Module 15 Inductive Proofs a brief introduction Rosen

Module #15 – Inductive Proofs § 3. 3: Mathematical Induction • A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large. • Essentially a “domino effect” principle. • Based on this inference rule: P(0) “The First Principle of Mathematical n 0 (P(n) P(n+1)) Induction” n 0 P(n) 9/18/2021 (c)2001 -2003, Michael P. Frank 2

Module #15 – Inductive Proofs The “Domino Effect” • Premise #1: Domino #0 falls. • Premise #2: For every n N, if domino #n falls, then so does domino #n+1. • Conclusion: All of the dominoes fall 2 down! 1 6 5 4 3 0 9/18/2021 (c)2001 -2003, Michael P. Frank 3

Module #15 – Inductive Proofs The “Domino Effect” • Premise #1: Domino #0 falls. • Premise #2: For every n N, if domino #n falls, then so does domino #n+1. • Conclusion: All of the dominoes fall 2 down! 1 0 9/18/2021 (c)2001 -2003, Michael P. Frank 6 4 … 5 Note: this works even if there are infinitely many dominoes! 3 4

Module #15 – Inductive Proofs • First premiss: the basis step (also: base case)

Module #15 – Inductive Proofs Validity of Induction Proof that k>0 P(k) is a valid conclusion: Consider any k>0. The 2 nd premiss n 0 (P(n) P(n+1)) implies that (P(0) P(1)) (P(1) P(2)) … (P(k 1) P(k)). Premiss #1 says that P(0). One application of Modus Ponens (in combination with (P(0) P(1))) gives us P(1); another gives us P(2) , and so on to P(k) (using k steps of Modus Ponens in total)). Thus k>0 P(k). ■ 9/18/2021 (c)2001 -2003, Michael P. Frank 6

Module #15 – Inductive Proofs The Well-Ordering Property • A more general way to prove the validity of the inductive inference rule is by using the fact that (N, ) is a well-ordering: – Every non-empty set of natural numbers has a least (“smallest”) element. – S N : m S : n S : m n • This implies that {n| P(n)} (if non-empty!) has a min. element m, but then the assumption that P(m− 1) P((m− 1)+1) would be contradicted. 9/18/2021 (c)2001 -2003, Michael P. Frank 7

Module #15 – Inductive Proofs have a fixed structure: • Suppose we want to prove n P(n)… – Basis step: Prove P(0). – Inductive step: Prove n P(n) P(n+1). • • • E. g. you could use a direct proof, as follows: Assume P(n). (inductive hypothesis = IH) Now, under this assumption, prove P(n+1). – If you manage to do all this, then the inductive inference rule gives us n P(n). 9/18/2021 (c)2001 -2003, Michael P. Frank 8

Module #15 – Inductive Proofs • Induction is like heridity: • “If Adam had big feet, and big feet are heriditary then every person (except possibly Eve) has/had big feet” • (Note: what if a person’s feet can be just slightly smaller than that of both of his parents? ) 9/18/2021 (c)2001 -2003, Michael P. Frank 9

Module #15 – Inductive Proofs Generalizing Induction slightly • Rule can also be used to prove n c P(n) for a given constant c Z, where maybe c 0. – In this circumstance, the base case is to prove P(c) rather than P(0), and the inductive step is to prove n c (P(n) P(n+1)). 9/18/2021 (c)2001 -2003, Michael P. Frank 10

Module #15 – Inductive Proofs Proof by induction (1 st princ): a very simple example • Theorem: x N (x 0). Proof: • Basis step: we show that 0 0. This is obvious, since is reflexive. • Inductive step: we show that x (x 0 x+1 0). Induction Hypothesis (IH): Suppose a 0. Clearly, a+1 a. Transitivity allows us to infer from these two propositions that a+1 0. QED 9/18/2021 (c)2001 -2003, Michael P. Frank 12

Module #15 – Inductive Proofs Find the error • “Theorem”: All sets of horses are ‘uniform’, in the sense that all their elements have the same colour. • Method: induction over the cardinality of the set • P(x): all sets of cardinality x are uniform “Proof”: • Basis step: P(1) (This is trivial) 9/18/2021 (c)2001 -2003, Michael P. Frank 13

Module #15 – Inductive Proofs Find the error • “Theorem”: All sets of horses are uniform • Induction Hypothesis: P(n). To prove: P(n+1). Proof: Consider a set S= {h 1, h 2, …, hn+1} of n+1 horses. Split S into its subsets A and B: A= {h 1, h 2, …hn} B= {h 2, h 3, …, hn+1} Induction Hyp implies that A is uniform and B is uniform. A B , so A B is uniform. QED 9/18/2021 (c)2001 -2003, Michael P. Frank 14

Module #15 – Inductive Proofs Second Example (1 st princ. ) • Prove that the sum of the first n odd positive integers is n 2. That is, prove: • Proof by induction. P(n) – Base case: Let n=1. The sum of the first 1 odd positive integer is 1, which equals 12. (Cont…) 9/18/2021 (c)2001 -2003, Michael P. Frank 15

Module #15 – Inductive Proofs Example cont. – Inductive step: Prove n 1: P(n) P(n+1). Let n 1, assume P(n) (IH) , and prove P(n+1). By inductive hypothesis P(n) 9/18/2021 (c)2001 -2003, Michael P. Frank 16

Module #15 – Inductive Proofs Proving the induction step of the first principle of m. i. To prove the universally quantified statement n 1: P(n) P(n+1) , you do this: – Taking an arbitrary n , you hypothesise P(n) – If this allows you to prove P(n+1) then you have proved n 1: P(n) P(n+1) 9/18/2021 (c)2001 -2003, Michael P. Frank 17

Module #15 – Inductive Proofs Second Principle of Induction A. k. a. “Strong Induction” • Characterized by another inference rule: P(0) P is true in all previous cases n 0: ( 0 k n P(k)) P(n+1) n 0: P(n) • The only difference between this and the 1 st principle is that: – the inductive step here makes use of the stronger hypothesis that P(k) is true for all smaller numbers k<n+1, not just for k=n. 9/18/2021 (c)2001 -2003, Michael P. Frank 18

Module #15 – Inductive Proofs Proving the induction step of the second principle of m. i. To prove the universally quantified statement n 0: (( 0 k n P(k)) P(n+1)) you do this: – Taking an arbitrary n 0 , you hypothesise 0 k n P(k) – If this allows you to prove P(n+1) then you have proved n 0: (( 0 k n P(k)) P(n+1)) 9/18/2021 (c)2001 -2003, Michael P. Frank 19

Module #15 – Inductive Proofs Example of Second Principle Theorem: Show that every n>1 can be written as a product pi = p 1 p 2…ps of prime numbers. – Let P(n)=“n has that property” • Definition: x is a prime number x is only dividable by 1 and by x • Base case: n=2, let s=1, p 1=2. • Inductive step: Let n 2. IH: assume 2 k n: P(k). Consider n+1. If it’s prime, then let s=1, p 1=n+1. Else …. . 9/18/2021 (c)2001 -2003, Michael P. Frank 20

Module #15 – Inductive Proofs Example of Second Principle Theorem: Show that every n>1 can be written as a product pi = p 1 p 2…ps of some series of s prime numbers. – Let P(n)=“n has that property” • Base case: n=2, let s=1, p 1=2. • Inductive step: Let n 2. IH: assume 2 k n: P(k). Consider n+1. If it’s prime, then let s=1, p 1=n+1. Else n+1=ab, where 1<a n and 1<b n. Then (by IH) a=p 1 p 2…pt and b=q 1 q 2…qu. Consequently, n+1 = p 1 p 2…pt q 1 q 2…qu, a product of s=t+u primes. 9/18/2021 (c)2001 -2003, Michael P. Frank 21

Module #15 – Inductive Proofs Application to the cardinality of sets • Theorem: for all finite sets A, if |A|=n then P(A) = 2 n. Proof: • Base step: if |A|=0 then P(A) = 20=1. Why? 9/18/2021 (c)2001 -2003, Michael P. Frank 22

Module #15 – Inductive Proofs Application to the cardinality of sets • Theorem: for all finite sets A, if |A|=n then P(A) = 2 n. Proof: • Base step: if |A|=0 then P(A) = 20=1. Why? P( )={ } and |{ }|=1. 9/18/2021 (c)2001 -2003, Michael P. Frank 23

Module #15 – Inductive Proofs Application to the cardinality of sets • Theorem: for all finite sets A, if |A|=n then P(A) = 2 n. Proof: • Base step: if |A|=0 then P(A) = 20=1. Why? P( )={ } and |{ }|=1. • I. H. : suppose for every set A such that |A|=n, it holds that P(A) = 2 n Then consider any set B such that |B|=n+1. To prove: |P(B)|=2 n+1 9/18/2021 (c)2001 -2003, Michael P. Frank 24

Module #15 – Inductive Proofs Application to the cardinality of sets • I. H. : suppose for every set A such that |A|=n, it holds that P(A) = 2 n. Then consider any set B such that |B|=n+1. To prove: |P(B)|=2 n+1 • Proof: B has twice as many subsets as A. (Think of B as consisting of a set B’ such that |B’|=|A|=n, plus one other element b. Every subset of B’ comes in two flavours: one with and one without b. ) Thus, |P(B)| = 2. |P(A)| = 2. 2 n (because of IH) = 2 n+1 9/18/2021 (c)2001 -2003, Michael P. Frank 25

Module #15 – Inductive Proofs • Mathematical Induction is valid on all wellordered structures (as we have proven). • We have seen examples where it is applied to the structure (N, ). More generally, ({x Z|x>n}, ). • Integers are everywhere, and this makes Mathematical Induction very powerful 9/18/2021 (c)2001 -2003, Michael P. Frank 26

Module #15 – Inductive Proofs Something else: `Nonmathematical` induction • We have talked about mathematical induction. • In physical sciences (and in ordinary life), people use a much less precise kind of inductive hypothesis to make inferences. • E. g. : “If the sun has come up every day up to a given day, then it will come up the next day”. • You can’t prove that! Non-mathematical induction may be highly plausible, but its induction step is never logically valid. 9/18/2021 (c)2001 -2003, Michael P. Frank 27