Appendix A: The Principle of Mathematical Induction We now present an important deductive method widely used in mathematics: the principle of mathematical induction. First, we provide some historical context for this topic. Later, we explore the logical underpinnings for this principle. Lastly, we look at some interesting results taken from a wide range of mathematical areas that are proved using this principle. Historical Preliminaries Some of the earliest mathematicians in Ancient Greece used pebbles and other small objects to investigate the remarkable properties of polygonal numbers. Famously, the Pythagoreans (5 th century BCE) found that the triangular numbers constructed as triangular and regular configurations of pebbles, where the first row contains one pebble and each subsequent row contains one more pebble than the previous one (see Figure 1 below) are generated by the sequence of partial sums of the positive integers: 1 1, 3 1, 6 1 3, 10 1 3 4, etc. Figure 1: The First Four Triangular Numbers. Moreover, they realized that the n th triangular number could be computed using the following well-known formula for the sum of the first n positive integers: T n nn1 1 3... n. (1) Hence, the sequence of triangular numbers is given by T n 1,3,6,10,15,..., n n 1 n, n 1,.... In additional to triangular numbers, the Pythagoreans also studied the properties of square numbers, pentagonal numbers, hexagonal numbers, oblong numbers, and many others. They also found interesting relationship between these types of numbers. For example, they observed that odd positive integers: S n, the n th square number, is the sum of the first n
S 135... n 1 n. () n Figure below shows how the result of formula () is consistent with the construction of square pebble arrays. Note how the red pebbles are all successive odd positive integers. Figure : The First Five Square Numbers. Square numbers are also the sum of two consecutive triangular numbers: n n n1 1 S T T n n n n. Today we may think of the result of these inquiries as rudimentary number theory. However, this was a significant break from the Egyptian and Babylonian tradition of using numbers mostly for the practical purposes of arithmetical computations and approximations. The investigations of the Pythagoreans with polygonal and, more broadly, figurate numbers constitutes one of the earliest instances in history of an inquiry into pure mathematics. The question of why formulae (1) or () should be true for all triangular or square numbers, respectively, is the focus of our present discussion. Why should anyone seeing this result for the first time be convinced that, say, the trillionth square number is the sum of the first trillion odd numbers? As it turns out, there is a powerful method in mathematics that validates the truth of statements like (1) or () for any positive integer n, no matter how large this variable gets. This is the Principle of Mathematical Induction. This method of proof is a rule of inference validated in predicate logic. As such, it is fully deductive (notwithstanding its name). As we shall see shortly, the principle of mathematical induction relies on only two steps (one of which is typically trivial) and is predicated on one fundamental axiom: the well-ordering of the positive integers. While the Pythagoreans did not use mathematical induction to prove their claims about figurate numbers, Euclid used this principle implicitly in his proof that there are infinitely many primes (Proposition 0 in Book IX of The Elements). In philosophy, Eubulides of Miletus (4 th century BC) advanced his famous heap paradox using an argument similar in spirit to induction, but counting down. He argued that if one agrees that a collection of 1,000,000 grains of sand is a heap, and that removing one grain of sand from the heap still makes it a heap, then eventually we must consider a single grain of sand
(or even none) as a heap! (This paradox illustrates the concept of vagueness associated with particular predicates used in language.) Later, the Islamic mathematician Al-Karaji (late 10 th century, Baghdad) also used a counting-down method to prove that the sum of the cubes of the first n positive integers is equal to the square of the sum of the first n positive integers: 3 3 3 3 n n1 1 3... n 1 3... n. (3) 4 Al-Karaji actually proved the statement for n 10, but his argument could clearly be applied in a more general setting. Using a 1...10 1...10 square, he first divided it into four pieces: a large 1...9 1...9 square, a smaller 10 10 square, and two 1...1010 rectangles. This is shown in Figure 3, below. Figure 3: Square Used in Al-Karaji s Induction Proof of Formula (3) By summing the areas of the pieces, he showed the following:
1...10 1... 9 10 1... 9 1... 9 10 9 10 10 10 1... 9 9 10 10 1... 9 10 3 Using an identical argument with a 1... 9 1... 8 9 3. 1...9 1...9 square, he showed that This then implied that 1...10 1... 8 9 3 10 3. Eventually, he worked his way down to the trivial case of n 1 and completed his proof. In the era of modern mathematics, Maurolycus was the first to state the principle in his Arithmetica (1575). Pierre de Fermat and Blaise Pascal (17 th century, France) were also amongst the first mathematicians to use this method with more rigor. Pascal explicitly states the principle of mathematical induction in his Traité du Triangle Arithmétique (1654), where he famously showcases the combinatorial properties of his eponymous triangle. Fermat used a similar variant of the principle called the method of infinite descent in his analysis of Diophantine equations. The formal statement of mathematical induction as a second-order axiom in predicate logic came much later with the advances in mathematical logic made by Peano, Dedekind and others in the later part of the 19 th century. The Logic of Mathematical Induction In this section we look more formally at the logical statement of the principle of mathematical induction. We prove it from the axiomatic principle of well-ordering, which asserts that every non-empty set of natural numbers has a least element (a number that is smaller than all the other numbers in the set). The Principle of Mathematical Induction. Let P(n) be a statement that makes an assertion involving the variable n, where n 1,,3,... Suppose the following two properties hold: (1) P(1) is true. () If P(k) is true, then P(k 1) is also true. Then P(n) is true for all positive integers n.
More formally, we write the following: P(1) k : P(k) P(k 1) n : P(n) This principle is essentially a recursive proof. To validate the truth of any statement ranging over the positive integers (or any ordered, countably infinite set), it then suffices to show the two steps enumerated above. The first step in this procedure is called the base case; it is often a trivial case of the statement. The second step is called the inductive step. The assumption that P(k) is true in the inductive step is called the induction hypothesis. Note that the base case need not start with 1. It could start with any positive integer m. In that case, P(n) is true for all n m and we have P(m) k m : P(k) P(k 1) n m : P(n) There is also a version of the principle called Strong Induction, in which the inductive step requires showing that P(k 1) is true whenever P(1),P(),...,P(k) are all assumed to be true. This version is equivalent to the Simple/Weak Induction stated at the top. Our next step is to show that this principle may be deduced from the Well-Ordering Principle, which states that every non-empty set of positive integers contains a least, or smallest element. This obvious property of the positive integers can be either derived from the principle of mathematical induction (which is then treated as an axiom), or treated conversely as an axiom from which the principle of induction can be deduced. We adopt the latter approach. Theorem. The Well-Ordering Principle Implies the Principle of Mathematical Induction. Proof. Suppose that the statement P(n) satisfies the two conditions for the principle of mathematical induction. So P(1) is true and whenever P(k) is true, P(k 1) is also true. Let S be the set of positive integers m for which P(m) is false. Then there are two possibilities: either S is empty or S is non-empty. If S is the empty set, then P(n) must be true for all positive integers and this concludes our proof. We therefore need to show if S has at least one element, then this leads to a contradiction. If S is non-empty, applying the well-ordering principle guarantees the existence of a positive integer m 0 in S such that m 0 is the least element in S (i.e. m 0 m, m S). Now, since P(1) is true, we know that m 0 1 m 0 1 0. By hypothesis, P(m 0 1) must be true since m 0 is the least element of S. But the second condition of the principle of mathematical induction implies that P(m 0 11) P(m 0 ) is true. We have thus reached our contradiction.
Proofs by Mathematical Induction Here we take a look at some note-worthy proofs by mathematical induction. The results are taken from a variety of areas in mathematics (algebra, combinatorics, number theory, geometry, etc.) and can be readily referenced from other sources. We leave some parts of the proofs as exercises. n n 1 1. The sum of the first n positive integers is equal to. This is Formula (1): 1 3... n nn 1. Classic proof. Omitted. The sum of the first n odd positive integers is equal to the n th square number. This is Formula (): 1 35...n 1 n Trivial. If n 1, then 1 1. k. Assume that 1 35... k 1 Then, 1 3 5...k 1 k 1 k k 1 k 1. 3. The sum of the cubes of the first n positive integers is equal to the square of the sum of the first n positive integers. This is Formula (3): 3 3 3 3 1 3... n 1 3... n Trivial. If n 1, then 1 3 1.
4. Every positive integer can be written as the sum of distinct powers of (e.g. 9 0 3 4 ). (Strong induction) Trivial since 1 0. Assume that all positive integers m, such that m k, can be written as distinct powers of. Then we need to show that k 1 can also be written as distinct powers of. There are two cases to consider: either k 1 is odd or k 1 is even. k 1 Suppose k 1 is even. Then, k and, by the inductive hypothesis, there exists distinct powers p i, i 1,,...,n, such that k 1 0 p 0 p 1... p n and p 0 p 1... p n. Multiplying by then yields k 1 p 0 p 1... p n k 1 p 0 1 p 1 1... p n 1 Suppose k 1 is odd. Then k is even and is, thus, expressible in terms of distinct positive powers of. In other words, k p 0 p 1...p n for 0 p 0 p 1... p n. This implies that k 1 0 p 0 p 1...p n is also expressible as a sum of distinct powers of. 5. The difference of a cube to itself is always a multiple of 3. In other words, n 3 n is divisible by 3. Trivial since 1 3 1 0 0 3.
n n 3 6. The number of diagonals in a convex n gon is equal to. Note that there exists no 1-gon, or -gon. A triangle (3-gon) has no diagonals, which is consistent with the formula: 3(3 3) 0. Use the following figure:
7. Suppose the USPS only had - and 5-cent stamps. Any package cost of 4 cents or more can be paid for with a combination of these two types of stamps. With two -cent stamps we can paid the 4 cents package. Suppose a package costing n cents, n 4, can be paid for with a combination of - and 4-cent stamps. Then we need to prove that we can pay the n 1 package with a combination of - and 4- cent stamps. There are two cases to consider: either our combination of stamps for the n cents package includes a 5-cent stamp or not (then it consists entirely of -cent stamps). In the first case, we can replace the 5-cent stamp by three -cent stamps to get n 1cents. In the second case, we can remove two -cent stamps and replace it with a 5-cent stamp to, again, get n 1cents. 8. n! n, n 4. 9. n 1 k 1 1 4 1 9... 1 n 1 n k1 10. A set with n elements has n subsets.