Mathematical Induction

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

Hanna Wedin

Examensarbete i matematik, 15 hp Handledare: Inger Sigstam Examinator: Veronica Crispin Quinonez Juni 2020

Department of Mathematics Uppsala University

Abstract Mathematical Induction is used in all ﬁelds of mathematics. In this thesis we will do an overview of mathematical induction and see 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 where the name mathematical induction came from. We will also look at diﬀerent types of induction, weak and strong induction. You can also do induction on other types of structures, like the length of propositions.

1 Contents

1 Introduction 3 1.1 History ...... 4 1.2 Paradoxes with induction ...... 5

2 Mathematical Induction 5 2.1 Axiom ...... 6 2.2 Examples ...... 6 2.2.1 The sum of i ...... 6 2.2.2 The sum of i2 ...... 7

3 Strong Induction 8 3.1 Examples ...... 9 3.1.1 Example ...... 9 3.1.2 The Fibonacci sequence ...... 10 3.1.3 Fundamental Theorem of Arithmetic ...... 10 3.1.4 The sum jk ...... 10

4 Induction on other structures 12 4.1 Examples ...... 13

5 More Examples 15 5.1 The Ackermann function ...... 15 5.1.1 Example 1 ...... 15 5.1.2 Example 2 ...... 16 5.1.3 Example 3 ...... 16 5.1.4 Example 4 ...... 17 5.2 Linear algebra ...... 18 5.3 Binomial coeﬃcients ...... 18 5.3.1 Binomial ...... 18 5.3.2 The Binomial Theorem ...... 19 5.4 Calculus ...... 20 5.4.1 The General Power Rule ...... 20 5.4.2 The Leibniz rule ...... 21 5.4.3 Taylor’s theorem ...... 22

6 Conclusion 24

2 1 Introduction

Mathematical induction has a big inﬂuence in mathematics. It is a way to prove mathematical statements about natural numbers. You start learn about mathematical induction and the principle of induction in the later upper secondary school in Sweden. You also learn about induction in the university if you study mathematics. 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 random value, 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 inductive step, it is true for every value. You can think of induction in an illustrating way, think of a ladder. In the Base case we check that we have a ﬁrst step to step on. Then in the induction step we go to an arbitrary step p and then we show that if we go to p there is a step after, p + 1. So if we go to an arbitrary step on the ladder we want to show that we can lift our leg and go to the next step. Why does this work? Because if our base case is p and then we can show that we can go to the second step p + 1, if we now let the second step be p we can show that we can go to the third step, p + 1, and then we can let the third step be p, and so it goes on and therefore 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

3 base step is that the ﬁrst domino will fall, and then if any domino fall then the domino 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 there are infinitely many primes, even though there is no evidence that he used it, but some writers think that he implied it without being stated directly. Some also thinks that Plato and Pappus used induction but there are no evidence that they used it[3]. The Italian mathematician Francesco Maurolico (1494 − 1575) did a noninductive proof concerning the sum of the first n natural numbers. Even though Maurolico did a noninductive proof there are people who think that Pascal got his inspiration for the induction principle from Maurolico, when Pascal in the 16th century showed by induction what the sum of the first n natural numbers is [3]. In [3] it says that sometimes induction is compared to the method of exhaustion, like Calivieres principle where he calculates the area of a circle. There he draws polygons that fits in the circle, because he knew how to calculate the area of a polygon, and then he draws larger and larger polygons, until he has drawn a polygon which is almost in the same size as the circle. The similarities with induction and the method of exhaustion is that you start with a guess, and to prove your guess you do infinitely many iterations which follows from earlier steps. There are some proofs that are used with the method of exhaustion that can be translated into an inductive proof. There was an Egyptian called ibn al-Haytham (969-1038) who used inductive reasoning to prove the formula for

n X n 1 1 1 i4 = + n n + (n + 1)n − . 5 5 2 3 i=1

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 induction where we ﬁrst do the base case and then the induction step to show that it hold for n to n + 1, Levi started with the induction step and then he showed for 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 named this method yet. According to [2] the ﬁrst ones who started to name induction was the Englishman John Wallis (1616 − 1703) and the Swiss Jacob Bernoulli (1655 − 1705). Wallis used a kind of induction called incomplete induction to

4 Pn 2 2 find the ratio between i=1 i and n (n + 1). Wallis incomplete induction both got bad and good criticism. Bernoulli was one of the ones who gave Wallis bad criticism and he introduced the principle argument from n to n + 1. Bernoulli criticized Wallis but he also thought that even though Wallis used incomplete induction it was a start to induction. Bernoulli showed the Binomial theorem with the argument when you go from n to n + 1. Georg Simon Klügel(1739 − 1812) explained the weakness of Wallis induction in his dictionary, he also explains Bernoullis proof from n to n + 1. Then in England Thomas Simpson (1710 − 1761) used the n to n + 1, but neither did he give 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 “demonstrative induction”. In 1833 Augustus De Morgan suggested a new name “successive induction”, but in the end of his article he used “mathematical induction”. Both the names “demonstrative induction” and “mathematical induction” are used by writers. But then the “demonstrative induction” get disused. In the USA induction was used but not called by its name, in Europe, the name “mathematical induction” was used. The Italian mathematician Giuseppe Peano (1858 − 1932) formulated the axiom system we call the Peano’s axiom in 1889. With Peano’s axiom we can construct all the natural numbers, and one of his axiom is the one we call the Induction axiom. You will see the axiom in section 2.1. There are a lot more of well-known mathematicians who has used mathematical induction, if you want to read more about it you can look in the history section in [3]. There are a lot more sources where you can find more information if you are interested.

1.2 Paradoxes with induction There are some paradoxes with induction and inductive reasoning. [3] has some fun examples of diﬀerent paradoxes. One example is about a teacher who told his students that next week they will get an unprepared test. And the students think that if they don’t have had the test on Thursday night they cannot have it on Friday because then they will know that the test must be on Friday and it will not be unprepared. So the test cannot be on Friday, by the same argument the test 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 head you are bald but if you only have one hair you will still be bald, so if you only have 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 start learn in school and use to prove simpler theorems. It is easy to think that mathematical induction is something you only use in logic and algebra, but the fact is

5 that you use mathematical induction in almost all ﬁelds of mathematics. Later we 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 talk about 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, to show that the statement holds for the smallest integer. Then you do the induction hypothesis and assume that the statement holds for some arbitrary positive integer p, and if you can show that the statement holds for p + 1 you can by the principle 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 X n(n + 1) i = 2 i=1

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

1 X i = 1 i=1 and then the right side, 1(1 + 1) = 1. 2 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 hypothesis: Assume the formula holds for some arbitrary positive integer p ≥ 1,

p X p(p + 1) i = . 2 i=1 The next step is now to show that the formula also holds for the following integer p + 1, p+1 X (p + 1)(p + 2) i = . 2 i=1 We start with the left side

p+1 p X X i = i + (p + 1), i=1 i=1 Pp and by the Induction hypothesis we see that we can rewrite i=1 i and get p(p + 1) p(p + 1) + 2(p + 1) p2 + 3p + 2 (p + 1)(p + 2) + (p + 1) = = = . 2 2 2 2 We see that the formula also holds for p+1. By the Induction axiom the formula holds for all integers n ≥ 1.

2.2.2 The sum of i2 n X n(n + 1)(2n + 1) i2 = 6 i=1 Proof. Base case: For n = 1 we have

n X i2 = 1 i=1 and 1(1 + 1)(2 + 1) 6 = = 1. 6 6

7 We see that both left side and right side are equal to 1, so the formula holds for n = 1. Induction hypothesis: Suppose it holds for some arbitrary positive integer p, i.e. p X p(p + 1)(2p + 1) i2 = . 6 i=1 Induction step: Now we want to show that the formula also holds for p + 1, i.e. p+1 X (p + 1)(p + 2)(2p + 3) i2 = . 6 i=1 We start with the left side

p+1 p X X i2 = i2 + (p + 1)2, i=1 i=1 and using the Induction hypothesis, we get

p(p + 1)(2p + 1) (p + 1)(p(2p + 1) + 6(p + 1)) + (p + 1)2 = = 6 6 (p + 1)(2p2 + 7p + 6) (p + 1)(p + 2)(2p + 3) = . 6 6 We see that the formula also holds for p + 1. By the Induction axiom the formula holds for all positive integers n.

3 Strong Induction

The first type of induction described in section 2 is sometimes called weak induction, in this section we will describe the type of induction called strong induction. Although this type is called strong induction doesn’t it mean that it actually is stronger then the other type. Strong induction is also called complete induction. This type of induction is like the first type of induction but here you have a different induction hypothesis. In weak induction you assume in the induction hypothesis that the formula/statement holds for some positive integer p, but in strong induction you assume that the formula/statement holds for all positive integer less than or equal to p. The two types of induction are 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.

8 Then all statements are true.

3.1 Examples 3.1.1 Example An easy example where we use strong induction instead of weak induction is taken from [4].

Let h : N −→ Q be deﬁned as follows

h(0) = 1,  h(1) = 1  h(n−1) 1 h(n) = 2 + 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 show that it holds for n = 0 and n = 1. We have by deﬁnition that h(0) = 1 and h(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 have that p + 1 > 1, so h(p) 1 h(p + 1) = + . 2 h(p − 1) The Induction hypothesis gives us

1 h(p) 1 1 ≤ < 1 and < ≤ 1, 2 2 2 h(p − 1) so h(p) 1 1 1 h(p + 1) = + > + = 1, 2 h(p − 1) 2 2 and h(p) 1 h(p + 1) = + < 1 + 1 = 2. 2 h(p − 1) We now have that 1 ≤ h(p + 1) < 2. According to the Induction axiom it holds for all positive integers n.

9 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 easily be 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 part of the Fundamental theorem of arithmetic. Theorem (Fundamental Theorem of Arithmetic). Every positive integer greater than 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 the statement 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 + 1 is not a prime number. Then p + 1 can be written as p + 1 = a · b, where a, b ∈ {2, 3, . . . , p}. By the Induction hypothesis a, b can be written as products of prime numbers, and therefore a · b also can be written as a product of prime numbers. 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 Pn Pn 2 j=1 i and j=1 i from 2.2.1 and 2.2.2. The theorem can also be used when

10 R a k you want to calculate 0 x dx. Then you use the Riemann sum and you get

n k Z a a X aj xk = lim n→∞ n n 0 j=1 n 1 X = ak+1 lim jk n→∞ nk+1 j=1 1 1 P (n) = ak+1 lim + + k−1 n→∞ k + 1 2n nk+1 ak+1 = . k + 1

Theorem. Let n ∈ N. Then n X nk+1 nk jk = + + P (n) k + 1 2 k−1 j=1 for every positive integer, where Pk−1(n) is some polynomial in n of degree ≤ k − 1. For every positive integer k.

Pn 2 If we look at i=1 i and we know that it can be written as

n X n(n + 1)(2n + 1) i2 = , 6 i=1 and now we can rewrite it n(n + 1)(2n + 1) 2n3 + 3n2 + n n3 n2 = = + + P (n). 6 6 3 2 2 Proof. Base case: k = 1

n X n(n + 1) n2 + n n1+1 n j1 = = = + + P (n). 2 2 1 + 1 2 0 j=1

Induction hypothesis:

n X nk+1 nk jk = + + P (n) k + 1 2 k−1 j=1 for k = 1, 2, . . . , m. Induction step: We want to show for m + 1

n X nm+2 nm+1 jm+1 = + + P (n). m + 2 2 m j=1

11 To get where we want we use the Binomial Theorem which gives us (m + 2)(m + 1) (j + 1)m+2 = jm+2 + (m + 2)jm+1 + jm + ... + 1, 2 thus (m + 2)(m + 1) (j + 1)m+2 − jm+2 = (m + 2)jm+1 + jm + ... + 1 2 for j = 1, 2, 3, ..., n. Thus by taking sums we get

n n n n X X (m + 2)(m + 1) X X ((j+1)m+2−jm+2) = (m+2) (jm+1)+ (jm)+...+ 1. 2 j=1 j=1 j=1 j=1 Left side telescopes and gives us

n n n X (m + 2)(m + 1) X X (n + 1)m+2 − 1 = (m + 2) (jm+1) + (jm) + ... + 1. 2 j=1 j=1 j=1

Pn m We expand the left side and use the Induction hypothesis on the term j=1 j on the right side, we get (m + 2)(m + 1) nm+2 + (m + 2)nm+1 + nm + ... + 1 − 1 = 2 n n X (m + 2)(m + 1) nm+1 nm X (m + 2) (jm+1) + + + P (n) + ... + 1. 2 m + 1 2 m−1 j=1 j=1

Pn m+1 We want j=1 j alone

n X nm+2 nm+1 jm+1 = + nm+1 − + P (n) m + 2 2 m j=1 which gives n X nm+2 nm+1 jm+1 = + + P (n). m + 2 2 m j=1

4 Induction on other structures

In the examples of induction above we only have talked about mathematical induction on positive integers, but you can do mathematical induction on other structures. The principle is the same as when you do on positive integers. We look at the principle by an example from logic. We will start to deﬁne a language for propositional logic, in [6] they deﬁne the language like this.

12 Deﬁnition 4.1. The language of propositional logic has an alphabet consisting of

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 there are 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 as usual try to show that the statement is true for propositions with length 1. In the induction hypothesis we will assume that the statement holds for propositions with length 1, 2, 3, . . . , p. Last but not least we will in the induction step show that it holds for propositions with length p + 1. As you can see it is the same principle 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 is equal to the number of right parenthesis. Proof. We call the number of left parenthesis for L(ϕ) and right parenthesis for R(ϕ). Base case: We ﬁrst want to show for a formula ϕ with length n = 1. ϕ = pi or ϕ = ⊥, V (ϕ) = H(ϕ) = 0. Induction hypothesis: We assume that it is true for all formulas of length n = 1, 2, 3, . . . , p. Induction step: Now we want to show that it also is true for formulas with length p + 1. Let ϕ be a formula with length p + 1. ϕ must contain at least one connective 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 all these forms. To make it easier we call ∧, ∨, →, ↔ for . We start with ϕ = (ψδ). ψ and δ must have length ≤ p. We know from the Induction hypothesis that L(ψ) = R(ψ) = l and L(δ) = R(δ) = j. The

13 formation of the string (ψδ) adds one more left parenthesis and one more right parenthesis, which gives us

L((ψδ)) = k + j + 1 = R((ψδ)).

So it is true for (ψδ). Now we want to show for ϕ = (¬ψ). ψ has the length ≤ p and we know from the Induction hypothesis that L(ψ) = R(ψ). The formation of (¬ψ) does add one more left parenthesis and one more right parenthesis which gives 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 can according to the Induction axiom say that it is true for all propositions with length n ≥ 1. Deﬁnition 4.3. The rank r(ϕ) of a proposition ϕ is deﬁned by  r(ϕ) = 0 for atomic ϕ,  r((ϕψ)) = max(r(ϕ), r(ψ)) + 1, r((¬ϕ)) = r(ϕ) + 1

Proposition 2. Let ϕ ∈ PROP and let C(ϕ) denote the number of occurrences of connectives in ϕ. Then r(ϕ) ≤ C(ϕ). Proof. Base case: We start to show for ϕ with length n = 1. Then ϕ is atomic and 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 with length p + 1. We want to show for (ψδ) and (¬ψ), where ψ and δ are formulas with length ≤ p. We start with (ψδ). We know that ψ and δ must have length ≤ p, so by the Induction hypothesis we know that r(ψ) ≤ C(ψ) and r(δ) ≤ C(δ). We have that r((ψδ)) = max(r(ψ), r(δ)) + 1 C((ψδ)) = C(ψ) + C(δ) + 1. Case 1: r(ψ) ≥ r(δ). Then we have that

r((ψδ)) = r(ψ) + 1 According to the Induction hypothesis we have ≤ C(ψ) + 1 ≤ C(ψ) + C(δ) + 1 = C((ψδ)). Case 2: r(δ) > r(ψ). Then we have that

r((ψδ)) = r(δ) + 1 According to the Induction hypothesis we have ≤ C(δ) + 1 ≤ C(ψ) + C(δ) + 1 = C((ψδ)).

14 So it is true for (ψδ). Now we want to show for (¬ψ), where ψ is a formula with length ≤ p. From the Induction hypothesis we have that r(ψ) ≤ C(ψ). We also 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 Induction axiom 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 recursive theory. It is a total computable function that is not primitive recursive. It is named after the German Wilhelm Ackermann (1896 − 1962). We will see that the function increases very quickly. The Ackermann function has some fun properties that we are going to prove with induction. Deﬁnition 5.1. 2n, if m = 0  0, if m ≥ 1, n = 0 A(m, n) = 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 = 1 A(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 equal to 2.

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

2 22 A(2, 4) = A(1,A(2, 3)) = A(1, 22 ) = 22

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

.2 .. T (n) = 22 , 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 T n(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 make a guess. A(3, n) = T n(1). We see that the result depends on the result from the previous number, and therefore it is easier to use strong induction.

Proof. Base case: n = 1, we have already seen that it holds for n = 1 cause A(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 equal to 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 induction to prove a theorem. Theorem 5.1. The determinant of an upper triangular matrix is the product of 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 matrix of order (n − 1) × (n − 1) Induction step: Consider an (n × n) upper triangular matrix.   a1,1 a1,2 a1,3 ··· a1,n  0 a2,2 a2,3 ··· a2,n     0 0 a3,3 ··· a3,n  An,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 that D1,1 is the determinant of an upper triangular matrix with a2,2, . . . , an,n on the diagonal, of order (n−1)×(n−1). According to the Induction hypothesis D1,1 = 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 n n + 1 + = k − 1 k k where n n! = . k k!(n − k)!

18 Proof. n n n! n! + = + k − 1 k (k − 1)!(n − k + 1)! k!(n − k)! k · n! (n − k + 1) · n! = + k(k − 1)!(n − k + 1)! k!(n − k)!(n − k + 1) k · n! + (n − k + 1) · n! = (k)!(n − k + 1)! n!(k + (n − k + 1)) n!(n + 1) (n + 1)! n + 1 = = = = . (k)!(n − k + 1)! (k)!(n − k + 1)! (k)!(n − k + 1)! k

5.3.2 The Binomial Theorem Theorem 5.3 (The Binomial Theorem).

n X n (a + b)n = an−kbk k k=0 Proof. Base case: n = 1 Left side: (a + b)1 = a + b Right side: 1 X 1 a1−kbk = a + b k k=0 Both of the statements are equal so the formula holds for n = 1. Induction hypothesis: Assume that the formula holds for an arbitrary positive integer p. p X p (a + b)p = ap−kbk k k=0 Induction step: We want to show that the formula also holds for p + 1

p+1 X p + 1 (a + b)p+1 = ap+1−kbk. k k=0 Consider the left side:

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

By the Induction hypothesis this equals,

p p p X p X p X p (a + b) ap−kbk = ap−k+1bk + ap−kbk+1. k k k k=0 k=0 k=0

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

p p+1 X p X p = ap+1−kbk + ap−k+1bk k k − 1 k=0 k=1 p p X p X p = ap+1 + ap+1−kbk + ap+1−kbk + bp+1 k k − 1 k=1 k=1 p X p p = ap+1 + + ap+1−kbk + bp+1, k k − 1 k=1 and according to lemma 5.2 this equals

p p+1 X p + 1 X p + 1 ap+1 + ap+1−kbk + bp+1 = ap+1−kbk. k k k=1 k=0

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 some examples taken from [1].

5.4.1 The General Power Rule The general power rule you learn already in upper secondary school and is very familiarly and often used. It is very easy to prove the theorem with mathematical induction. Theorem 5.4. If f(x) = xn, then f 0(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, we have 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 us D(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 similarities with the formulation and proof of the Binomial theorem 5.3.2, both in the way they are deﬁned and proved. Theorem 5.5 (The Leibniz rule). n X n (fg)(n) = f (n−k)g(k) k k=0 Proof. Base case: n = 1 Left side: (fg)0 = f 0g + fg0 Right side: 1 X 1 f (1−k)g(k) = f 0g + fg0 k k=0 Both sides are equal so the formula holds for n = 1. Induction hypothesis: Assume that the Leibniz rule holds for a positive integer p, p X p (fg)(p) = f (p−k)g(k). k k=0 Induction step: We want to show that the theorem also holds for p + 1 p+1 X p + 1 (fg)(p+1) = f (p+1−k)g(k). k k=0 We consider the left side, where we use the Induction hypothesis in the second step p ! d d X p (fg)(p+1) = (fg)(p) = f (p−k)g(k) , dx dx k k=0 and the product rule gives us that this equals p p X p X p f (p+1−k)g(k) + f (p−k)g(k+1) k k k=0 k=0 p p+1 X p X p = f (p+1−k)g(k) + f (p+1−k)g(k) k k − 1 k=0 k=1 p X p p =f (p+1)g0 + + f (p+1−k)g(k) + f 0g(p+1). k k − 1 k=1

21 According to Lemma 5.2 we have

p X p + 1 f (p+1)g0 + f (p+1−k)g(k) + f 0g(p+1) k k=1 p+1 X p + 1 = f (p+1−k)g(k). k k=0

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

f(x) = Pn(x) + En(x) where f (n)(a) P (x) = f(a) + f 0(a)(x − a) + ··· + (x − a)n n n! and f (n+1)(X) E (x) = (x − a)n+1 n (n + 1)! for some X between a and x. To prove Taylor’s theorem we need the Mean Value Theorem (MVT) and the general mean value theorem (GMVT). Theorem 5.7 (The mean value theorem). If f is continuous on the closed interval [a, b], and diﬀerentiable on the open interval (a, b), then there exists a point c in (a, b) such that

f(b) − f(a) = f 0(c). b − a Theorem 5.8 (The general mean value theorem). If f and g are continuous on [a, b], diﬀerentiable on (a, b) and g0(x) 6= 0 on (a, b), then there exists a point c in (a, b) such that f(b) − f(a) f 0(c) = . g(b) − g(a) g0(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 and x such that f(x) − f(a) = f 0(X), x − a which gives us f(x) − f(a) = f 0(X)(x − a)

22 so 0 f(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 x applies h(x) = Pk−1(x) + Ek−1(x) where hk(X) E (x) = (x − a)k k−1 k! for some X between a and x. Induction step: Now we want to show for n = k. Assume a < x. Apply the k+1 GMVT on Ek(t) and g(t) = (t − a) on [a, x], and Ek(t) = f(t) − Pk(t). So there is a number c between a and x such that E (x) − E (a) E0 (c) k k = k , g(x) − g(a) g0(c) so E (x) − 0 E0 (c) k = k . (x − a)k+1 (k + 1)(c − a)k We now have that E0 (c) E (x) = k (x − a)k+1. (1) k (k + 1)(c − a)k

0 Now we want to see what Ek(c) is. We have that

Ek(t) = f(t) − Pk(t) = f 00(a) f (k)(a) = f(t) − f(a) + f 0(a)(t − a) + (t − a)2 + ··· + (t − a)k . 2! k!

We diﬀerentiate and get

f 00(a) f k(a) E0 (t) = f 0(t) − 0 + f 0(a) + 2(t − a) + ··· + k(t − a)k−1 . k 2! k!

Let h(t) = f 0(t), then

hk−1(a) E0 (t) = h(t) − h(a) + h0(a)(t − a) + ··· + k(t − a)k−1 k (k − 1)!

= h(t) − Pk−1(t).

We can now put t = c 0 Ek(c) = h(c) − Pk−1(c).

23 According to the Induction hypothesis the theorem holds for h and Pk−1 which gives us h(k)(X) E0 (c) = (c − a)k. k k! By (1) we have

E0 (c) E (c) = k (x − a)k+1 k (k + 1)!(c − a)k 1 h(k)(X) = (c − a)k(x − a)k+1 (k + 1)(c − a)k k! h(k)(X) = (x − a)k+1 = [h = f 0] (k + 1)! f (k+1)(X) = (x − a)k+1. (k + 1)!

6 Conclusion

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

24 References

[1] Robert Alexander Adams and Christopher Essex. Calculus: a complete course. Vol. 9. Pearson Canada, Toronto, 2016. [2] Florian Cajori. “Origin of the name “mathematical induction.””. In: The american mathematical monthly 25.5 (1918), pp. 197–201. [3] David S Gunderson. Handbook of mathematical induction: Theory and applications. 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.