SupposeSis a set of natural numbers that has the following two properties:
A. The number 1 is inS.
B. Whenever a natural number is inS, the next natural number is also inS.
The second property can be stated a little more formally: Ifkis a natural number andkis inS, thenkC1is inS.
What can we say about a setSthat has those two properties? Since 1 is inS(by property A), it follows from property B that 2 is inS. Since 2 is inS, it follows from property B that 3 is inS. Since 3 is inS, 4 is inS. Then 5 is inS, 6 is inS, 7 is inS, and so on. It seems clear thatSmust contain every natural number. That is, the only set of natural numbers with the above two properties is the set of all natural numbers. We state this formally:
The Principle of Mathematical Induction 2.1.1. IfS is any set of natural num- bers with the properties that
A. 1 is inS, and
B. kC1is inSwheneverkis any number inS, thenSis the set of all natural numbers.
10 2 Mathematical Induction
We gave an indication above of why the Principle of Mathematical Induction is true. A more formal proof can be based on the following more obvious fact, which we assume as an axiom.
The Well-Ordering Principle 2.1.2. Every set of natural numbers that contains at least one element has a smallest element in it.
We can establish the Principle of Mathematical Induction from the Well-Ordering Principle as follows. Suppose that the Well-Ordering Principle holds for all sets of natural numbers. Let S be any set of natural numbers and suppose that S has properties A and B of the Principle of Mathematical Induction. To prove the Principle of Mathematical Induction, we must prove that the only such setS is the set of all natural numbers. We will do this by showing that it is impossible that there is any natural number that is not inS. To see this, suppose thatSdoes not contain all natural numbers. Then letT denote the set of all natural numbers that are not inS. Assuming thatSis not the set of all natural numbers is equivalent to assuming that T has at least one element. If this were the case, then well-ordering would imply thatT has a smallest element. We will show that this is impossible.
Suppose thatt was the smallest element ofT. Since 1 is inS, 1 is not inT. Therefore,t is larger than 1, sot1is a natural number. Sincet1is less than the smallest numbert inT,t1cannot be inT. SinceT contains all the natural numbers that are not inS, it follows thatt1is inS. This, however, leads to the following contradiction. SinceS has property B,.t 1/C1 must also be inS. But this ist, which is inT and therefore not inS. This shows that the assumption that there is a smallest element ofT is not consistent with the properties ofS. Thus, there is no smallest element ofTand, by well-ordering, there is therefore no element inT. This proves thatSis the set of all natural numbers.
The way mathematical induction is usually explained can be illustrated by considering the following example. Suppose that we wish to prove, for every natural numbern, the validity of the following formula for the sum of the firstn natural numbers:
1C2C3C C.n1/CnD n.nC1/
2
One way to prove that this formula holds for every nis the following. First, the formula does hold forn D 1, for in this case the left-hand side is just 1 and the right-hand side is1.12C1/, which is equal to 1. To prove that the formula holds for all n, we will establish the fact that whenever the formula holds for any given natural number, the formula will also hold for the next natural number. That is, we will prove that the formula holds forn D kC1whenever it holds forn D k. (This passage fromktokC1is often called “the inductive step.”) If we prove this fact, then, since we know that the formula does hold fornD1, it would follow from this fact that it holds for the next natural number, 2. Then, since it holds forn D 2, it holds for the natural number that follows 2, which is 3. Since it holds for 3, it holds for 4, and then for 5, and 6, and so on. Thus, we will conclude that the formula holds
2.1 The Principle of Mathematical Induction 11
for every natural number. (This is really just the Principle of Mathematical Induction as we formally stated it above. IfSis the set of allnfor which the formula for the sum is true, showing thatShas properties A and B leads to the conclusion thatSis the set of all natural numbers.)
To prove the formula in general, then, we must show that the formula holds for n D kC1whenever it holds forn D k. Assume that the formula does hold for nDk, wherekis any fixed natural number. That is, we assume the formula
1C2C3C C.k1/CkD k.kC1/
2
We want to derive the formula fornDkC1from the above equation. That is easy to do, as follows. Assuming the above formula, addkC1to both sides, getting
1C2C3C C.k1/CkC.kC1/D k.kC1/
2 C.kC1/
We shall see that a little algebraic manipulation of the right-hand side of the above will produce the formula fornDkC1. To see this, simply note that
k.kC1/
2 C.kC1/D k.kC1/
2 C 2.kC1/
2 D k.kC1/C2.kC1/
2 D .kC2/.kC1/
2 D .kC1/.kC2/
2 D .kC1/
.kC1/C1 2
Thus,
1C2C3C C.k1/CkC.kC1/D .kC1/
.kC1/C1 2
This equation is the same as that obtained from the formula by substitutingkC1 for n. Therefore we have established the inductive step, so we conclude that the formula does hold for alln.
There are many very similar proofs of similar formulas.
Theorem 2.1.3. For every natural numbern,
12C22C32C Cn2 D n.nC1/.2nC1/
6
12 2 Mathematical Induction
Proof. LetS be the set of all natural numbers for which the theorem is true. We want to show thatScontains all of the natural numbers. We do this by showing that Shas properties A and B.
For property A, we need to check that 12 D 1.1C1/.261C1/. This is true, so S satisfies property A. To verify property B, letkbe inS. We must show thatkC1is inS. Sincekis inS, the theorem holds fork. That is,
12C22C32C Ck2D k.kC1/.2kC1/
6
Using this formula, we can prove the corresponding formula forkC1as follows.
Adding.kC1/2to both sides of the above equation, we get 12C22C32C Ck2C.kC1/2D k.kC1/.2kC1/
6 C.kC1/2 Now we do some algebraic manipulations to the right-hand side to see that it is what we want:
k.kC1/.2kC1/
6 C.kC1/2D k.kC1/.2kC1/C6.kC1/2 6
D .kC1/
k.2kC1/C6.kC1/
6 D .kC1/
.2k2Ck/C.6kC6/
6 D .kC1/.2k2C7kC6/
6
D .kC1/.kC2/.2kC3/
6
The last equation is the formula in the case when n D k C 1, so k C 1 is inS. Therefore,S is the set of natural numbers by the Principle of Mathematical
Induction. ut
Sometimes one wants to prove something by induction that is not true for all natural numbers, but only for those bigger than a given natural number. A slightly more general principle that we can use in such situations is the following.
The Generalized Principle of Mathematical Induction 2.1.4. Letmbe a natural number. IfSis a set of natural numbers with the properties that
A. mis inS, and
B. kC1is inSwheneverkis inSand is greater than or equal tom, thenScontains every natural number greater than or equal tom.
2.1 The Principle of Mathematical Induction 13
The Principle of Mathematical Induction is the special case of the generalized principle whenm D 1. The generalized principle states that we can use induction starting at any natural number, not just at 1.
For example, consider the question: which is larger,nŠor2n? Recall thatnŠ D n.n1/.n2/ 321. FornD1, 2, and 3, we see that
1ŠD1 < 21D2 2ŠD21D2 < 22D22D4 3ŠD321D6 < 23D222D8 But whennD4, the inequality is reversed, since
4ŠD4321D24 > 24D2222D16 WhennD5,
5ŠD54321D120 > 25D22222D32
If you think about it a bit, it is clear why eventually nŠis much bigger than2n. In both expressions we are multiplyingnnumbers together, but for2nwe are always multiplying by 2, whereas the numbers we multiply to buildnŠget larger and larger.
While it is not true thatnŠ > 2nfor every natural number (since it is not true fornD 1, 2, and 3), we can, as we now show, use the more general form of mathematical induction to prove that it is true for all natural numbers greater than or equal to 4.
Theorem 2.1.5. nŠ > 2nforn4.
Proof. We use the Generalized Principle of Mathematical Induction withm D 4.
LetSbe the set of natural numbers for which the theorem is true. As we saw above, 4Š > 24. Therefore,4is inS. Thus, property A is satisfied. For property B, assume thatk 4and thatkis inS; i.e.,kŠ > 2k. We must show that.kC1/Š > 2kC1. Multiplying both sides of the inequality fork(which we have assumed to be true) bykC1gives
.kC1/.kŠ/ > .kC1/2k
The left-hand side is just.kC1/Š; therefore we have the inequality .kC1/Š > .kC1/2k
Sincek4,kC1 > 2. Therefore, the right-hand side of the inequality,.kC1/2k, is greater than22kD2kC1. Combining this with the above inequality, we get
.kC1/Š > .kC1/2k> 2kC1
14 2 Mathematical Induction
Thus,kC1 is in S, which verifies property B. By the Generalized Principle of Mathematical Induction,S contains all natural numbers greater than or equal to 4.
u t The following is an example where mathematical induction is useful in establish- ing a geometric result. We will use the word “tromino” to denote an L-shaped object consisting of three squares of the same size. That is, a tromino looks like this:
Another way to think of a tromino is that it is the geometric figure obtained by taking a square that is composed of four smaller squares and removing one of the smaller squares.
We are going to consider what geometric regions can be covered by trominos, all of which have the same size and that do not overlap each other. As a first example, start with a square made up of 16 smaller squares (i.e., a square that is “4 by 4”) and remove one small square from a corner of the square:
Can the region that is left be covered by trominos (each made up of three small squares of the same size as the small squares in the region) that do not overlap each other? It can:
We can use mathematical induction to prove the following.
Theorem 2.1.6. For each natural numbern, consider a square consisting of 22n smaller squares. (That is, a2n2nsquare.) If one of the smaller squares is removed from a corner of the large square, then the resulting region can be completely covered by trominos (each made up of three small squares of the same size as the small squares in the region) in such a way that the trominos do not overlap.
Proof. To begin a proof by mathematical induction, first note that the theorem is certainly true fornD1; the region obtained after removing a small corner square is a tromino, so it can be covered by one tromino.
Suppose that the theorem is true forn D k. That is, we are supposing that if a small corner square is removed from any2k2ksquare consisting of22ksmaller
2.1 The Principle of Mathematical Induction 15
squares, then the resulting region can be covered by trominos. The proof will be established by the Principle of Mathematical Induction if we can show that the same result holds forn D kC1. Consider, then, any2kC12kC1square consisting of smaller squares. Remove one corner square to get a region that looks like this:
The region can be divided into four “medium-sized” squares that are each2k2k, like this:
16 2 Mathematical Induction
Now place a tromino in the middle of the region, as illustrated below.
The four “medium-sized” squares of the region are each2k2k and, because of the tromino in the middle, the “medium-sized” squares remaining to be covered each have one corner covered or missing.
By the inductive hypothesis, trominos can be used to cover the rest of each of the four “medium-sized” squares. This leads to a covering of the entire2kC12kC1 square, thus finishing the proof by mathematical induction. ut