Mathematical Induction

Motivation Mathematics uses 2 kinds of arguments: deductive inductive P(n): 1 + 2 + … + n = n(n + 1)/2. Observe that P(1), P(2), P(3), & P(4). Conjecture: nN, P(n). Mathematical induction is a finite proof pattern for proving nN, P(n).

The Principle of Mathematical Induction Let P(n) be a proposition function: n N, P(n) is either true or false. To prove nN P(n), it suffices to prove: P(1) is true. n ( P(n)  P(n + 1) ). We are applying modus ponens, ad infinitum. It is a framework for constructing a finite proof that n P(n).

Proving P(3) Given P(1)  n  1 (P(n)  P(n + 1)). Proof: 1. P(1). [premise 1] 2. P(1)  P(2). [U.S. of premise 2] 3. P(2). [step 1, 2, & modus ponens] 4. P(2)  P(3). [[U.S. of premise 2] 5. P(3). [step 2, 3, & modus ponens] Construct a finite proof for P(1,999,765).

Mathematical Induction as the Domino Principle If the 1st domino falls over and the nth domino falls over implies that the (n + 1)st domino falls over then domino n falls over for all n  N.

Mathematical Induction as the Domino Principle 1 2 3 n n +1

A rule of inference A tautological implication is a rule of inference. The Principle of Mathematical Induction is a rule of inference: ( P(1)  n (P(n)  P(n + 1)) )  n P(n)

The 3-Step Method Typically, the proof of step 2 is direct. Thus, the framework has 3 steps: Prove P(1). [called the basis] Assume P(n) [called the induction hypothesis] Prove P(n + 1) [called the inductive step] The last 2 steps are for arbitrary n  N.

Induction as a Creative Process Mathematical induction is similar to, but not identical to, scientific induction. In both cases, a “theory” is created. Look at specific cases; perceive a pattern. Hypothesizing a pattern, a theory, is a creative process (only people who are bad at mathematics say otherwise - sour grapes). With mathematical induction, a “theory” can be proved.

Scientific theories cannot be proved. They can be disproved. Mathematical models of scientific theories are amenable to mathematical proof. Like axioms, the relationship between: the mathematical model the physical reality cannot be proven correct.

Example 1 = 1 3 = 1 + 2 6 = 1 + 2 + 3 10 = 1 + 2 + 3 + 4 What is general formula, if any, for 1 + 2 + … + n? F(n): 1 + 2 + . . . + n. F(n + 1) = F(n) + (n + 1).

A Geometric Interpretation 1: 2: 3: Put these blocks, which represent numbers, together to form sums: 1 + 2 = 1 + 2 + 3 =

n n Area is n2/2 + n/2 = n(n + 1)/2

Mathematical Induction Proof Assume F(n) = n(n + 1)/2 Show F(n + 1) = (n + 1)(n + 2)/2. F(n + 1) = 1 + 2 + . . . + n + (n + 1) = F(n) + n + 1 = n(n + 1)/2 + n + 1 [Induction hyp.] = n(n + 1)/2 + (n + 1)(2/2) = (n + 1)(n + 2)/2.

In proving the formula by mathematical induction, we focused on the: similarities differences for successive values of n. Very roughly speaking: Isolate difference for successive values of n. Find a pattern in this sequence of differences.

Example: 13 + 23 + . . . + n3 = ? Let F(n) = 13 + 23 + . . . + n3. What is a formula for F(n)? 1 = 13 9 = 13 + 23 36 = 13 + 23 + 33 100 = 13 + 23 + 33 + 43 Do you see a pattern?

Try to prove that n F(n) = [n(n + 1)/2]2 2. Assume F(n) = [n(n + 1)/2]2 .  I.H. 3. Prove F(n + 1) = [(n + 1)(n + 2)/2]2 . F(n + 1) = 13 + 23 + . . . + n3 + (n+1)3  Defn. of F(n + 1) = F(n) + (n+1)3  Enable use of I.H. = [n(n + 1)/2]2 + (n+1)3  Use I. H. = (n+1)2[(n/2)2 + (n+1)] = (n+1)2[n 2 /4 + (4/4)(n+1)] = (n+1)2[(n 2 + 4n + 4)/4] = [(n+1) (n+2)/2]2.

Translating the starting point Suppose we know P(n) is false for 1  n  9 but think P(n) is true for n > 9. Define Q(n) = P(n + 9). Use mathematical induction to show that n N, Q(n). Thus, we can start the induction at any integer, not just 1.

Example: Stamps Suppose the US Post Office prints only 5 & 9 cent stamps. Prove n > 34, you can make postage for n cents, using only 5 & 9 cent stamps. S(n): you can make postage for n cents 1. Basis: For n = 35, use seven 5 cent stamps. 2. I.H.: S(n).

3. Prove S(n + 1). Case 1: Zero 9-cent stamps are used for S(n): only 5 cent stamps are used for S(n). Then at least 7 such stamps were used. Replace those with four 9 cent stamps. Case 2: At least one 9 cent stamp is used for S(n). Replace it with two 5 cent stamps.

Can we start from a negative number? Can we use this technique for something that is true only for positive odd integers? How general is this? (Think of composing functions)

1 = 12 1 + 3 = 22 1 + 3 + 5 = 32 What is the generalization? What is the proposition function?

Proof by induction Basis: P(0): 2(0) + 1 = (0 +1)2 Let P(n) be 1 + 3 + 5 + … + (2n + 1) = (n + 1)2 Induction: Show P(n)  P(n + 1). Assume 1 + 3 + 5 + … + (2n + 1) = (n + 1)2 Show 1 + 3 + 5 + … + (2n + 1) + (2(n+1) + 1) = (n + 2)2 1 + 3 + 5 + … + (2n + 1) + (2(n+1) + 1) = (n + 1)2 + (2(n+1) + 1) = n2 + 2n + 1 + 2n + 3 = (n + 2) 2

Generalizing the Basis To prove nN, P(n), if suffices to show: P(1)  P(2). n ( ( P(n)  P(n + 1) )  P(n + 2) ). That is, if: We can push over the first 2 dominos AND Pushing over any 2 adjacent dominos implies pushing over the domino following those. then we can push over all the dominos.

The Fibonacci Formula Let F(0) = 0, F(1) = 1, F(n) = F(n-1) + F(n-2). Prove F(n) = 5-1/2 ([(1 + 51/2)/2]n - [(1 - 51/2)/2]n ). Basis: F(0) = 5-1/2 ([(1 + 51/2)/2]0 - [(1 - 51/2)/2]0 ) = 0. F(1) = 5-1/2 ([(1 + 51/2)/2]1 - [(1 - 51/2)/2]1 ) = (5-1/2/2) (1 + 51/2 - 1 + 51/2 ) = 1.

Assume Show F(n) = 5-1/2 ([(1 + 51/2)/2]n - [(1 - 51/2)/2]n ) and 1.Let a = (1 + 51/2)/2 and b = (1 - 51/2)/2. 2. Then, by the induction hypothesis, F(n) = 5-1/2 (an - bn) and F(n+1) = 5-1/2 (an+1 - bn+1) .

3. F(n+2) = F(n+1) + F(n), by definition. 4. F(n+2) = 5-1/2 (an+1 - bn+1) + 5-1/2 (an - bn) = 5-1/2 (an + an+1 - bn - bn+1) = 5-1/2 ( an(a+1) - bn (b+1) ) 5. But, a + 1 = a2 and b + 1 = b2. 6. F(n+2) = 5-1/2 (an+2 - bn+2) .

Generalizing this ... If P(1)  P(2)  . . . P(k) and n ( (P(n+1)  P(n+2)  . . . P(n+k))  P(n+k+1) ) then, n P(n).

Strong Mathematical Induction If P(1)  P(2)  . . . P(k) and for n  k, if P(i) is true, for 1  i  n then P(n+1) then, n P(n).

Example: Fundamental Theorem of Arithmetic Prove that all natural numbers  2 can be represented as a product of primes. Basis: 2: 2 is a prime. Assume that 1, 2, . . ., n can be represented as a product of primes.

Show that n+1can be represented as a product of primes. Case n + 1 is a prime: It can be represented as a product of 1 prime, itself. Case n + 1 is composite: n + 1 = ab, for some a,b < n + 1. By induction hypothesis, a = p1p2 . . . pk & b = q1q2 . . . ql, where the pis & qis are primes. Then, n + 1 = p1p2 . . . pkq1q2 . . . ql.