U.U.D.M. Project Report 2020:21

Examensarbete i matematik, 15 hpHandledare: Inger SigstamExaminator: Veronica Crispin QuinonezJuni 2020

Department of MathematicsUppsala University

Mathematical Induction

Hanna Wedin

Abstract

Mathematical Induction is used in all fields of mathematics. Inthis thesis we will do an overview of mathematical induction andsee how we can use it to prove statements about natural numbers.We will take a look at how it has been used in history and wherethe name mathematical induction came from. We will also look atdifferent types of induction, weak and strong induction. You canalso do induction on other types of structures, like the length ofpropositions.

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Contents

1 Introduction 31.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Paradoxes with induction . . . . . . . . . . . . . . . . . . . . . . 5

2 Mathematical Induction 52.1 Axiom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 The sum of i . . . . . . . . . . . . . . . . . . . . . . . . . 62.2.2 The sum of i2 . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Strong Induction 83.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.1.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93.1.2 The Fibonacci sequence . . . . . . . . . . . . . . . . . . . 103.1.3 Fundamental Theorem of Arithmetic . . . . . . . . . . . . 103.1.4 The sum jk . . . . . . . . . . . . . . . . . . . . . . . . . . 10

4 Induction on other structures 124.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

5 More Examples 155.1 The Ackermann function . . . . . . . . . . . . . . . . . . . . . . . 15

5.1.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 155.1.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.1.3 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . 165.1.4 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5.2 Linear algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.3 Binomial coefficients . . . . . . . . . . . . . . . . . . . . . . . . . 18

5.3.1 Binomial . . . . . . . . . . . . . . . . . . . . . . . . . . . 185.3.2 The Binomial Theorem . . . . . . . . . . . . . . . . . . . 19

5.4 Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205.4.1 The General Power Rule . . . . . . . . . . . . . . . . . . . 205.4.2 The Leibniz rule . . . . . . . . . . . . . . . . . . . . . . . 215.4.3 Taylor’s theorem . . . . . . . . . . . . . . . . . . . . . . . 22

6 Conclusion 24

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1 Introduction

Mathematical induction has a big influence in mathematics. It is a way to provemathematical statements about natural numbers. You start learn about math-ematical induction and the principle of induction in the later upper secondaryschool in Sweden. You also learn about induction in the university if you studymathematics.

The principle of Mathematical Induction consist of three steps:

1. Base case, show that it holds for the first value.

2. Induction step: Here you assume that the statements holds for a randomvalue, and then you show that it also holds for the value after that.

3. Conclusion, because the statement holds for the base and for the inductivestep, it is true for every value.

You can think of induction in an illustrating way, think of a ladder. In theBase case we check that we have a first step to step on. Then in the inductionstep we go to an arbitrary step p and then we show that if we go to p there is astep after, p+ 1. So if we go to an arbitrary step on the ladder we want to showthat we can lift our leg and go to the next step. Why does this work? Becauseif our base case is p and then we can show that we can go to the second stepp + 1, if we now let the second step be p we can show that we can go to thethird step, p+ 1, and then we can let the third step be p, and so it goes on andtherefore we have show that we can go to any step on the ladder.

Figure 1: A ladder, an illustration of induction.

You can also think of mathematical induction like a domino effect. The

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base step is that the first domino will fall, and then if any domino fall then thedomino after also will fall. So all dominoes will fall. [4]

1.1 History

Mathematical induction has been used in mathematics way back in history.Some people think that even Euclid used induction when he proved that thereare infinitely many primes, even though there is no evidence that he used it, butsome writers think that he implied it without being stated directly. Some alsothinks that Plato and Pappus used induction but there are no evidence thatthey used it[3].

The Italian mathematician Francesco Maurolico (1494− 1575) did a nonin-ductive proof concerning the sum of the first n natural numbers. Even thoughMaurolico did a noninductive proof there are people who think that Pascal gothis inspiration for the induction principle from Maurolico, when Pascal in the16th century showed by induction what the sum of the first n natural numbersis [3].

In [3] it says that sometimes induction is compared to the method of ex-haustion, like Calivieres principle where he calculates the area of a circle. Therehe draws polygons that fits in the circle, because he knew how to calculate thearea of a polygon, and then he draws larger and larger polygons, until he hasdrawn a polygon which is almost in the same size as the circle. The similaritieswith induction and the method of exhaustion is that you start with a guess, andto prove your guess you do infinitely many iterations which follows from earliersteps. There are some proofs that are used with the method of exhaustion thatcan be translated into an inductive proof.

There was an Egyptian called ibn al-Haytham (969-1038) who used inductivereasoning to prove the formula for

n∑i=1

i4 =

(n

5+

1

5

)n

(n+

1

2

)[(n+ 1)n− 1

3

].

Levi ben Gerson (1288-1344), used mathematical induction, he called the method“rising step by step without end”. Comparing to how we are used to use induc-tion where we first do the base case and then the induction step to show thatit hold for n to n+ 1, Levi started with the induction step and then he showedfor the base. He used induction to show that

(1 + 2 + · · ·+ n)2 = n3 + (1 + 2 + · · ·+ (n− 1))2.

Levi also did an inductive proof where he went from n to n− 1 [5].As you can see mathematicians in history have used mathematical induction

and inductive reasoning for a long time, but there were no one who had namedthis method yet. According to [2] the first ones who started to name inductionwas the Englishman John Wallis (1616 − 1703) and the Swiss Jacob Bernoulli(1655 − 1705). Wallis used a kind of induction called incomplete induction to

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find the ratio between∑n

i=1 i2 and n2(n+ 1). Wallis incomplete induction both

got bad and good criticism. Bernoulli was one of the ones who gave Wallis badcriticism and he introduced the principle argument from n to n + 1. Bernoullicriticized Wallis but he also thought that even though Wallis used incompleteinduction it was a start to induction. Bernoulli showed the Binomial theoremwith the argument when you go from n to n+ 1.

Georg Simon Klugel (1739− 1812) explained the weakness of Wallis induc-tion in his dictionary, he also explains Bernoullis proof from n to n+1. Then inEngland Thomas Simpson (1710− 1761) used the n to n+ 1, but neither did hegive it a name. In the early part of the nineteenth century George Peacock, 1830,used induction, and the n to n + 1 argument and he called it “demonstrativeinduction”. In 1833 Augustus De Morgan suggested a new name “successive in-duction”, but in the end of his article he used “mathematical induction”. Boththe names “demonstrative induction” and “mathematical induction” are usedby writers. But then the “demonstrative induction” get disused. In the USAinduction was used but not called by its name, in Europe, the name “mathe-matical induction” was used.

The Italian mathematician Giuseppe Peano (1858 − 1932) formulated theaxiom system we call the Peano’s axiom in 1889. With Peano’s axiom we canconstruct all the natural numbers, and one of his axiom is the one we call theInduction axiom. You will see the axiom in section 2.1.

There are a lot more of well-known mathematicians who has used mathe-matical induction, if you want to read more about it you can look in the historysection in [3]. There are a lot more sources where you can find more informationif you are interested.

1.2 Paradoxes with induction

There are some paradoxes with induction and inductive reasoning. [3] has somefun examples of different paradoxes. One example is about a teacher who toldhis students that next week they will get an unprepared test. And the studentsthink that if they don’t have had the test on Thursday night they cannot have iton Friday because then they will know that the test must be on Friday and it willnot be unprepared. So the test cannot be on Friday, by the same argument thetest cannot be on Thursday, and either on Wednesday, or Tuesday or Monday.So they will not have a test next week.

Another example is about being bald, if you don’t have any hair on you headyou are bald but if you only have one hair you will still be bald, so if you onlyhave one hair more you will still be bald, therefore everyone is bald.

2 Mathematical Induction

The most common induction is induction over integers. This is the one you startlearn in school and use to prove simpler theorems. It is easy to think that math-ematical induction is something you only use in logic and algebra, but the fact is

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that you use mathematical induction in almost all fields of mathematics. Laterwe are going to show some fun examples from different parts of mathematics,like calculus and linear algebra.

2.1 Axiom

In [4] Peano’s axioms are formulated as follows, where s is the successor function,s(n) is the immediate successor of n.

Peano’s axioms:

1. There is a positive integer called 1.

2. For all x there is a unique number s(x).

3. For all x, s(x) 6= 1.

4. For all x and y, if s(x) = s(y), then x = y.

5. Let M be a subset of positive integers such that

(a) 1 is in M

(b) If x is in M , then s(x) is in M .

Then you can make the conclusion that M is the set of all positive integers.

The fifth axiom is the Induction Axiom, and the one we refer to when we talkabout the induction axiom.

In [4] they formulate the principle of induction like this:The principle of induction: Assume S1, S2, S3, . . . are statements such that

1. S1 is true,

2. For all positive integers p the implication Sp =⇒ Sp+1 holds.

Then all statements are true.

To prove something by mathematical induction you first do the base case, toshow that the statement holds for the smallest integer. Then you do the induc-tion hypothesis and assume that the statement holds for some arbitrary positiveinteger p, and if you can show that the statement holds for p+ 1 you can by theprinciple by induction say that the statement holds for all positive integers.

2.2 Examples

2.2.1 The sum of i

An easy example on how to use the principle of induction is to show

n∑i=1

i =n(n+ 1)

2

6

Proof. Base case: First we start with the base case, here we shall show thatthe formula holds for n = 1. We start with the left side:

1∑i=1

i = 1

and then the right side,1(1 + 1)

2= 1.

We see that the left side is equal to the right side, so the formula holds for n = 1.Induction step: In the induction step we first start with our Induction hy-pothesis: Assume the formula holds for some arbitrary positive integer p ≥ 1,

p∑i=1

i =p(p+ 1)

2.

The next step is now to show that the formula also holds for the following integerp+ 1,

p+1∑i=1

i =(p+ 1)(p+ 2)

2.

We start with the left side

p+1∑i=1

i =

p∑i=1

i+ (p+ 1),

and by the Induction hypothesis we see that we can rewrite∑p

i=1 i and get

p(p+ 1)

2+ (p+ 1) =

p(p+ 1) + 2(p+ 1)

2=p2 + 3p+ 2

2=

(p+ 1)(p+ 2)

2.

We see that the formula also holds for p+1. By the Induction axiom the formulaholds for all integers n ≥ 1.

2.2.2 The sum of i2

n∑i=1

i2 =n(n+ 1)(2n+ 1)

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Proof. Base case: For n = 1 we have

n∑i=1

i2 = 1

and1(1 + 1)(2 + 1)

6=

6

6= 1.

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We see that both left side and right side are equal to 1, so the formula holds forn = 1.Induction hypothesis: Suppose it holds for some arbitrary positive integer p,i.e.

p∑i=1

i2 =p(p+ 1)(2p+ 1)

6.

Induction step: Now we want to show that the formula also holds for p + 1,i.e.

p+1∑i=1

i2 =(p+ 1)(p+ 2)(2p+ 3)

6.

We start with the left side

p+1∑i=1

i2 =

p∑i=1

i2 + (p+ 1)2,

and using the Induction hypothesis, we get

p(p+ 1)(2p+ 1)

6+ (p+ 1)2 =

(p+ 1)(p(2p+ 1) + 6(p+ 1))

6=

(p+ 1)(2p2 + 7p+ 6)

6=

(p+ 1)(p+ 2)(2p+ 3)

6.

We see that the formula also holds for p + 1. By the Induction axiom theformula holds for all positive integers n.

3 Strong Induction

The first type of induction described in section 2 is sometimes called weak in-duction, in this section we will describe the type of induction called stronginduction. Although this type is called strong induction doesn’t it mean thatit actually is stronger then the other type. Strong induction is also called com-plete induction. This type of induction is like the first type of induction buthere you have a different induction hypothesis. In weak induction you assumein the induction hypothesis that the formula/statement holds for some posi-tive integer p, but in strong induction you assume that the formula/statementholds for all positive integer less than or equal to p. The two types of inductionare equivalent to each other, but sometimes the strong induction is easier to use.

In [4] they formulate the principle of Strong induction: Assume S1, S2, S3, . . .is a sequence of statements such that

1. S1 are true,

2. For all positive integers p the implication S1 ∧ S2 ∧ · · · ∧ Sp =⇒ Sp+1

holds.

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Then all statements are true.

3.1 Examples

3.1.1 Example

An easy example where we use strong induction instead of weak induction istaken from [4].

Let h : N −→ Q be defined as followsh(0) = 1,

h(1) = 1

h(n) = h(n−1)2 + 1

h(n−2) , n > 1

with help of strong induction we will show 1 ≤ h(n) < 2 for all n ∈ N.

Proof. Base case: We start as usual with the base case and we want to showthat it holds for n = 0 and n = 1. We have by definition that h(0) = 1 andh(1) = 1, and we are done.Induction hypothesis: Assume that it holds for j = 0, 1, 2, . . . , p, where p ≥ 1

1 ≤ h(j) < 2.

Induction step: Now we want to show that it also holds for p + 1. We havethat p+ 1 > 1, so

h(p+ 1) =h(p)

2+

1

h(p− 1).

The Induction hypothesis gives us

1

2≤ h(p)

2< 1 and

1

2<

1

h(p− 1)≤ 1,

so

h(p+ 1) =h(p)

2+

1

h(p− 1)>

1

2+

1

2= 1,

and

h(p+ 1) =h(p)

2+

1

h(p− 1)< 1 + 1 = 2.

We now have that1 ≤ h(p+ 1) < 2.

According to the Induction axiom it holds for all positive integers n.

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3.1.2 The Fibonacci sequence

The principle of induction can also be used to define sequences of numbers [7].The Fibonacci sequence is defined by induction and properties of it can easilybe established with strong induction.

The Fibonacci sequence f(1), f(2), . . . is defined recursively as follows:f(1) = 1,

f(2) = 1

f(n) = f(n− 1) + f(n− 2), n ≥ 3

We show that f(n) < 2n for all n ≥ 1.

Proof. Base case: n = 1, we have f(1) = 1 < 21. n = 2, we have f(2) = 1 < 21.Induction hypothesis: Assume that f(j) < 2j is true for all j = 1, 2, . . . , p,where p ≥ 2.Induction step: Now we want to show for p+ 1, we have

f(p+ 1) = f(p) + f(p− 1) < 2p + 2p−1

< 2p + 2p = 2p(1 + 1) = 2p · 2 = 2p+1.

3.1.3 Fundamental Theorem of Arithmetic

On good example of strong induction is how you can prove the existence partof the Fundamental theorem of arithmetic.

Theorem (Fundamental Theorem of Arithmetic). Every positive integer greaterthan 1 can be written as a product of prime numbers in a unique way.

Proof. (Existence part of the fundamental theorem of arithmetic) Let Sn be thestatement that n can be written as a product of prime numbers.Base case: n = 2, it is pretty clear because 2 is a prime number.Induction hypothesis: Assume that S2, S3, . . . , Sp are true, where p ≥ 2.Induction step: We now want to show that Sp+1 is true. There are two cases.First case: p + 1 is a prime number, then Sp+1 is true. Second case: p + 1is not a prime number. Then p + 1 can be written as p + 1 = a · b, wherea, b ∈ {2, 3, . . . , p}. By the Induction hypothesis a, b can be written as productsof prime numbers, and therefore a · b also can be written as a product of primenumbers. So Sp+1 can be written as a product of prime numbers.

3.1.4 The sum jk

In this example we will show a generalization of earlier examples on the sums∑nj=1 i and

∑nj=1 i

2 from 2.2.1 and 2.2.2. The theorem can also be used when

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you want to calculate∫ a

0xkdx. Then you use the Riemann sum and you get∫ a

0

xk = limn→∞

a

n

n∑j=1

(aj

n

)k

= ak+1 limn→∞

1

nk+1

n∑j=1

jk

= ak+1 limn→∞

(1

k + 1+

1

2n+Pk−1(n)

nk+1

)=ak+1

k + 1.

Theorem. Let n ∈ N. Then

n∑j=1

jk =nk+1

k + 1+nk

2+ Pk−1(n)

for every positive integer, where Pk−1(n) is some polynomial in n of degree≤ k − 1. For every positive integer k.

If we look at∑n

i=1 i2 and we know that it can be written as

n∑i=1

i2 =n(n+ 1)(2n+ 1)

6,

and now we can rewrite it

n(n+ 1)(2n+ 1)

6=

2n3 + 3n2 + n

6=n3

3+n2

2+ P2(n).

Proof. Base case: k = 1

n∑j=1

j1 =n(n+ 1)

2=n2 + n

2=n1+1

1 + 1+n

2+ P0(n).

Induction hypothesis:

n∑j=1

jk =nk+1

k + 1+nk

2+ Pk−1(n)

for k = 1, 2, . . . ,m.Induction step: We want to show for m+ 1

n∑j=1

jm+1 =nm+2

m+ 2+nm+1

2+ Pm(n).

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To get where we want we use the Binomial Theorem which gives us

(j + 1)m+2 = jm+2 + (m+ 2)jm+1 +(m+ 2)(m+ 1)

2jm + ...+ 1,

thus

(j + 1)m+2 − jm+2 = (m+ 2)jm+1 +(m+ 2)(m+ 1)

2jm + ...+ 1

for j = 1, 2, 3, ..., n. Thus by taking sums we get

n∑j=1

((j+1)m+2−jm+2) = (m+2)

n∑j=1

(jm+1)+(m+ 2)(m+ 1)

2

n∑j=1

(jm)+...+

n∑j=1

1.

Left side telescopes and gives us

(n+ 1)m+2 − 1 = (m+ 2)

n∑j=1

(jm+1) +(m+ 2)(m+ 1)

2

n∑j=1

(jm) + ...+

n∑j=1

1.

We expand the left side and use the Induction hypothesis on the term∑n

j=1 jm

on the right side, we get

nm+2 + (m+ 2)nm+1 +(m+ 2)(m+ 1)

2nm + ...+ 1− 1 =

(m+ 2)

n∑j=1

(jm+1) +(m+ 2)(m+ 1)

2

(nm+1

m+ 1+nm

2+ Pm−1(n)

)+ ...+

n∑j=1

1.

We want∑n

j=1 jm+1 alone

n∑j=1

jm+1 =nm+2

m+ 2+ nm+1 − nm+1

2+ Pm(n)

which givesn∑

j=1

jm+1 =nm+2

m+ 2+nm+1

2+ Pm(n).

4 Induction on other structures

In the examples of induction above we only have talked about mathematicalinduction on positive integers, but you can do mathematical induction on otherstructures. The principle is the same as when you do on positive integers. Welook at the principle by an example from logic.

We will start to define a language for propositional logic, in [6] they definethe language like this.

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Definition 4.1. The language of propositional logic has an alphabet consistingof

1. proposition symbols: p0, p1, p2, . . .

2. connectives: ∨,∧,←,¬, ,↔,⊥

3. auxiliary symbols: (, ).

Definition 4.2. We define the set of all propositions (PROP ).

1. ⊥ ∈ PROP and pi ∈ PROP for every i ∈ N

2. If ϕ,ψ ∈ PROP then (ϕ ∧ ψ), (ϕ ∨ ψ), (ϕ→ ψ), (ϕ↔ ψ) ∈ PROP

3. If ϕ ∈ PROP then (¬ϕ) ∈ PROP .

The propositions in the base are called atomic propositions. To see that thereare no difference between induction over natural numbers and on propositions,we will do induction over the length n, where n ∈ N. In the base case we will asusual try to show that the statement is true for propositions with length 1. In theinduction hypothesis we will assume that the statement holds for propositionswith length 1, 2, 3, . . . , p. Last but not least we will in the induction step showthat it holds for propositions with length p + 1. As you can see it is the sameprinciple as in strong induction and it is built on the same axiom, Peano’s axiom,as we have seen before.

4.1 Examples

Proposition 1. Let ϕ ∈ PROP . Then the number of left parenthesis in ϕ isequal to the number of right parenthesis.

Proof. We call the number of left parenthesis for L(ϕ) and right parenthesis forR(ϕ).Base case: We first want to show for a formula ϕ with length n = 1. ϕ = pior ϕ = ⊥,

V (ϕ) = H(ϕ) = 0.

Induction hypothesis: We assume that it is true for all formulas of lengthn = 1, 2, 3, . . . , p.Induction step: Now we want to show that it also is true for formulas withlength p+ 1. Let ϕ be a formula with length p+ 1. ϕ must contain at least oneconnective and need to be on one of these forms:

(¬ψ), (ψ ∧ δ), (ψ ∨ δ), (ψ → δ), (ψ ↔ δ)

where ψ and δ are formulas. We need to show that the statement is true for allthese forms. To make it easier we call ∧,∨,→,↔ for �.

We start with ϕ = (ψ�δ). ψ and δ must have length ≤ p. We know fromthe Induction hypothesis that L(ψ) = R(ψ) = l and L(δ) = R(δ) = j. The

13

formation of the string (ψ�δ) adds one more left parenthesis and one moreright parenthesis, which gives us

L((ψ�δ)) = k + j + 1 = R((ψ�δ)).

So it is true for (ψ�δ). Now we want to show for ϕ = (¬ψ). ψ has the length ≤ pand we know from the Induction hypothesis that L(ψ) = R(ψ). The formationof (¬ψ) does add one more left parenthesis and one more right parenthesis whichgives us

L((¬ψ)) = L(ψ) = R(ψ) = R((¬ψ)).

So it is true for ϕ = (¬ψ). Because the statement is true for all cases we canaccording to the Induction axiom say that it is true for all propositions withlength n ≥ 1.

Definition 4.3. The rank r(ϕ) of a proposition ϕ is defined byr(ϕ) = 0 for atomic ϕ,

r((ϕ�ψ)) = max(r(ϕ), r(ψ)) + 1,

r((¬ϕ)) = r(ϕ) + 1

Proposition 2. Let ϕ ∈ PROP and let C(ϕ) denote the number of occurrencesof connectives in ϕ. Then r(ϕ) ≤ C(ϕ).

Proof. Base case: We start to show for ϕ with length n = 1. Then ϕ is atomicand we have that r(ϕ) = 0, and C(ϕ) = 1 or C(ϕ) = 0, so r(ϕ) < C(ϕ).Induction hypothesis: Assume that it is true for formulas with length 1, 2, . . . , p.Induction step: Now we want to show that it also is true for formulas withlength p+ 1. We want to show for (ψ�δ) and (¬ψ), where ψ and δ are formulaswith length ≤ p.We start with (ψ�δ). We know that ψ and δ must have length ≤ p, so by theInduction hypothesis we know that r(ψ) ≤ C(ψ) and r(δ) ≤ C(δ). We havethat

r((ψ�δ)) = max(r(ψ), r(δ)) + 1

C((ψ�δ)) = C(ψ) + C(δ) + 1.

Case 1: r(ψ) ≥ r(δ). Then we have that

r((ψ�δ)) = r(ψ) + 1 According to the Induction hypothesis we have

≤ C(ψ) + 1

≤ C(ψ) + C(δ) + 1 = C((ψ�δ)).

Case 2: r(δ) > r(ψ). Then we have that

r((ψ�δ)) = r(δ) + 1 According to the Induction hypothesis we have

≤ C(δ) + 1

≤ C(ψ) + C(δ) + 1 = C((ψ�δ)).

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So it is true for (ψ�δ). Now we want to show for (¬ψ), where ψ is a formulawith length ≤ p. From the Induction hypothesis we have that r(ψ) ≤ C(ψ). Wealso have C((¬ψ)) = C(ψ) + 1, so

r((¬ψ)) = r(ψ) + 1 ≤ C(ψ) + 1 = C((¬ψ)).

Because the statement is true for all the cases we can according to the Inductionaxiom say that it is true for all formulas with length n ≥ 1.

5 More Examples

5.1 The Ackermann function

The Ackermann function is used in the part of mathematics called recursivetheory. It is a total computable function that is not primitive recursive. Itis named after the German Wilhelm Ackermann (1896 − 1962). We will seethat the function increases very quickly. The Ackermann function has some funproperties that we are going to prove with induction.

Definition 5.1.

A(m,n) =

2n, if m = 0

0, if m ≥ 1, n = 0

2, if m ≥ 1, n = 1

A(m− 1, A(m,n− 1), if m ≥ 1, n ≥ 2

5.1.1 Example 1

Show that A(1, n) = 2n for all positive integers n.

Proof. Base case: n = 1A(1, 1) = 2 = 21

Induction hypothesis: Assume that it holds for some positive integer p.

A(1, p) = 2p

Induction step: We want to show that it also holds for p+ 1

A(1, p+ 1) = 2p+1

Consider the left side:

A(1, p+ 1) = A(0, A(1, p),

and according to the Induction hypothesis this equals

A(0, 2p) = 2 · 2p = 2p+1.

15

5.1.2 Example 2

Show that A(m, 2) = 4 for all positive integers m.

Proof. Base case: m = 1

A(1, 2) = A(0, A(1, 1)) = 2 ·A(1, 1) = 2 · 2 = 4

Induction hypothesis: Assume that it holds for some arbitrary integer p ≥ 1.

A(p, 2) = 4

Induction step: We want to show that it also holds for p+ 1

A(p+ 1, 2) = 4

Consider the left side:

A(p+ 1, 2) = A(p,A(p+ 1, 1) = A(p, 2)

According to the Induction hypothesis is it equal to 4.

5.1.3 Example 3

To find some more properties we look at what happens if m always was equalto 2.

A(2, 1) = 2

A(2, 2) = A(1, A(2, 1)) = A(1, 2) = 22

A(2, 3) = A(1, A(2, 2)) = A(1, 22) = 222

A(2, 4) = A(1, A(2, 3)) = A(1, 222

) = 2222

We start to see a pattern and make a guess that A(2, n) = T (n), where

T (n) = 22...2

,

where the number of 2 is n. We try to prove it with induction.

Proof. Base case: n = 1,

A(2, 1) = 21 = T (1)

Induction hypothesis: Assume that A(2, p) = T (p) is true for some arbitrary

16

integer p ≥ 1.Induction step: Now we want to show for n = p+ 1

A(2, p+ 1) = A(1, A(2, p)) = 2A(2,p),

Now we use the induction hypothesis and get

= 2T (p) = T (p+ 1).

5.1.4 Example 4

We try and see what happens if m = 3.

A(3, 1) = 2 = T (1)

A(3, 2) = A(2, A(3, 1)) = T (T (1)) = T (2) = 4 = T (T (1)) = T 2(1)

A(3, 3) = A(2, A(3, 2)) = T (T (2)) = T (4) = 256 = T (T (T (1))) = T 3(1)

A(3, 4) = A(2, A(3, 3)) = T (T (4)) = T (256) = T 4(1)

Here Tn(1) means n applications of T on the number 1. We see that T (T (4))is extremely big, it is 2 raised to 2, 256 times. We start to see a pattern a makea guess.

A(3, n) = Tn(1).

We see that the result depends on the result from the previous number, andtherefore it is easier to use strong induction.

Proof. Base case: n = 1, we have already seen that it holds for n = 1 causeA(3, 1) = 2 = T (1).Induction hypothesis: We assume that the formula hold for n = 1, 2, 3, . . . , p,where p ≥ 1.

A(3, p) = T p(1)

Induction step: Now we want to show that it also holds for p+ 1.

A(3, p+ 1) = A(2, A(3, p)).

We know from previous example and by the Induction hypothesis that it is equalto

T (T p(1)) = T p+1(1).

17

5.2 Linear algebra

From linear algebra we are going to show an example on when you use inductionto prove a theorem.

Theorem 5.1. The determinant of an upper triangular matrix is the productof the elements on the diagonal.

Proof. Base case: It is pretty clear that it holds for a matrix of order (1× 1).Induction hypothesis: Assume that it holds for an upper triangular matrixof order (n− 1)× (n− 1)Induction step: Consider an (n× n) upper triangular matrix.

An,n =

a1,1 a1,2 a1,3 · · · a1,n

0 a2,2 a2,3 · · · a2,n0 0 a3,3 · · · a3,n...

......

. . ....

0 0 0 · · · an,n

.

By expanding column 1 we get, where D is the determinant of the minor.

det(A) = a1,1D1,1 − a2,1D2,1 + a3,1D3,1 − · · · ± an,1Dn,1

det(A) = a1,1D1,1 − 0 ·D2,1 + 0 ·D3,1 − · · · ± 0 ·Dn,1

det(A) = a1,1D1,1.

Note thatD1,1 is the determinant of an upper triangular matrix with a2,2, . . . , an,non the diagonal, of order (n−1)×(n−1). According to the Induction hypothesisD1,1 = a2,2×, · · · × an,n.

5.3 Binomial coefficients

5.3.1 Binomial

To prove some of the following examples we will need the following lemma.

Lemma 5.2. Let 1 ≤ k ≤ n, then(n

k − 1

)+

(n

k

)=

(n+ 1

k

)where (

n

k

)=

n!

k!(n− k)!.

18

Proof.(n

k − 1

)+

(n

k

)=

n!

(k − 1)!(n− k + 1)!+

n!

k!(n− k)!

=k · n!

k(k − 1)!(n− k + 1)!+

(n− k + 1) · n!

k!(n− k)!(n− k + 1)

=k · n! + (n− k + 1) · n!

(k)!(n− k + 1)!

=n!(k + (n− k + 1))

(k)!(n− k + 1)!=

n!(n+ 1)

(k)!(n− k + 1)!=

(n+ 1)!

(k)!(n− k + 1)!=

(n+ 1

k

).

5.3.2 The Binomial Theorem

Theorem 5.3 (The Binomial Theorem).

(a+ b)n =

n∑k=0

(n

k

)an−kbk

Proof. Base case: n = 1Left side:

(a+ b)1 = a+ b

Right side:1∑

k=0

(1

k

)a1−kbk = a+ b

Both of the statements are equal so the formula holds for n = 1.Induction hypothesis: Assume that the formula holds for an arbitrarypositive integer p.

(a+ b)p =

p∑k=0

(p

k

)ap−kbk

Induction step: We want to show that the formula also holds for p+ 1

(a+ b)p+1 =

p+1∑k=0

(p+ 1

k

)ap+1−kbk.

Consider the left side:

(a+ b)p+1 = (a+ b)(a+ b)p.

By the Induction hypothesis this equals,

(a+ b)

p∑k=0

(p

k

)ap−kbk =

p∑k=0

(p

k

)ap−k+1bk +

p∑k=0

(p

k

)ap−kbk+1.

19

To get what we want we put k to k − 1 in the second sum,

=

p∑k=0

(p

k

)ap+1−kbk +

p+1∑k=1

(p

k − 1

)ap−k+1bk

= ap+1 +

p∑k=1

(p

k

)ap+1−kbk +

p∑k=1

(p

k − 1

)ap+1−kbk + bp+1

= ap+1 +

p∑k=1

[(p

k

)+

(p

k − 1

)]ap+1−kbk + bp+1,

and according to lemma 5.2 this equals

ap+1 +

p∑k=1

(p+ 1

k

)ap+1−kbk + bp+1 =

p+1∑k=0

(p+ 1

k

)ap+1−kbk.

5.4 Calculus

In calculus you can prove some of the theorems with mathematical induction.Calculus is a lot of derivatives and integrals, here we are going to show someexamples taken from [1].

5.4.1 The General Power Rule

The general power rule you learn already in upper secondary school and is veryfamiliarly and often used. It is very easy to prove the theorem with mathemat-ical induction.

Theorem 5.4. If f(x) = xn, then f ′(x) = nxn−1, for all positive integers n.

Proof. Let D(f) denote the derivative of the function f .Base case: Show that it holds for n = 1

D(x1) = D(x) = 1

Induction hypothesis: Assume that for an arbitrary positive integer p, wehave

D(xp) = pxp−1

Induction step: We want to show that it also holds for p+ 1,

D(xp+1) = (p+ 1)xp.

Consider the left side:D(xp+1) = D(xp · x),

product rule gives usD(xp) · x+D(x) · xp,

20

now we use the Induction hypothesis and get

p · xp−1 · x+ xp = p · xp + xp = (p+ 1)xp.

5.4.2 The Leibniz rule

The Leibniz rule is a generalization of the product rule. We observe the simi-larities with the formulation and proof of the Binomial theorem 5.3.2, both inthe way they are defined and proved.

Theorem 5.5 (The Leibniz rule).

(fg)(n) =

n∑k=0

(n

k

)f (n−k)g(k)

Proof. Base case: n = 1Left side:

(fg)′ = f ′g + fg′

Right side:1∑

k=0

(1

k

)f (1−k)g(k) = f ′g + fg′

Both sides are equal so the formula holds for n = 1.Induction hypothesis: Assume that the Leibniz rule holds for a positiveinteger p,

(fg)(p) =

p∑k=0

(p

k

)f (p−k)g(k).

Induction step: We want to show that the theorem also holds for p+ 1

(fg)(p+1) =

p+1∑k=0

(p+ 1

k

)f (p+1−k)g(k).

We consider the left side, where we use the Induction hypothesis in the secondstep

(fg)(p+1) =d

dx

((fg)(p)

)=

d

dx

(p∑

k=0

(p

k

)f (p−k)g(k)

),

and the product rule gives us that this equalsp∑

k=0

(p

k

)f (p+1−k)g(k) +

p∑k=0

(p

k

)f (p−k)g(k+1)

=

p∑k=0

(p

k

)f (p+1−k)g(k) +

p+1∑k=1

(p

k − 1

)f (p+1−k)g(k)

=f (p+1)g0 +

p∑k=1

[(p

k

)+

(p

k − 1

)]f (p+1−k)g(k) + f0g(p+1).

21

According to Lemma 5.2 we have

f (p+1)g0 +

p∑k=1

(p+ 1

k

)f (p+1−k)g(k) + f0g(p+1)

=

p+1∑k=0

(p+ 1

k

)f (p+1−k)g(k).

5.4.3 Taylor’s theorem

Theorem 5.6 (Taylor’s Theorem). If f has n + 1 derivatives in an intervalcontaining a and x, then

f(x) = Pn(x) + En(x)

where

Pn(x) = f(a) + f ′(a)(x− a) + · · ·+ f (n)(a)

n!(x− a)n

and

En(x) =f (n+1)(X)

(n+ 1)!(x− a)n+1

for some X between a and x.

To prove Taylor’s theorem we need the Mean Value Theorem (MVT) andthe general mean value theorem (GMVT).

Theorem 5.7 (The mean value theorem). If f is continuous on the closedinterval [a, b], and differentiable on the open interval (a, b), then there exists apoint c in (a, b) such that

f(b)− f(a)

b− a= f ′(c).

Theorem 5.8 (The general mean value theorem). If f and g are continuouson [a, b], differentiable on (a, b) and g′(x) 6= 0 on (a, b), then there exists a pointc in (a, b) such that

f(b)− f(a)

g(b)− g(a)=f ′(c)

g′(c).

Proof of Taylor’s theorem. Base case: n = 0. We want to show f(x) = P0(x)+E0(x). We have P0(x) = f(a). According to MVT there is an X between a andx such that

f(x)− f(a)

x− a= f ′(X),

which gives usf(x)− f(a) = f ′(X)(x− a)

22

sof(x) = f(a) + f ′(X)(x− a) = P0(x) + E0(x).

So the formula is true for n = 0.Induction hypothesis: Assume the formula is true for k− 1. In other words,for every function h with k derivatives on an interval which contains a and xapplies

h(x) = Pk−1(x) + Ek−1(x)

where

Ek−1(x) =hk(X)

k!(x− a)k

for some X between a and x.Induction step: Now we want to show for n = k. Assume a < x. Apply theGMVT on Ek(t) and g(t) = (t − a)k+1 on [a, x], and Ek(t) = f(t) − Pk(t). Sothere is a number c between a and x such that

Ek(x)− Ek(a)

g(x)− g(a)=E′k(c)

g′(c),

soEk(x)− 0

(x− a)k+1=

E′k(c)

(k + 1)(c− a)k.

We now have that

Ek(x) =E′k(c)

(k + 1)(c− a)k(x− a)k+1. (1)

Now we want to see what E′k(c) is. We have that

Ek(t) = f(t)− Pk(t) =

= f(t)−(f(a) + f ′(a)(t− a) +

f ′′(a)

2!(t− a)2 + · · ·+ f (k)(a)

k!(t− a)k

).

We differentiate and get

E′k(t) = f ′(t)−(

0 + f ′(a) +f ′′(a)

2!2(t− a) + · · ·+ fk(a)

k!k(t− a)k−1

).

Let h(t) = f ′(t), then

E′k(t) = h(t)−(h(a) + h′(a)(t− a) + · · ·+ hk−1(a)

(k − 1)!k(t− a)k−1

)= h(t)− Pk−1(t).

We can now put t = cE′k(c) = h(c)− Pk−1(c).

23

According to the Induction hypothesis the theorem holds for h and Pk−1 whichgives us

E′k(c) =h(k)(X)

k!(c− a)k.

By (1) we have

Ek(c) =E′k(c)

(k + 1)!(c− a)k(x− a)k+1

=1

(k + 1)(c− a)kh(k)(X)

k!(c− a)k(x− a)k+1

=h(k)(X)

(k + 1)!(x− a)k+1 = [h = f ′]

=f (k+1)(X)

(k + 1)!(x− a)k+1.

6 Conclusion

We have now seen how mathematical induction is used in history, how Peano’saxiom is formulated and a lot of different examples of when we can use inductionin logic, algebra, linear algebra and calculus. There are different paradoxes withinduction if you do it in the wrong way but the Peano’s axioms is the way we candefine all the natural numbers, and has gained great impact in mathematics.

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References

[1] Robert Alexander Adams and Christopher Essex. Calculus: a completecourse. Vol. 9. Pearson Canada, Toronto, 2016.

[2] Florian Cajori. “Origin of the name “mathematical induction.””. In: Theamerican mathematical monthly 25.5 (1918), pp. 197–201.

[3] David S Gunderson. Handbook of mathematical induction: Theory and ap-plications. CRC Press, 2014.

[4] Johan Jonasson and Stefan Lemurell. Algebra och diskret matematik. Stu-dentlitteratur, 2004.

[5] Victor J Katz. A history of mathematics. Pearson/Addison-Wesley, 2004.

[6] Dirk Van Dalen. Logic and structure. Springer, 2004.

[7] Anders Vretblad and Kerstin Ekstig. Algebra och geometri. Gleerup, 2006.

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