonacci” Sequence (yes, seriously)

The Tribonacci sequence is defined as: T0 = T1 = T2 = 1 andTn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3. Prove that, for alln ∈ N, Tn < 2n+1


Errors in proofs by mathematical induction




This section

We will present some examples of non-sensical statements provenvia the principle of induction, or of correct statements withincorrect proofs!

The teaching staff assumes no responsibility for studentsbelieving those to be valid statements or proofs!

All examples courtesy of materials taken from MATH 347,Summer ’14, UIUC.


http://www.math.illinois.edu/~ajh/347.summer14/

http://www.math.illinois.edu/~ajh/347.summer14/


Non-sensical proof #1Sum of first n non-zero integers

∀n ∈ N∗,n∑

i=1

i =1

2(n +

1

2)2

Proof (By weak induction on n).

Let r ∈ N∗ be a generic particular and P (n) the statement we want to solve. We proceed inductively:

1 Inductive base: For r = 1, P (1) holds.

2 Inductive hypothesis: Assume that P (r) holds ∀r ≥ 1, i.er∑

i=1

i =1

2(r +

1

2)2

3 Inductive step: We want to prove P (r + 1), i.e

r+1∑i=1

i =1

2((r + 1) +

1

2)2. Beginning from the LHS

we have:

r+1∑i=1

i =r∑

i=1

i + (r + 1) =1

2(r +

1

2)2

+ (r + 1) (By breaking apart the sum and by I.H)

=1

2

[(r

2+ r +

1

4) + (2r + 2)

]=

1

2

[(r +

3

2)2 − 3r −

9

4+ r +

1

4+ (2r + 2)

](By algebra)

=1

2

[(r + 1) +

1

2

]2⇒ P (r + 1) holds. (By algebra)

Since r was chosen arbitrarily from N∗, we conclude that the result must hold for all positive integers.



Non-sensical proof #2

Raising 2 to a non-negative integer equals 1

∀n ∈ N, 2n = 1

Proof (by strong induction on n).

Let P (n) be the proposition that we’re trying to prove and r ∈ N be a generic particular.We proceed inductively:

1 Inductive base: For r = 0, 2r = 1. So P (0) = 1.

2 Inductive hypothesis: For r > 0, we assume that ∀i ∈ {0, 1, . . . , r}, P (i) holds, i.ethat 2i = 1.

3 Inductive step: We want to prove that P (r + 1) holds, i.e that 2r+1 = 1. We beginfrom the LHS:

2r+1 =21 · 2r (By properties of exponentiation)

=1 · 1 (By inductive hypothesis)

= 1

Therefore, P (r + 1) holds.

Since r was arbitrarily chosen within N, the result holds ∀n ∈ N.



Non-sensical proof #3

LOL WUT

All real numbers are equal.

Proof.

Equivalently, we want to prove that, ∀k ∈ N, a1 = a2 = · · · = ak, whereai ∈ R. We will proceed via strong induction on k. Let r ∈ N∗ be ageneric particular and P (k) be the proposition we’re trying to solve.We proceed inductively:



“All reals are equal” proof, contd.

Proof.

1 Inductive base: For r = 1, the statement is trivially true: a1 = a1.

2 Inductive hypothesis: For r > 1, we will assume that ∀i ∈ {0, 1, . . . , r − 1},P (i) is true, i.e any i reals are true: a1 = a2 = · · · = ai.

3 Inductive step: We want to prove that P (r) is true, i.e thata1 = a2 = · · · = ar. Consider the first r − 1 reals. Since r − 1 < r, from the I.Hwe can deduce:

a1 = a2 = · · · = ar−1 (1)

The same can be deduced for the last r − 1 reals:

a2 = a3 = · · · = ar (2)

From (1) and (2) we conclude that a1 = a2 = · · · = ar−1 = ar, and P (r) is true.

Since r was arbitrarily chosen, the result holds ∀k ∈ N.



True statement, incorrect proof

Parity of the factorial

∀n > 1, n! is even

Proof (via strong induction on n).

Let r ∈ {2, 3, . . . } be a generic particular and P (n) be the proposition we’re tryingto prove. We proceed inductively:

1 Inductive base: For r = 2, 2! = 2 · 1 = 2. 2 is even, therefore P (2) holds.

2 Inductive hypothesis: For r > 2, we assume P (i) ∀i ∈ {2, 3, . . . , r − 1}, i.e i!is even.

3 Inductive step: We want to prove P (r + 1), i.e that (r + 1)! is even. By therecursive definition of the factorial, (r + 1)! = (r + 1) · r! From the inductivehypothesis, we know that r! is even. From a known theorem, we know that theproduct of two integers, where at least one is even, is also even. Therefore,(r + 1)! is even and P (r + 1) holds.

Since r was chosen arbitrarily, the result holds for every integer n ≥ 2.



Vote!

Proof (via strong induction on n).

Let r ∈ {2, 3, . . . } be a generic particular and P (n) be the proposition we’re tryingto prove. We proceed inductively:

1 Inductive base: For r = 2, 2! = 2 · 1 = 2. 2 is even, therefore P (2) holds.

2 Inductive hypothesis: For r > 2, we assume P (i) ∀i ∈ {2, 3, . . . , r − 1}, i.e i!is even.

3 Inductive step: We want to prove P (r + 1), i.e that (r + 1)! is even. By therecursive definition of the factorial, (r + 1)! = (r + 1) · r! From the inductivehypothesis, we know that r! is even. From a known theorem, we know that theproduct of two integers, where at least one is even, is also even. Therefore,(r + 1)! is even and P (r + 1) holds.

Since r was chosen arbitrarily, the result holds for every integer n ≥ 2.

This proof has an error in:

The Ind. Base The Ind. Hypothesis The Ind. Step Somewhere else


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