Chapter 2. The Principle of Mathematical Induction Section 1. Introductory Investigations In this chapter, we will be making conjectures concerning the positive integers, and in order to make our conjectures, we will usually look at a number of examples to see if a specific pattern emerges. We will need a method of proof to verify that our hunches are correct. The method we will use will rely on the Principle of Mathematical Induction. Example 1. Recall that for each positive integer n, n! = 1⋅2 ⋅3!n . So n! is the product of the first n positive integers. We will attempt to find a formula for

1⋅(1!)+ 2 ⋅(2!)+!+ n ⋅(n!) where n is an arbitrary positive integer. To do this, we will first compute 1!, 2!, 3!, 4!, and 5!, and then we will compute the sums 1.(1!), 1.(1!) + 2.(2!), 1.(1!) + 2.(2!) + 3.(3!), 1.(1!) + 2.(2!) + 3.(3!) + 4.(4!), and 1.(1!) + 2.(2!) + 3.(3!) + 4.(4!) + 5.(5!) to see if we are able to recognize a pattern. For simplicity of notation, for each positive integer n, we’ll define Sn by

Sn = 1⋅(1!)+ 2 ⋅(2!)+!+ n ⋅(n!) . (We used the letter S and the subscript n to remind ourselves that we are computing the sum of the first n terms.) Let's compile our results in a table.

n n! n ⋅(n!) Sn 1 1 1 1 2 2 4 5 3 6 18 23 4 24 96 119 5 120 600 719

If we compare Columns 2 and 4, we see that each entry in Column 4 is 1 less than the next entry in Column 2. So a reasonable conjecture would be that for each positive integer n, Sn = (n +1)!−1 . Later, we will determine if our formula is correct. Example 2. Let's see if we can make any conjectures concerning the numbers n2 + n + 41 , where n runs over the positive integers. As before, our strategy will be to begin by computing n2 + n + 41 for several values of n to see if we can detect any interesting common properties.

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n n2 + n + 41 1 43 2 47 3 53 4 61 5 71 6 83

All of the numbers appearing in the second column are primes. (Recall that an integer p is a prime if p > 1 and the only positive divisors of p are 1 and itself.) So it seems reasonable to conjecture that for all positive integers n, the expression n2 + n + 41 is a prime.

Remarks. Just recognizing a pattern isn't enough. In fact, one of the conjectures we made above is true and the other one is false. We will discuss both examples further in the next section.

Example 3. We would like to find a general formula for the sum of the first n odd positive integers (where n is a positive integer). To see if we can identify a pattern, we'll start by computing the sums 1, 1 + 3, 1 + 3 + 5, 1 + 3 + 5 + 7, and 1 + 3 + 5 + 7 + 9. In our discussion below we will use the fact that the nth odd positive integer is 2n− 1. (Note: we know that 2n− 1 is odd since 2n− 1 = 2(n− 1) + 1.) Once again, let's look at a table.

n 1+ 3+ 5 +!+ (2n −1) 1 1 2 4 3 9 4 16 5 25

For each positive integer n, can we find a “simpler” formula for

1+ 3+ 5 +!+ (2n −1) where we do not need to list all of the n terms 1, 3, 5,…, 2n −1 each time? In the above computations, we see that when n = 1, the sum is 1= 12 , when n = 2, the sum is 4 = 22 , when n = 3, the sum is 9 = 32 , when n = 4, the sum is 16 = 42 , and when

, the sum is 25 = 52 . So it seems reasonable to conjecture that for all positive integers n, 1+ 3+ 5 +!+ (2n −1) = n2 . How would we prove our conjecture? Here’s one strategy that is a good one to try on this problem (and on others).

n = 5

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For each positive integer n, let's define Sn by

Sn = 1+ 3+ 5 +!+ (2n −1) .

If we put in a few more terms near the end of the expression for the sum, we see that

Sn = 1+ 3+ 5 +!+ (2n −1) = 1 + 3 + 5 + !+ (2n − 5)+ (2n − 3)+ (2n −1). Let's pair 1 with (2n− 1), 3 with (2n− 3), 5 with (2n − 5) , etc. For example, when n = 6, S6 = 1+ 3+ 5 + 7 + 9 +11 and we will group 1 with 11, 3 with 9, and 5 with 7. Notice that in this specific example, the sum of the elements in each pair is 12. In the general case, each time we pair up the elements in this fashion, and add the elements within each pair, the sum is 2n. How many pairs do we have? The answer depends upon whether n is even or odd. If n is even, then we have n2 pairs, each of

which sums to 2n. So, when n is even, Sn = n2 (2n) = n2 . But what happens when n is

odd? Do we still have a sum of n2 ? For example, when n = 7, our strategy tells us to pair 1 with 13, 3 with 11, and 5 with 9:

S7 = 1+ 3+5+ 7 + 9+11+13= (1+13)+ (3+11)+ (5+ 9)+ 7.

This time the term in the "middle" of the expression for S7 has no mate. What is the middle term and how many remaining pairs do we have? In the case where n = 7 (and in the general case where n is odd), we can determine the number of pairs we have and a formula for the middle term. However, to do that is more complicated than in the case where n is even. But this line of reasoning suggests a way to proceed. If we could guarantee that each term had a mate, we could more easily compute the sum. Notice that one way of ensuring that every term has a mate is to compute Sn + Sn (instead of Sn ) where we use in the proof below the fact that

Sn = 1+ 3+5+!+ (2n−5)+ (2n− 3)+ (2n−1)= (2n−1)+ (2n− 3)+ (2n−5)+!+5+ 3+1= (2n−1)+ (2n− 3)+ (2n−5)+!+ 2n− (2n−5)( ) + 2n− (2n− 3)( ) + 2n− (2n−1)( ).

So the odd integers between 1 and 2n−1 are all of the form 2n− k where k runs through the odd integers between 1 and 2n−1 as well. In our proof below, we will pair each odd integer k with its “mate” 2n− k and notice that the sum in each pair is 2n. Let's now write our proof.

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Proof. For each positive integer n, let Sn = 1+ 3+5+!+ (2n−5)+ (2n− 3)+ (2n−1) . We first note that Sn may also be written as follows:

Sn = 1+ 3+5+!+ (2n−5)+ (2n− 3)+ (2n−1)= (2n−1)+ (2n− 3)+ (2n−5)+!+5+ 3+1= (2n−1)+ (2n− 3)+ (2n−5)+!+ 2n− (2n−5)( ) + 2n− (2n− 3)( ) + 2n− (2n−1)( ).

So each odd integer k between 1 and 2n−1 has precisely one “mate ” of the form 2n− k , and the sum of k and its “mate” is always 2n. Therefore,

2Sn = Sn + Sn

= 1 + 3 + 5 + ! + (2n − 5)+ (2n − 3)+ (2n −1)↓ ↓ ↓ ↓ ↓ ↓

+ (2n −1)+ (2n − 3) + (2n − 5)+ ! + 5 + 3 + 1

= 1+ (2n −1)( ) + 3+ (2n − 3)( ) +!+ (2n − 3)+ 3( ) + (2n −1)+1( )

= (2n)+ (2n)+ (2n)+!+ (2n)= n ⋅(2n) = 2n2 . So 2Sn = 2n2 and therefore Sn = n2 .

u

Remarks. 1. Our proof above is somewhat informal with our use of the word “mate.” In the next section we will give a more “formal” proof. 2. Throughout the rest of this chapter, we will be asking you to make conjectures and to find “simpler” ways of expressing a sum or product, similar to the formula Sn = n2 we found in Example 3. (Authors often call that formula a closed-form formula.)

Exercise 1. Compute the sums 1, 1 + 2, 1 + 2 + 3, 1 + 2 + 3 + 4, and 1 + 2 + 3 + 4 + 5. For each positive integer n, let Sn = 1+ 2 +!+ n . Can you find a formula that relates

Sn+1 to Sn ? Find a “simpler” form for Sn in terms of n where we do not list all of the terms 1, 2, 3, …, n. Exercise 2. Compute the sums 1 + 2, 1 + 2 + 22 , 1 + 2 + 22 + 23 , and 1 + 2 + 22 + 23 + 24 . For each positive integer n, let Sn = 1+ 2 + 2

2 + 23 +!+ 2n . Can you find a formula that relates Sn+1 to Sn ? Find a “simpler” form for Sn in terms of n where we do not list all of the terms 1, 2, 22 , …, 2n .

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Section 2. The Principle of Mathematical Induction In this section we would like to develop a systematic way of proving our conjectures from Section 1. In that section we defined a statement pn for each positive integer n and we wanted to prove that our statements were true for all n. In a given problem, suppose we also know that

(i) p1 is true, and (ii) pm+1 is true whenever pm is true.

If we apply (ii) with m = 1, then we know that p2 is true since we also know that p1 is true by Property (i). Now if we apply (ii) this time with m = 2, we may deduce that p3 is true. Similarly we may deduce that p4 and p5 are true. So it seems reasonable to conjecture that pn is true for all positive integers n. That is, in fact, the assertion of the Principle of Mathematical Induction. Principle of Mathematical Induction. Suppose that for each positive integer n we have a statement pn . Suppose further that

(i) p1 is true, and (ii) for all m in N, pm+1 is true whenever pm is true.

Then pn is true for all positive integers n.

Remarks. 1. We will not prove that the Principle of Mathematical Induction is true. Indeed, the Principle of Mathematical Induction is one of the defining axioms of the set N of natural numbers. (Recall from Chapter 0 (Preliminaries) that the term “natural number” is just another name for “positive integer.”) 2. Note that some books use the following equivalent formulation of the Principle of Mathematical Induction as their starting point:

Let S be a subset of the set N of natural numbers such that

(i) 1∈S , and (ii) for all m in N, m+1∈S whenever m∈S .

Then S = N.

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How can we use the Principle of Mathematical Induction to prove our mathematical statements? Our strategy will be to first identify the statements pn that we wish to prove. We will then try to prove that p1 is true, and that pm+1 is true whenever pm is true. If we can prove both of those statements, the Principle of Mathematical Induction will allow us to conclude that pn is true for all positive integers n. The proof that p1 is true is often called the initial step or base step. The proof that pm+1 is true whenever pm is true is called the inductive step. Usually our argument in the inductive step will start off with the assumption that m is a natural number for which pm is true, an assumption known as the inductive assumption or inductive hypothesis. We will then use the information provided by that assumption to deduce that pm+1 must be true.

Remarks. It is important to note that in the inductive step, we are not assuming that pm is true for all m. (If we did that, we would be assuming what we are trying to prove!) Rather, in that step, our strategy will be to prove that pm+1 is true whenever we have an m for which pm is true.

Let's now use this technique on some of the conjectures we made in Section 1. Note that in several of the problems that follow, we will begin with a discussion to illustrate the background work we did when thinking through the problem. We will then follow the discussion with a formal explanation or proof. (We note that most formal proofs you will see in advanced textbooks will not have discussion sections preceding the proofs.) Example 4. In Example 1 of Section 1, for each positive integer n, we defined Sn by

Sn = 1⋅(1!)+ 2 ⋅(2!)+!+ n ⋅(n!) . We then conjectured that Sn = (n +1)!−1 for all n. Discussion. If we want to prove our conjecture using the Principle of Mathematical Induction, then for each positive integer n, we should let pn be the statement Sn = (n +1)!−1 . So p1 is the statement 1(1!) = (1+1)! − 1, pm is the statement Sm = (m +1)!−1 , and pm+1 is the statement Sm+1 = (m +1)+1( )! −1 . In the initial step, we will use the fact that 1! = 1 and 2! = 2 to prove that p1 is true. In the inductive step, we will start by assuming that for some fixed m, pm is true. In other words, we will assume that Sm = (m +1)!−1 for some fixed m. We will then use that information to derive the fact that Sm+1 = (m +1)+1( )! −1. Proof. For each positive integer n, let pn be the statement:

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Sn = (n +1)!−1 . We see that p1 is true since 1(1!) = 1= 2−1= 2!−1= (1+1)!−1. Assume that for some fixed m, pm is true. Then Sm = (m +1)!−1 by assumption. By definition,

Sm+1 = 1(1!)+ 2(2!)++ (m +1)(m +1)!= 1(1!)+ 2(2!)++m(m!)( ) + (m +1)(m +1)!.

Since Sm = 1⋅(1!)+ 2 ⋅(2!)++m ⋅(m!) , we see that

Sm+1 = Sm + (m +1)(m +1)!. By our inductive assumption, Sm = (m +1)!−1 . Therefore,

Sm+1 = Sm + (m +1)(m +1)!= (m +1)! −1[ ] + (m +1)(m +1)!

= 1⋅(m +1)!+ (m +1)(m +1)![ ]−1= 1+ (m +1)[ ](m +1)! −1= (m + 2)(m +1)! −1.

Since (m + 1)! is the product of the first m + 1 positive integers, (m + 2)(m + 1)! is the product of the first m + 2 positive integers. Therefore (m + 2)(m + 1)! = (m + 2)! and consequently,

Sm+1 = Sm + (m +1)(m +1)!= (m + 2)(m +1)! −1= (m + 2)! −1= (m +1)+1( )! −1.

We have proven that p1 is true, and that pm+1 is true whenever pm is true. The Principle of Mathematical Induction tells us that pn is true for all positive integers n and so Sn = (n +1)!−1 whenever n is a positive integer.

u

Remarks. In problems using the Principle of Mathematical Induction, a common mistake that students make in the inductive step is to assume that pm+1 is true and to then use that fact to deduce a statement that we know is true. In the inductive

8

step our approach will be to assume that pm , not pm+1 , is true and to then deduce that pm+1 must be true.

Example 5. In Section 1 we conjectured that for each positive integer n, n2 + n + 41 is a prime. How would we prove that assertion using the Principle of Mathematical Induction? Discussion. For each positive integer n, let pn be the statement n2 + n + 41 is a prime. So pm+1 asserts that (m +1)2 + (m +1)+ 41 is prime. In our previous discussion, we already noticed that p1 is true. Let's consider the inductive step where we assume that pm is true and prove that pm+1 is true. Suppose we know that m2 +m + 41 is a prime.

What can we conclude about (m +1)2 + (m +1)+ 41? We know that

(m +1)2 + (m +1)+ 41= m2 + 2m +1+m +1+ 41= (m2 +m + 41)+ (2m + 2).

The first term, m2 +m + 41 , in the above sum is a prime by assumption. But to compute (m +1)2 + (m +1)+ 41 , we have added the prime m2 +m + 41 to the expression 2m + 2. Does that allow us to conclude that (m +1)2 + (m +1)+ 41 is a prime? It is not clear how to get the desired result from our information that m2 +m + 41 is prime. But that doesn't mean our conjecture is false, only that we do not know how to give a proof of the result using the Principle of Mathematical Induction. Let's stop and reexamine the original conjecture. Is it possibly false? In other words, is it possible to find a positive integer n such that n2 + n + 41 can be factored in a nontrivial way? Now, the term 41 can be factored as 1⋅41 . So if each of the other two terms has a factor of 41, then we should be able to find a positive integer n for which n2 + n + 41 is not a prime. Let's ask ourselves the following question: Can we choose n so that both n2 and n have the common factor 41? One way of doing that is to choose n to be 41 itself. With that choice of n,

n2 + n + 41= 412 + 41+ 41= 41(41+1+1)= 41⋅43.

Therefore when n = 41, n2 + n + 41 is not a prime. So our conjecture is indeed false. Exercise 3. Find another integer (other than 41) that we can use to show that the conjecture in Example 5 is false.

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Example 6. We've already given a proof that the sum of the first n odd positive integers is n2 . Let's give a proof of that result using the Principle of Mathematical Induction. Discussion. Once again, for each positive integer n, let’s define

Sn = 1+ 3+ 5 +!+ (2n −1) . In our proof, we will let pn be the statement Sn = n

2 . So p1 is the statement S1 = 12 , pm

is the statement Sm = m2 , and pm+1 is the statement Sm+1 = (m +1)2 . In the inductive step, we will begin by assuming that Sm = m2 is true and we will then use that information to deduce that Sm+1 = (m +1)2 is true. Proof. For each positive integer n, define

Sn = 1+ 3+ 5 +!+ (2n −1) , and let pn be the statement:

Sn = n2 .

Notice that p1 is true since S1 = 1= 1

2 . Now assume that pm is true for a fixed positive integer m. In other words, assume that Sm = m2 for a fixed positive integer m. By definition,

Sm+1 = 1+ 3+ 5 ++ 2(m +1)−1( ) , and so

Sm+1 = 1+ 3+ 5 ++ (2m −1)( ) + 2(m +1)−1( ) . Since Sm = 1+ 3+ 5 ++ (2m −1) , we have that

Sm+1 = Sm + 2(m +1)−1( ). Our inductive assumption says that Sm = m2 . Consequently,

Sm+1 = Sm + 2(m +1)−1( )= m2 + 2(m +1)−1( )= m2 + 2m +1= (m +1)2.

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Therefore pm+1 is true whenever pm is true. The Principle of Mathematical Induction then tells us that pn is true for all positive integers n, that is, Sn = n

2 for all positive integers n.

u Example 7. In Investigation 6 at the end of this chapter, we will examine the sequence an n=1

∞ defined as follows:

Let a1 = 1, and for each positive integer n, let an+1 = 3an +1 . So

a1 = 1,

a2 = 3a1 +1 = 3⋅1+1 = 4 = 2 ,

a3 = 3a2 +1 = 3⋅2 +1 = 7 ≈ 2.65 ,

a4 = 3a3 +1 = 3⋅ 7 +1 ≈ 2.99 , etc. Let's conjecture that each value is at most 4 and see if we can prove that conjecture using the Principle of Mathematical Induction. In our inductive step we will assume that am ≤ 4 and we will use that information to prove that am+1 ≤ 4 . Proof. For each positive integer n, let pn be the statement:

an ≤ 4 . Then p1 is true since a1 = 1≤ 4 . Assume that pm is true, that is, assume that am ≤ 4 , for some positive integer m. Since am ≤ 4 , we have that 3am +1≤ 3⋅4 +1 and so

am+1 = 3am +1

≤ 3⋅4 +1

= 13≤ 4.

Hence pm+1 is true whenever pm is true. By the Principle of Mathematical Induction, pn is true for all positive integers n. In other words, an ≤ 4 whenever n is a positive integer.

u In number theory we are frequently interested in questions concerning divisibility. The Principle of Mathematical Induction is often helpful in such settings.

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Recall that a nonzero integer n is said to divide an integer m (or m is a multiple of n) if there exists an integer c such that m = nc.

Notice again that we have stated the given property in terms of multiplication and not division! Example 8. Prove that for each positive integer n, 3 divides n3 + 5n + 6 . Proof. For each positive integer n, let pn be the statement:

3 divides n3 + 5n + 6 . We see that p1 is true since 13 + 5 ⋅1+ 6 = 12 = 3⋅4 and 4 is an integer. Now assume that pm is true, that is, assume that 3 divides m3 + 5m + 6 , for some positive integer m. Then

there exists an integer c such that m3 + 5m + 6 = 3c . (Let's use that assumption to prove that pm+1 is true, where pm+1 is the statement 3 divides (m +1)3 + 5(m +1)+ 6 .) Now

(m +1)3 + 5(m +1)+ 6 = m3 + 3m2 + 3m +1+ 5m + 5 + 6= (m3 + 5m + 6)+ (3m2 + 3m + 6)= 3c + 3(m2 +m + 2)= 3(c +m2 +m + 2).

Since c and m are integers, c +m2 +m + 2 is an integer as well and hence (m +1)3 + 5(m +1)+ 6 is a multiple of 3. We have shown that pm+1 is true whenever pm is true. Therefore, by the Principle of Mathematical Induction, pn is true for all positive integers n. In other words, for each positive integer n, 3 divides n3 + 5n + 6 .

u

Remark. In the above proof, we wanted to use the information we had about the expression m3 + 5m + 6 , and so we did not simplify the expression (m +1)3 + 5(m +1)+ 6 by grouping like terms.

Exercise 4. Is it true that 3 divides n3 + 5n + 6 for each integer n where n ≥ −2 ? How would you prove or disprove your conjecture? Example 9. Let’s prove that for each positive integer n, 6 divides n3 − n . Discussion. For each positive integer n, let pn be the statement 6 divides n3 − n . Then pm+1 is the statement (m +1)3 − (m +1) is divisible by 6. Let's first look at the inductive

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step where we assume that pm is true and try to deduce that pm+1 is true. The assumption that pm is true means that there exists an integer c such that m3 −m = 6c . So

(m +1)3 − (m +1) = (m3 + 3m2 + 3m +1)− (m +1)= (m3 −m)+ (3m2 + 3m)= 6c + (3m2 + 3m)

= 6 c + m2 + m2

⎛⎝⎜

⎞⎠⎟.

Notice that as in Example 8, we did not combine coefficients to simplify the expression (m +1)3 − (m +1) , but rather grouped the terms m3 and m and used our information concerning their difference.

We would like to state that c + m2 + m2

is an integer in order to conclude that 6 divides

(m +1)3 − (m +1) . But to do that we would need to know that m2 + m2

is an integer.

Now, the expression m2 + m2

is the quotient of two integers and quotients of integers do

not always yield integers. So why is this particular quotient an integer?

To say that m2 + m2

is an integer is the same as saying that m2 +m is even. (Recall that

an integer k is even if k is twice some integer.) We know that m2 +m is even when m = 1, 2, 3, or 4. Let's conjecture that, in general, m2 +m is even. There are many ways of proving that m2 +m is even. Perhaps the simplest approach is to examine two cases, namely, the case where m is even, and the other case where m is

odd. If we can deduce that in each case m2 +m is even, we will have that m2 + m2

is

always an integer. We'll leave it as an exercise to prove that m2 +m is even by examining those two cases. But since we are studying the Principle of Mathematical Induction, in the proof below, we will use induction to prove that m2 +m is even whenever m is a positive integer. (Note that in our first step in the proof, we will use the notation n rather than m and prove by induction that n2 + n is even for all positive integers n.) We will then have all the pieces we need to write up our formal proof for Example 9. The proof will then proceed in two steps.

Step 1: Use induction to prove that n2 + n is even for all positive integers n.

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Step 2: Proceed, as in the above Discussion, with the proof that 6 divides n3 − n , using the result of Step 1 at the appropriate point.

Proof. We will first prove that n2 + n is even whenever n is a positive integer. For each positive integer n, let rn be the statement:

n2 + n is even. Notice that r1 is true since 12 +1= 2 = 2 ⋅1 and 1 is an integer. Assume that rm is true for some positive integer m. Then m2 +m is even and so there exists an integer a such that m2 +m = 2a . We now rewrite (m +1)2 + (m +1) as follows:

(m +1)2 + (m +1) = (m2 + 2m +1)+ (m +1)= (m2 +m)+ (2m + 2)= 2a + (2m + 2)= 2(a +m +1).

Since a and m are integers, a + m + 1 is an integer as well and therefore (m +1)2 + (m +1) is even. Since r1 is true, and rm+1 is true whenever rm is true, the Principle of Mathematical Induction tells us that rn is true for all positive integers n. So n2 + n is even whenever n is a positive integer. Now let's proceed with the proof that 6 divides n3 − n whenever n is a positive integer. For each positive integer n, let pn be the statement:

6 divides n3 − n . p1 is true since 13 −1= 0 = 6 ⋅0 and 0 is an integer. Suppose that pm is true for some positive integer m. Then there exists an integer c such that m3 −m = 6c . So

(m +1)3 − (m +1) = (m3 + 3m2 + 3m +1)− (m +1)= (m3 −m)+ (3m2 + 3m)= 6c + (3m2 + 3m)

= 6 c + m2 + m2

⎛⎝⎜

⎞⎠⎟.

By the first part of our proof, we know that m2 +m is even, and so there exists an integer d such that m2 +m = 2d . Then,

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(m +1)3 − (m +1) = 6 c + m2 + m2

⎛⎝⎜

⎞⎠⎟

= 6 c + 2d2

⎛⎝⎜

⎞⎠⎟

= 6(c + d).

Since c and d are integers, so is c + d. Therefore (m +1)3 − (m +1) is a multiple of 6. We have now shown that p1 is true, and that pm+1 is true whenever pm is true. Hence by the Principle of Mathematical Induction, 6 divides n3 − n for all positive integers n.

u Exercise 5. Prove that n2 + n is even for all integers n by examining two cases: (i) the case where n is even, and (ii) the case where n is odd. Exercise 6. Use the Principle of Mathematical Induction to prove each of the following statements.

a) For each positive integer n, 3 divides 4n −1 . (Hint: Use the fact that 4n+1 −1= 4 ⋅4n −1 in the inductive step.) b) For each positive integer n, 4 divides 21n −1 . c) 21 ⋅23 ⋅25!22n−1 = 2n

2

for all n in N. In our statement of the Principle of Mathematical Induction, we started with the base case with n = 1. There are times when we would like to start the inductive process with a different integer. For example, suppose we wish to prove that 3n >1+ 2n whenever n is a positive integer with n ≥ 2 . (It is certainly not the case that 31 >1+ 21 .) To prove this assertion, we’ll use the following statement that is equivalent to the Principle of Mathematical Induction. Principle of Mathematical Induction II. Let n0 be a fixed integer. Suppose that for each integer n with n ≥ n0 we have a statement pn . Suppose further that

(i) pn0 is true, and (ii) for all integers m with m ≥ n0 , pm+1 is true whenever pm is true.

Then pn is true for all integers n with n ≥ n0 .

Why does this version of the Principle of Mathematical Induction make sense? If we apply (ii) with n = n0 , we may conclude that pn0+1 is true. Now if we apply (ii) again, this time with n = n0 +1 , we see that pn0+2 is true. Continuing in this manner, we deduce

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that pn0+3 is true, pn0+4 is true, etc. So it seems reasonable to assert that pn is true for all integers n ≥ n0 . It is possible to give a formal proof that version two of the induction principle follows from the original version of the principle. Let’s map out our strategy. For each positive integer n, let rn be the statement pn0+n−1 . So r1 is the statement pn0+1−1 . In other words, r1 is the statement pn0 . Similarly, r2 is the statement pn0+1 , r3 is the statement pn0+2 , etc. Notice that as n runs through the positive integers, n0 + n −1 runs through all the integers that are greater than or equal to n0 . So if we assume that (i) and (ii) in the second version of the Principle of Mathematical Induction are true, we can then use that information to show that rn is true for all positive integers n. That will then allow us to deduce that pn is true for all integers n with n ≥ n0 . (We will leave the proof that rn is true for all positive integers n as an exercise.) Exercise 7. Notice that in the statement of the Principle of Mathematical Induction II, we did not make the requirement that n0 is a positive integer. Are there examples that you can think of where a statement pn is true for all integers n with n ≥ 0 ? Now give an example where the statement pn is true for all integers n with n ≥ −1. We will use the Principle of Mathematical Induction II in much the same way as we used the original version of the principle. Example 10. Let’s prove that for all integers n with n ≥ 2 , 3n >1+ 2n . Proof. For each integer n with n ≥ 2 , let pn be the statement:

3n >1+ 2n . Since 32 = 9 > 5 = 1+ 22 , we see that p2 is true. Assume that pm is true, i.e., assume that 3m >1+ 2m is true, for some integer m where m ≥ 2 . Then

3m+1 = 3⋅3m

> 3⋅(1+ 2m )= 3+ 3⋅2m

>1+ 3⋅2m

>1+ 2 ⋅2m

= 1+ 2m+1.

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So 3m+1 >1+ 2m+1 and therefore pm+1 is true whenever pm is true. Consequently, by the Principle of Mathematical Induction II, pn is true for all integers n with n ≥ 2 .

u Exercise 8. Prove that each of the following statements is true:

a) 2n ≥ 3 for all integers n with n ≥ 2 . b) 3n ≥ 2n + n for all nonnegative integers n.

Section 3. The Principle of Complete Induction and its Application to the Factorization of Integers One of the most important properties of the integers is given in the following theorem: Unique Factorization Theorem (also known as the Fundamental Theorem of Arithmetic). Let n be a positive integer with n > 1. Then there exist primes

p1, p2 ,..., pr such that n = p1 p2! pr . Moreover if q1,q2 ,...,qs are primes for which

n = q1q2!qs , then r = s and, after rearrangement, pi = qi for all i. The Unique Factorization Theorem is often restated as follows:

Every positive integer greater than 1 can be factored uniquely as a product of primes.

For example, we can factor 6 as 2 ⋅3 and also as 3⋅2 . Notice that we have used the same primes in both factorizations and have used those primes the same number of times in both. So, up to rearrangement, our factorization of 6 is “unique.” In this section (Example 11), we will use the Principle of Complete Induction, a principle that is equivalent to the Principle of Mathematical Induction, to prove the first assertion of the Unique Factorization Theorem. In the next section we will use the method of contradiction to give a different proof of that fact. We will then prove the “uniqueness” portion of the Unique Factorization Theorem in Chapter 6 (Further Properties of the Integers) after we have developed some additional tools. Example 11. Let’s prove that if n is a positive integer with n ≥ 2 , then n is a product of primes. Initial Discussion. First of all, let’s recall our working definition of a prime. The integer p is said to be a prime if p > 1 and the only positive divisors of p are 1 and p itself. Suppose we are given a positive integer n where n is greater than one. What does it mean to say that n is not a prime? (Such integers are said to be composite.) If n is not a prime,

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then we can find a positive integer a having the properties that a divides n, a is greater than 1 and a≠ n. Since a divides n, by definition, there exists an integer b such that ab = n. As a and n are positive, so is b. Moreover, since a ≠ n , b ≠ 1. (Indeed, if b = 1, then n = ab = a, a contradiction of the assumption that a ≠ n .) So we know that a and b must both be greater than 1. Therefore, a < ab = n and b < ab = n as well. In other words, if n is an integer such that n > 1 and n is not prime, then there exist integers a and b such that n = ab, 1 < a < n, and 1 < b < n. (By the way, did you notice how we “snuck in” an argument by contradiction in our discussion?) Conversely, if we can factor n as ab where a and b satisfy the inequalities 1 < a < n and 1 < b < n, then n has positive divisors (namely, a and b) other than 1 and n itself. So n is not a prime. The above observations will be important in our proof and will be important in the next section as well. So we’ll record them below.

Lemma. Let n be a positive integer with n > 1. The integer n is not a prime (that is, the integer n is composite) if and only if there exist positive integers a and b with 1 < a < n, 1 < b < n, and ab = n.

Remark. In mathematics, the term lemma refers to an intermediate result that is used in the proof of a more substantial result such as a theorem.

Before attempting our proof of the fact that each integer greater than 1 can be factored as a product of primes, we also need to understand what it means for a prime such as 2 or 13 or … to be a “product” of primes. We will use the standard convention that a prime itself is a "product" of primes (with only one factor used in the product). Continuation of our discussion of Example 11. Now, how can we go about using the Principle of Mathematical Induction to prove our result in Example 11? We should start by defining, for each positive integer n with n ≥ 2 , the statement pn to be the assertion that n is a product of primes. Given our convention that each prime is a product of primes, it’s easy to see that p2 is true. So, let’s look at the inductive step. Suppose that pm is true, that is, suppose that m is a product of primes, for some integer m where m is greater than or equal to 2. Can we deduce that pm+1 is true? In other words, can we deduce that m + 1 is a product of primes? Well, if m + 1 is itself a prime, then by our convention, m + 1 is a product of primes. Suppose then that m + 1 is not a prime. Then we may factor m + 1 as ab, where a and b are integers with 1< a < m +1 and 1< b < m +1 . Since a and b are integers, we must also have that 2 ≤ a ≤ m and 2 ≤ b ≤ m . Now, if a and b are themselves products of primes, then ab is also a product of primes. But do we know that a and b can each be factored as a product of primes? Well, if a is either 2 or m, and b is either 2 or m, our inductive assumption (and our convention that 2 is a “product” of primes) tells us that a

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and b are products of primes. But we can't guarantee that we are that lucky! All we know is that both a and b are between 2 and m. Fortunately, there is another version of the Principle of Mathematical Induction that will allow us to deal with this case. Principle of Complete Induction (also known as the Principle of Strong Induction). Let n0 be a fixed integer. Suppose that for each integer n with n ≥ n0 we have a statement pn . Suppose further that

(i) pn0 is true, and (ii) for all integers m with m ≥ n0 , pm+1 is true whenever all of the statements pn0 , pn0+1 , ..., pm are true.

Then pn is true for all integers n with n ≥ n0 .

Remarks. 1. It is possible to prove that the Principle of Complete Induction follows from our first version of the induction principle, but we will not do so here. 2. It is also possible to use a slight modification of Property (ii) and state the Principle of Complete Induction without including Property (i) (the base case).

Example 11 continued. We now have all the results we need to prove that if n is a positive integer with n ≥ 2 , then n is a product of primes. Proof. For each positive integer n with n ≥ 2 , let pn be the statement:

n is a product of primes. Then p2 is true since 2 is a prime. (Recall that we are using our convention that a prime is a product of primes.) Suppose that m is a fixed integer with m ≥ 2 and suppose that p2 , p3 , ..., pm are true. So we are assuming that all of the positive integers between 2

and m (inclusive) are products of primes. If m + 1 is a prime, then pm+1 is true. So we may now assume that m + 1 is not a prime. By our lemma, we can factor m + 1 as ab where a and b are positive integers with 2 ≤ a ≤ m and 2 ≤ b ≤ m . By assumption, the statements pa and pb are true and so a and b must themselves be products of primes. Therefore ab is a product of primes as well, that is, m+1 is a product of primes. We have now shown that pm+1 is true whenever p2 , p3 , ..., pm are true. It follows from the Principle of Complete Induction that pn is true for all integers n with n ≥ 2 .

u

Remarks. There is an analogue of the Unique Factorization Theorem that applies to polynomials with real coefficients. One version of that theorem states that each

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polynomial with real coefficients having degree at least one can be factored as a product of linear polynomials and quadratic polynomials. (Here the degree of a nonzero polynomial is the highest power of x that has a nonzero coefficient.) That result is the starting point in calculus for the method of partial fractions, a technique used to integrate quotients of polynomials.

In the next section we will continue with our discussion of principles that are equivalent to the Principle of Mathematical Induction. Section 4. The Well Ordering Principle and Further Applications to Properties of the Integers We conclude this chapter with a discussion about the Well Ordering Principle, an important principle that is equivalent to the Principle of Mathematical Induction. Well Ordering Principle. If S is any nonempty subset of the set N of positive integers, then S has a smallest element.

First of all, what does it mean for S to have a smallest element? We need to be careful here. We do not want to assert the existence of an element s0 in S such that s0 < s for all s in S. Indeed, that condition would be violated if we were to substitute s0 for s in the inequality s0 < s . So instead, we will say that S has a smallest element if there exists an element s0 in S such that s0 ≤ s for all s in S. We will not give a proof that the Well Ordering Principle is equivalent to the Principle of Mathematical Induction. However, in a moment, we will use the Well Ordering Principle (and its extension) to prove two important results concerning the integers. But before doing so, let’s delve more deeply into the Well Ordering Principle. When discussing the Principle of Mathematical Induction, we noticed that we didn't need to start with the base case n = 1. Can the same be said for the Well Ordering Principle? For example, can we assert that if S is any nonempty subset of the set of nonnegative integers, then S has a smallest element? The answer is yes. There are two cases to consider. First of all, if S contains the integer 0, then 0 is the smallest element of S. What happens in the other case, the case where 0 is not an element of S? Then S is a nonempty subset of the set of positive integers and so by the Well Ordering Principle, we know that S has a smallest element. Therefore, in either case we see that S has a smallest element. We’ll record that result below as the Extended Well Ordering Principle.

Extended Well Ordering Principle. If S is any nonempty subset of the set N∪{0} of nonnegative integers, then S has a smallest element.

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Exercise 9. Suppose that S is a nonempty subset of the set {n∈Z: n ≥ −1} . Prove that S has a smallest element. (Recall that Z is the set of integers.) How far "down" can we continue with this argument? In particular, is it true that every nonempty subset of the set of integers has a smallest element? We have seen that some nonempty subsets of the set of integers certainly do contain a smallest element. For example, 1 is the smallest element of N and −2 is the smallest element of

−2,0,2,4,6,8,…{ } . But the set of integers itself is nonempty and contains no smallest element. (See Exercise 10 below.) So the answer to the question "Is it true that every nonempty subset of the set of integers has a smallest element?" is no. Exercise 10. Use the method of contradiction to show that the set Z of integers has no smallest element. In the previous section we proved the first part of the Unique Factorization Theorem using the Principle of Complete Induction. Now let’s give a proof using the Well Ordering Principle and the method of proof by contradiction. Proof (by contradiction) that every integer greater than 1 is a product of primes. Suppose not, that is, suppose there exists a positive integer n with n > 1 such that n is not a product of primes. Let

S ={t ∈N : t >1 and t is not a product of primes}. Then S ⊆ N . Moreover, since n∈S , S ≠ ∅ . By the Well Ordering Principle, S has a smallest element that we’ll denote by n0 . Since n0 ∈S we know that n0 >1 . We also know that n0 is not a product of primes. By our convention, every prime is a product of primes. So n0 is not a prime. Therefore (by our lemma from Section 3), n0 = ab where a and b are integers with 1< a < n0 and 1< b < n0 . Since both a and b are less than n0 , neither a nor b is an element of S. So as a is greater than 1 and b is greater than 1, a must be a product of primes and b must be a product of primes. But then n0 = ab is a product of primes as well. That contradicts the fact that n0 is an element of S. Consequently, every positive integer greater than one is a product of primes.

u One of the most important results concerning the set of integers is the following: Division Algorithm for Integers. Suppose that a and b are integers with b ≠ 0 . Then there exist integers q and r with

(i) a = bq + r, and (ii) 0 ≤ r < | b | .

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Remarks. The Division Algorithm has many applications. For example, in Chapter 1 (Language, Logic, Sets, and Methods of Proof), we applied the Division Algorithm with b = 2 to prove that every integer is either even or odd. In Chapter 4 (Equivalence Relations), we will use the Division Algorithm to describe an important algebraic structure known as the ring of integers modulo n. Then in Chapter 6 (Further Properties of the Integers), we will use the Division Algorithm to develop an algorithm for finding the greatest common divisor of two integers, even when the integers are very large and very hard to factor.

The Division Algorithm will be our focus for the rest of this section. We’ll start by discussing a strategy for proving that the Division Algorithm is true. Discussion. In the statement of the Division Algorithm, if q and r have the desired properties, then a − bq = r ≥ 0 . So in order to find r, it is natural to consider the set S obtained by including all possible choices of r, that is, the set S defined by

S ={a − bq : q is an integer and a − qb ≥ 0} . We seek that element of S which is also less than | b | . To determine the elements of S, we will proceed as follows. First compute a− bq, where q ranges over all of the integers, and put the resulting differences in S whenever they are nonnegative. For example, suppose a = 31 and b = 6. Let’s consider the following table in which we compute 31− 6q for a few values of q to see if we can gather any information that will help us move forward in our proof.

q 31 − 6q −3 49 − 2 43 −1 37 0 31 1 25 2 19 3 13 4 7 5 1 6 −5 7 −11 8 −17

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It appears from the table that when q is negative, then 31− 6q is an element of S. Indeed, if q is a negative integer, then 6q ≤ −6 and so 31− 6q ≥ 31− (−6) = 37 ≥ 0 . Now, if we let q range over the nonnegative integers and compute 31− 6q , the only elements we are allowed to include in S are 31− 6(0) = 31 , 31− 6(1) = 25 , 31− 6(2) = 19 ,

31− 6(3) = 13 , 31− 6(4) = 7 , and 31− 6(5) = 1. In the statement of the Division Algorithm where a = 31 and b = 6, we require that 0 ≤ r < | b | = 6 . Therefore the only appropriate choice for r in this example is r = 1. Notice that this value is, in fact, the smallest element of S. The above observation provides the approach we will use to give a proof of the Division Algorithm when b > 0. We will then use that result to prove that the Division Algorithm holds when b < 0. Proof of the Division Algorithm. Let’s first assume that b > 0. Define S by

S ={a − bq : q is an integer and a − qb ≥ 0} . Notice that a− bq is an integer whenever q is an integer, and so S is a subset of the set of nonnegative integers. To apply the Extended Well Ordering Principle to S, we need to know that S is nonempty. If a ≥ 0 , then a − b ⋅0 is in S. On the other hand, if a < 0, then a − ba = (1− b)a ≥ 0 since 1− b ≤ 0 and a < 0. So in this case, a− ba is an element of S. Consequently in either case, S is nonempty and therefore contains a smallest element. Let's denote that element by r. We know that r is an element of S, r ≥ 0 and r = a− bq for some integer q. In other words, a = bq + r where q and r are integers with r ≥ 0. In order to complete the proof of the Division Algorithm when b > 0, we must show that r < b . We will assume that r ≥ b and show that this assumption leads to a contradiction. Assume that r ≥ b . Then r − b ≥ 0 . Moreover, r − b = (a − bq)− b = a − b(q +1) . As q is an integer, so is q +1 , and therefore r − b is of the form a − bx for some integer x. We also have that r − b ≥ 0 . By the definition of S, we see that r − b is an element of S. So r − b must be greater than or equal to the smallest element of S, that is, r − b ≥ r . But b > 0 and therefore r − b < r . We have shown that the inequalities r − b < r and r − b ≥ r hold at the same time, a contradiction. So r must be less than b. Consequently the Division Algorithm holds when b > 0. Suppose now that b < 0. Then − b > 0 and so by the above, there exist integers q and r such that a = (−b)q + r and 0 ≤ r < −b = | b | . Then − q and r are integers satisfying

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a = b(−q)+ r and 0 ≤ r < | b | . Therefore the Division Algorithm holds when b < 0 as well.

u

Remarks. The symbols q and r used in the statement of the Division Algorithm reflect their “functions.” Indeed, in many computations, q is the integral part of the quotient of a upon division by b, and r is the integral part of the remainder of that quotient. That is, q and r are integers satisfying the equation ab = q + r

b . Example 12. Let’s find q and r when a = 50 and b = 3. The usual strategy is to find integers q and r satisfying

ab = q + r

b and then to verify that those choices “work.” In our particular example, we can write 503 as 16 + 23 and so we’ll choose q to be 16 and r to be 2.

Now let’s see that those choices satisfy the conditions in the Division Algorithm. Since

a = 50 = 3⋅16+ 2 and 0 ≤ 2 < 3 , 16 and 2 are indeed appropriate choices for q and r, respectively. Example 13. We need to be careful when ab < 0 . For example, if we take a to be −50

and b to be 3, and if we compute −503 , we obtain −503 = −16 + −23 . Although

−50 = 3(−16)+ (−2)

is a correct equation, −2 is not an appropriate choice for r since −2 does not satisfy the inequality 0 ≤ r < 3 . To correct for this problem, we will decrease the coefficient, −16 , of 3 in the equation −50 = 3(−16)+ (−2) by 1, and we will increase −2 (our first “stab” at a remainder) by 3 in order to maintain the equality. We then have

−50 = 3( −17 ) + 1. Notice that −17 and 1 are appropriate choices for q and r, respectively, since −50 = 3( −17 ) + 1 and 1 satisfies the inequality 0 ≤ r < 3 . Exercise 11. Compute q and r in each of the following.

a) a = 1000 and b = 7 b) a = 1000 and b = −7 c) a = −1000 and b = −7 d) a = −1000 and b = 7

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Important Extension of the Division Algorithm Notice that we computed just one pair of integers q and r corresponding to each pair of our integers a and b in Examples 12 and 13. That’s because the integers q and r in the statement of the Division Algorithm are unique. Let’s now prove that assertion. Proof. Let a and b be integers with b ≠ 0 . Suppose that a = bq + r, where q and r are integers with 0 ≤ r < | b | , and suppose as well that a = bs + t, where s and t are integers with 0 ≤ t < | b | . We will prove that q = s and r = t. We know that either r ≤ t or t ≤ r , and so we will consider two cases in our proof.

Case 1. r ≤ t In the first case, we have that

0 ≤ | t − r | = t − r ≤ t < | b | . Now, if q ≠ s , then | q − s | is a positive integer and so

|t − r | = |(a − bs)− (a − bq) | = |bq − bs | = |b | |q − s | ≥ |b | . But | t − r | < | b | , and so we have reached a contradiction. We now know that q and s must be equal if r ≤ t . As t − r = bq − bs = b(q − s) = 0 , we can also conclude that t = r in Case 1.

Case 2. t ≤ r The proof that t = r and q = s in Case 2 is similar to the argument we gave in Case 1.

u Exercise 12. Prove that t = r and q = s in Case 2 of our proof above. Let’s record our result for use in subsequent chapters. Extended Division Algorithm for Integers. Suppose that a and b are integers with b ≠ 0 . Then there exist unique integers q and r with

(i) a = bq + r, and (ii) 0 ≤ r < | b | .

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End of Chapter Problems I. Proofs and Counterexamples (Note that in the statements below, N, as usual, denotes the set of natural numbers.) 1. Prove that each of the following propositions is true.

a) 3+ 6 +12 +!+ 3⋅2n−1 = 3(2n −1) for all n in N. b) 2 + 2n ≤ 2 ⋅2n for all n in N. c) 1

3 + 23 +!+ n3 = 1+ 2 +!+ n( )2 for all n in N. d) 5 divides 6n+1 −1 for all n in N∪{0} . e) 3 divides 10n+1 +10n +1 for all n in N∪{0} . f) 8 divides 5n + 2 ⋅3n−1 +1 for all n in N. g) 1+

11( ) 1+ 12( ) 1+ 13( )! 1+ 1n( ) = n +1 for all n in N.

h) 1−122( ) 1− 132( ) 1− 142( )! 1− 1n2( ) = n+1

2n for all positive integers n with n ≥ 2 .

i) x − y divides xn − yn for all n in N whenever x and y are integers with x ≠ y .

j) 1n<1.1 for all n in N.

k) There exists a positive integer n such that 2n > n2 . 2. For each positive integer n, let

Sn = log10 1+ 11( ) + log10 1+ 1

2( ) + log10 1+ 13( ) +!+ log10 1+ 1

n( ) .

a) Prove that Sn = log10(n+1) for all n in N. b) What does our formula from part a tell us when n = 9? c) What does our formula from part a tell us when n = 99?

3. a) Define an for each positive integer n as follows:

a1 = 1, and an+1 = an +1 for all n in N.

First prove that an ≥1 whenever n is a positive integer. Then prove that an < an+1 for all n in N. b) We will define a sequence Fn n=0

∞ of nonnegative integers as follows:

F0 = 0 , F1 = 1 , and for each integer n with n ≥ 2 , let Fn = Fn−1 + Fn−2 .

So F2 = F1 + F0 = 1 , F3 = F2 + F1 = 2 , F4 = F3 + F2 = 3 , F5 = F4 + F3 = 5 ,

26

F6 = F5 + F4 = 8 , etc. (The integers Fn are called Fibonacci numbers.1) Let m denote a fixed positive integer. Use the Principle of Complete Induction to prove that for each positive integer n, Fm+n = FnFm+1 + Fn−1Fm . (Hint: you may want to first prove the assertion when n = 1 , and then use the Principle of Complete Induction with n0 = 2 .) Then prove that Fm divides F2m whenever m is a positive integer.

4. Prove that each of the following propositions is true.

a) For each n in N and for each positive real number x, ln(xn ) = n ln(x) . (Here, for each n in N, let pn be the statement ln(xn ) = n ln(x) for all positive real numbers x. In the inductive step, you may use the property that for all positive real numbers a and b, ln(a ⋅b) = ln(a)+ ln(b) .)

b) ddx

xn( ) = nxn−1 for all positive integers n with n ≥ 2 . (You may use the

product rule for derivatives, but not the power rule for derivatives, in the

inductive step. You may also use the fact that ddx

x( ) = 1 for all

real numbers x, and the fact that for all positive integers n and for all real numbers t, t n+1 = t nt .) c) cosθ + isinθ( )n = cosnθ + isinnθ for all n in N and for all real numbers θ . (Recall that i2 = −1 . You may use the trigonometric identities for cos(a + b) and for sin(a + b) in your proof.) d) (xy)n = xnyn for all n in N and for all real numbers x and y. (As in part b above, in the inductive step, you may use the fact that for all positive integers n and for all real numbers t, t n+1 = t nt .)

5. Recall that if m is an integer, then m is said to be odd if m = 2s + 1 for some integer s.

a) Prove that if a and b are odd integers, then ab is odd. b) Prove that if n is a positive integer, then 5n is odd. (You should first prove that 51 is odd by writing 51 in the form 2s + 1 for some integer s.)

6. Prove or give a counterexample for each of the statements below.

a) 2n ≥ 2 + n for all positive integers n with n ≥ 2 . b) 3 divides n2 − 3n + 2 for all n in N.

1 Leonardo Fibonacci was an Italian mathematician who has been described as the “greatest European mathematician of the Middle Ages.” His book, Book of Calculation, described many of the properties of the integers we use today. The Fibonacci sequence that we defined in Problem 3 has surprising “appearances” in nature.

27

7. Define a function f by f (x) = xex for all real numbers x.

a) Prove that ′f (x) = (x +1)ex . b) In general, prove that f

(n) (x) = (x + n)ex for all n in N. (Here, f(n) denotes

the n-th order derivative of the function f.) 8. Prove that each of the following propositions is true.

a) For each positive integer n where n ≥ 3 , the sum of the interior angles of any convex polygon having precisely n sides is (n − 2)180

! . b) If n is a positive integer with n ≥ 2 and if P1,P2,…,Pn are n distinct points in the plane, then the total number of line segments having endpoints in the set

P1,P2,…,Pn{ } is n2 − n2

.

9. For each positive integer n, let pn denote the statement:

2 divides 2n + 1.

a) Prove that the conditional statement:

pm ⇒ pm+1

is true for all m in N. b) Why can’t you use the Principle of Mathematical Induction to show that pn is true for all n in N? Where does the proof break down?

II. Incorrect Arguments In each of the following, the given "proof" is incorrect. Find all of the errors in the proof. Determine if the "claim" is correct or incorrect. (Be sure to justify your answer.) 1. Claim: All positive integers are equal.

Proof. For each positive integer n, let pn be the statement that the positive integers between 1 and n inclusive are equal, that is, pn is the statement 1 = 2 = != n. Then p1 is the statement that the positive integers between 1 and 1 inclusive are equal, that is, p1 is the statement 1 = 1. Clearly p1 is true. Now assume that pm is true. So we are assuming that 1 = 2 = != m. In order to show that pm+1 is true, we need to show that 1= 2 == m = m +1. Since we have already assumed that 1= 2 == m we only need to show that m = m + 1. From 1 = 2 = = m, we have that 1 = 2 = != m−1 = m. In particular m−1= m and therefore, ( m−1) + 1 = m + 1. Consequently m = m + 1 and so 1 = 2 = ! = m =

28

m + 1. Hence pm+1 is true whenever pm is true and so by the Principle of Mathematical Induction, pn is true for all positive integers n. Now we apply the above result to prove the claim. Suppose that m and n are positive integers. We know that either m ≤ n or n ≤ m. In the first case, since pn is true, m and n are equal. In the second case since pm is true, then n = m as

well. This completes the proof of the claim. 2. Claim: All horses are the same color.

Proof. For each positive integer n, let pn be the statement that all the horses in any given collection having precisely n horses are all the same color. If we take any collection containing only one horse, then clearly all the horses in that collection are the same color. So p1 is certainly true. Now assume that pm is true, that is, assume that whenever we are given a collection containing precisely m horses, all the horses in that collection are the same color. Consider a collection containing m+1 horses, and let's label the horses as H#1, H#2, … , H#m and H#(m+1). If we remove H#1 from the collection, we are left with the m horses, H#2, H#3, … , H#(m+1). By hypothesis, all of the horses in that collection are the same color. So if we can argue that H#1 is the same color as those horses, we will have shown that all horses in our original collection of m + 1 horses are the same color. We will in fact show that H#1 is the same color as H#2. Since we know that H#2 is the same color as H#3, H#4, etc., this will give us that H#1 is the same color as the rest. To show that H#1 and H#2 are the same color, let's take our original collection of horses, H#1, H#2, … , H#(m+1) and this time remove H#(m+1). We now have a collection of m horses. So we know by our inductive assumption that all the horses in this collection are the same color. In particular, H#1 and H#2 are the same color. So pm+1 is true whenever pm is true and therefore by the Principle of Mathematical Induction, pn is true for all positive integers n. Let's apply the above result to prove our claim. There are only finitely many horses in the world. Let ′n denote that number. Having just proven that all the horses in any given collection containing ′n horses are the same color, we now conclude that all the horses in the world are the same color!

3. Claim: All positive integers are both even and odd.

Proof. Assume that m is both even and odd. We will show that m + 1 is both even and odd.

Since m is even, we may write m = 2d for some integer d. Then m + 1 = 2d + 1, which tells us that m + 1 is odd. But since m is also odd, we know that m = 2c +1 for some integer c. Therefore m + 1 = 2c + 2 = 2(c + 1) where c + 1 is an integer. In other words, m + 1 is also even. We now know that m + 1 is both even and odd and so our claim is true.

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4. Claim: n2 + n is odd for all positive integers n.

Proof. 1 is odd and so the assertion holds when n = 1. Assume that m2 +m is odd. Then there exists an integer d such that m2 +m = 2d +1 . So

(m +1)2 + (m +1) = (m2 + 2m +1)+ (m +1)

= (m2 +m)+ (2m + 2)= (2d +1)+ (2m + 2)= 2(d +m +1)+1.

Since d and m are integers, so is d + m + 1. Consequently (m +1)2 + (m +1) is odd whenever m2 +m is odd. Therefore by the Principle of Mathematical Induction, n2 + n is odd for all positive integers n.

III. Proofs without Claims In each problem below, we provide a “proof” (with several incomplete lines). Your job is to determine the incomplete lines and an appropriate claim! 1. Claim:

Proof. For each positive integer n, let pn be the statement:

Then p1 is true since 13 = 1 and

1(1+1)2( )2

= 22( )2

= 12 = 1.

Now suppose that pm is true for some natural number m, that is, assume that

Then

13 + 23 +!+ (m +1)3 = 13 + 23 +!+m3( ) + (m +1)3

= m(m+1)2( )2 + (m +1)3

= m2 (m+1)24 + (m +1)3

= (m +1)2 m2

4 + (m +1)( )= (m +1)2 m2+4(m+1)

4( )= (m +1)2 m2+4m+4

4( )

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= (m +1)2 (m+2)24( )

= (m +1)2(m+1)+1( )2

4⎛⎝⎜

⎞⎠⎟

=(m+1) (m+1)+1( )

2⎛⎝⎜

⎞⎠⎟2

.

Consequently, by the Principle of Mathematical Induction,

2. Let’s define a sequence an n=1

∞ of real numbers as follows:

a1 = 1 and for each positive integer n, an+1 = 4an +1 .

Claim:

Proof. For each positive integer n, let pn be the statement: Then p1 is true since a1 = 1 and 1< 5 . Now suppose that pm is true for some natural number m, that is, assume that Then

an+1 = 4an +1

< 4 ⋅5+1

= 21< 5.

Consequently, by the Principle of Mathematical Induction, IV. Investigations 1. In Exercise 1 of Section 1, we defined Sn (for each positive integer n) by:

Sn = 1+ 2 +!+ n .

a) Find a formula that relates Sn+1 to Sn .

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b) For each positive integer n, find a “simpler” form for Sn in terms of n where we do not list all of the terms 1, 2, 3, …, n. (Hint: The "pairing" technique we used in Example 3 of Section 1 might be helpful here.) c) Use the Principle of Mathematical Induction to prove that your formula in part b is correct.

2. For each positive integer n, let Sn = 2

0 + 21 + 22 +!+ 2n .

a) Find a formula that relates Sn+1 to Sn . b) For each positive integer n, find a “simpler” form for Sn in terms of n where we do not list all of the terms 20 , 21 , …, 2n . c) Use the Principle of Mathematical Induction to prove that your formula in part b is correct.

3. For each positive integer n, let Tn = 2

0 ⋅21 ⋅22!2n .

a) Write each of the products 20 ⋅21 , 20 ⋅21 ⋅22 , 20 ⋅21 ⋅22 ⋅23 , 20 ⋅21 ⋅22 ⋅23 ⋅24 , and 20 ⋅21 ⋅22 ⋅23 ⋅24 ⋅25 in the form 2k where k is an integer. b) For each positive integer n, find a “simpler” form for Tn in terms of n where we do not list all of the factors 20 , 21 , …, 2n . c) Use the Principle of Mathematical Induction to prove that your formula in part b is correct.

4. For each integer n where n ≥ 2 , let Bn = 1− 12( ) 1− 13( )! 1− 1n( ) .

a) Calculate B2 , B3 , and B4 . b) Find a “simpler” formula for Bn in terms of n where we do not list all of the

n−1 factors 1− 12( ) , 1− 13( ) , …, 1− 1n( ) . c) Use the second version of the Principle of Mathematical Induction to prove your conjecture from part b.

5. For each positive integer n, let pn be the statement 2n > n2 .

a) For which positive integers is pn true? b) Use the Principle of Mathematical Induction II to prove your result in part a.

6. Suppose that the first number a1 in the sequence an n =1

∞ is defined to be 1, and for

each positive integer n, let an+1 = 3an +1 . So the next three values are computed as follows:

a2 = 3a1 +1 = 3⋅1+1 = 4 = 2 ,

a3 = 3a2 +1 = 3⋅2 +1 = 7 ≈ 2.65 , and

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a4 = 3a3 +1 = 3⋅ 7 +1 ≈ 2.99 .

a) Are the successive values in the sequence an n =1

∞ always increasing, always

decreasing, or sometimes increasing and sometimes decreasing? What happens in the long run? What conjectures would you make? b) What happens if you begin with a1 = 10 and still define an+1 to be 3an +1 for each positive integer n? c) Try several other values for a1 . What conjectures would you make? d) In Example 7 of Section 2, we proved that if we start with a1 = 1, then an ≤ 4 for all n in N. Let's conjecture that an ≤ 3 for all n. Try to give a proof of that conjecture using the Principle of Mathematical Induction. Where does your proof break down? Is the conjecture true nonetheless? Justify your answer.

7. Suppose a1 is a positive real number and for each positive integer n, let an+1 = an( )13 .

a) If we choose a1 to be 8, are the successive values always increasing, always decreasing, or sometimes increasing and sometimes decreasing? What happens in the long run? b) What happens if we begin with a1 = 18 ? c) Try several other values for a1 . What conjectures would you make? d) Find a formula that relates an to a1 . Prove that your formula is correct using the Principle of Mathematical Induction. e) Use your formula from part d to determine what happens to an in the long run. Can you formulate a theorem that describes the long run behavior of an for each value of a1?