Mathematical Induction: Proofs and Examples, Slides of Discrete Mathematics

Download Mathematical Induction: Proofs and Examples and more Slides Discrete Mathematics in PDF only on Docsity! Induction and recursion Chapter 5 Edited by Shih-Tsung Liang Chapter Summary Mathematical Induction Strong Induction Well-Ordering Recursive Definitions Structural Induction Principle of Mathematical Induction Principle of Mathematical Induction: To prove that P(n) is true for all positive integers n, we complete these steps: — Basis Step: Show that P(1) is true. — Inductive Step: Show that P(k) > P(k +1) is true for all positive integers k. To complete the inductive step, assuming the inductive hypothesis that P(k) holds for an arbitrary integer k, show that must P(k + 1) be true. Climbing an Infinite Ladder Example: — BASIS STEP: By (1), we can reach rung 1. — INDUCTIVE STEP: Assume the inductive hypothesis that we can reach rung k. Then by (2), we can reach rung k + 1. Hence, P(k) — P(k + 1) is true for all positive integers k. We can reach every rung on the ladder. < Important Points About Using Mathematical Induction ¢ Mathematical induction can be expressed as the rule of inference (P(1) A Vk (P(k) > P(k +1))) ~ Wn P(n), where the domain is the set of positive integers. * Ina proof by mathematical induction, we don’t assume that P(k) is true for all positive integers! We show that if we assume that P(k) is true, then P(k + 1) must also be true. * Proofs by mathematical induction do not always start at the integer 1. In such a case, the basis step begins at a starting point b where b is an integer. We will see examples of this soon. Validity of Mathematical Induction * Mathematical induction is valid because of the well ordering ( =) property, which states that every nonempty subset of the set of positive integers has a least element (see Section 5.2 and Appendix 1). Here is the proof: Suppose that P(1) holds and P(k) > P(k + 1) is true for all positive integers k. Assume there is at least one positive integer n for which P(n) is false. Then S={n€Z*| P(n) is false} is nonempty. By the well-ordering property, S has a least element, say m. (P(m) is false) We know that m can not be 1 since P(1) holds. Since mis positive and greater than 1, m- 1 must be a positive integer. Since m - 1 < m, it is not in S, so P(m - 1) must be true. But then, since the conditional P(k) > P(k +1) for every positive integer k holds, P(m) must also be true. This contradicts P(m) being false. Hence, P(n) must be true for every positive integer n. Conjecturing and Proving Correct a Summation Formula Example: Conjecture and prove correct a formula for the sum of the first n positive odd integers. Then prove your conjecture. Solution: We have: 1=1,1+3=4,14+3+4+5=9, 14+3454+7=161+3+54+7+9=25. — We can conjecture that the sum of the first n positive odd integers is n?, 14+34+54+:+(2n —1)4+(2n+1) =n’. — We prove the conjecture is proved correct with mathematical induction. — BASIS STEP: P(1) is true since 12 = 1. — INDUCTIVE STEP: P(k) > P(k + 1) for every positive integer k. Assume the inductive hypothesis holds and then show that P(k+1) holds has well. Inductive Hypothesis: 1+ 3 +5 +++ (2k —1) =k? So, assuming P(k), it follows that: 14345 4+ (2k —1) + (2k41) =[1 +345 +--+ (2k —1)] + (2k4+1) =k? + (2k+1) (by the inductive hypothesis) =k+2k+1 =(k+1)? Hence, we have shown that P(k + 1) follows from P(k). Therefore the sum of the first n positive odd integers is n?. <4 Proving Inequalities Example: Use mathematical induction to prove that n< 2” for all positive integers n. Solution: Let P(n) be the proposition that n < 2". — BASIS STEP: P(1) is true since 1 < 2! = 2. — INDUCTIVE STEP: Assume P(k) holds, i.e., k < 2‘, for an arbitrary positive integer k. — Must show that P(k + 1) holds. Since by the inductive hypothesis, k < 2k, it follows that: k+1<2k4+1 $2k +2k =2-2k = 21 Therefore n < 2” holds for all positive integers n. < Proving Inequalities Example: Use mathematical induction to prove that 2”< nl, for every integer n2 4. Solution: Let P(n) be the proposition that 2” <n!. — BASIS STEP: P(4) is true since 24 =16 <4! = 24. — INDUCTIVE STEP: Assume P(k) holds, i.e., 2k <k! for an arbitrary integer k > 4. To show that P(k + 1) holds: Dk+1 = D+Dk <2-k! (by the inductive hypothesis) < (k + 1)k! =(k+1)! Therefore, 2” <n! holds, for every integer n> 4. < Note that here the basis step is P(4), since P(0), P(1), P(2), and P(3) are all false. Tiling Checkerboards Example: Show that every 2” x2" checkerboard with one square removed can be tiled using right triominoes. A right triomino(=§hX) is an L-shaped tile which covers three squares at a time. Solution: Let P(n) be the proposition that every 2” x2” checkerboard with one square removed can be tiled using right triominoes. Use mathematical induction to prove that P(n) is true for all positive integers n. — BASISSTEP: P(1) is true, because each of the four 2 x2 checkerboards with one square removed can be tiled using one right triomino. — INDUCTIVE STEP: Inductive Hypothesis: Assume that P(k) is true, i.e., every 2‘ x2 checkerboard, for some positive integer k, with one square removed can be tiled using right triominoes. continued > Tiling Checkerboards It must be shown that under the assumption of the inductive hypothesis, P(k+1) must also be true; thatis, every 24! x2‘*1 checkerboard with one square removed can be tiled using right triominoes. * Considera 2’! x21 checkerboard with one square removed. Split this checkerboard into 4 checkerboards of size 2 x2‘ by dividingit in half in both directions. a F, * Remove a square from one of the four 2‘ x2‘ checkerboards. By the inductive hypothesis, this board can be tiled. Also by the inductive hypothesis, the other three boards can be tiled with the square from the corner of the center of the original board removed. We can then cover the three adjacent squares with a triominoe. * Hence, the entire 241 x2‘! checkerboard with one square removed can be tiled using right triominoes. This completes the inductive step. — We have completed the basis step and the inductive step. Therefore, by <4 mathematical induction P(n) is true for all positive integers n. An Incorrect “Proof” by Mathematical Induction (Reading Exercise) Example: Let P(n) be the statement that every set of n lines in the plane, no two of which are parallel, meet ina common point. Here is a “proof” that P(n) is true for all positive integers n > 2. — BASIS STEP: The statement P(2) is true because any two lines in the plane that are not parallel meet in a common point. — INDUCTIVE STEP: The inductive hypothesis is the statement that P(k) is true for the positive integer k = 2, i.e., every set of k lines in the plane, no two of which are parallel, meet in a common point. — We must show that if P(k) holds, then P(k + 1) holds, i.e., if every set of k lines in the plane, no two of which are parallel, k => 2, meet ina common point, then every set of k + 1 lines in the plane, no two of which are parallel, meet ina common point. continued > Strong Induction(sB ER Well-Ordering( 5 Section 5.2 A725) and aX YF) Strong Induction ¢ Strong Induction: To prove that P(n) is true for all positive integers n, where P(n) is a propositional function, complete two steps: — Basis Step: Verify that the proposition P(1) is true. — Inductive Step: Show the conditional statement [P(1) A P(2) A: A P(k)] — P(k + 1) holds for a positive integers k. Strong Induction is sometimes called the second principle of mathematical induction or complete induction. Strong Induction and the Infinite Ladder Strong induction tells us that we can reach all rungs if: 1. We can reach the first rung of the ladder. 2. For every integer k, if we can reach the first k rungs, then we can reach the (k + 1)st rung. To conclude that we can reach every rung by strong induction: * BASIS STEP: P(1) holds * INDUCTIVE STEP: Assume P(1) A P(2) Ax A P(k) holds for an arbitrary integer & and show that P(k + 1) must also hold. We will have then shown by strong induction that for every positive integer n, P(n) holds, i.e., we can a reach the nth rung of the ladder. “ Proof using Strong Induction Example: Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps. Solution: Let P(n) be the proposition that postage of n cents can be formed using 4-cent and 5-cent stamps. — BASIS STEP: P(12), P(13), P(14), and P(15) hold. * P(12) uses three 4-cent stamps. * P(13) uses two 4-cent stamps and one 5-cent stamp. * P(14) uses one 4-cent stamp and two 5-cent stamps. « P(15) uses three 5-cent stamps. — |NDUCTIVE STEP: The inductive hypothesis states that P(/) holds for 12 <j<k, where k215, Assuming the inductive hypothesis, it can be shown that P(k + 1) holds. — Using the inductive hypothesis, P(k — 3) holds sincek — 3 >12. To form postage of k + 1 cents, add a 4-cent stamp to the postage for k — 3 cents. Hence, P(n) holds for alln > 12. 4 Proof of Same Example using Mathematical Induction Example: Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps. Solution: Let P(n) be the proposition that postage of n cents can be formed using 4-cent and 5-cent stamps. — BASIS STEP: Postage of 12 cents can be formed using three 4-cent stamps. — INDUCTIVE STEP: The inductive hypothesis P(k) for any positive integer k is that postage of k cents can be formed using 4-cent and 5-cent stamps. To show P(k+ 1) where k > 12, we consider two cases: * Ifat least one 4-cent stamp has been used, then a 4-cent stamp can be replaced with a 5-cent stamp to yield a total of k + 1 cents. * Otherwise, no 4-cent stamp have been used and at least three 5-cent stamps were used. Three 5-cent stamps can be replaced by four 4-cent stamps to yield a total of k + 1 cents. Hence, P(n) holds for all n = 12. » Well-Ordering ( &) Property Well-ordering property: Every nonempty set of nonnegative integers has a least element. The well-ordering property is one of the axioms of the positive integers listed in Appendix 1. The well-ordering property can be used directly in proofs, as the next example illustrates. The well-ordering property can be generalized. — Definition: A set is well ordered if every subset has a least element. * Nis well ordered under <. * The set of finite strings over an alphabet using lexicographic ordering is well ordered. — We will see a generalization of induction to sets other than the integers in the next section. Recursively Defined Functions Definition: A recursive or inductive definition of a function consists of two steps. — BASIS STEP: Specify the value of the function at zero. — RECURSIVE STEP: Give a rule for finding its value at an integer from its values at smaller integers. * A function f(n) is the same as a sequence ay, Q,,..., a, where f(i) = a;. This was done using recurrence relations in Section 2.4. Recursively Defined Functions Example: Suppose f is defined by: f(0) = 3, f(n +1) = 2f(n) +3 Find f(1), f(2), f(3), (4) Solution: * f(1)=2f(0)+3=2:34+3=9 * f(2)=2f(1)}+3=2:94+3=21 * f(3)=2f(2)+3=2:214+3=45 * f(4) = 2f(3)+3=2:4543=93 Example: Give a recursive definition of the factorial function n!: Solution: f(0)=1 f(n + 1) =(n + 1): f(n) Recursively Defined Functions Example: Give a recursive definition of: Saag. k=0 Solution: The first part of the definition is 0 So a = ao- k=0 The second part is Sa _ (>: «) + Ont k=0 k=0 (a, b) = (b, r) = At Aa KT AI AD BR? 19| 2013] 102 += Bho a= bq+r? BAT (a,b) = (br) 2}-* + Bd Ma bwHRAMs HMA ab MA 5 d 48H BL r= a— bq BK dy, HH 18 %- 30 + FILA dy My b,r HAM» LM br HH By hy ABR > BBL a= bgt rH dh thy meperss eaten am: corte \ Lame’s Theorem(fi/##2 Lamé’s Theorem: Let a and b be positive integers with a > b. Then the number of divisions used by the Euclidian algorithm(l 6 3 583) to find gcd(a,b) is less than or equal to five times (5{74) the number of decimal digits in b (bf 221717 20). xz Proof: When we use the Euclidian algorithm to find gcd(a,b) with a > b, Hy (1795-1870) ¢ ndivisions are used to obtain (with a = ro,b =r, ): * Since each quotient q1, 42 5 --»Gp-1 iS at least 1 and q,, > 2: fo =the 0<n<n, Nh =hgntrs 0<1r3<h, ry 21=hf2, ; a 22r,22f,=fs tna =lni9nittn OSN< tay fy2 2 Vnat thatstfo=Sfa, Tana ="nQn- : 22 34g 2 frattn2=hn b=n>= nth, 2 fat Sfaa=Snev continued > Lamé’s Theorem (#11 #87e FE) It follows that if n divisions are used fn the Euclidian algorithm to find gcd(a,b) witha > b,then b = f,,,. By Example 4, f,,; > a”~ 1, for n> 2, where a=(1+ Je) Therefore, b> a"~1. Because log,) a ~ 0.208 > 1/5, log;, b > (n—1) log;y a > (n—1)/5 . Hence, n—1 <5 ‘logy) b. HT UY Suppose that b has k decimal digits. Then b < 10‘ and log) b<k. It follows that n — 1 <5k and since k is an integer, n<_5k. As aconsequence of Lamé’s Theorem, O(log b) divisions are used by the Euclidian algorithm to find gcd(a,b) whenever a > b. Recursively Defined Sets and Structures Recursive definitions of sets have two parts: — The basis step specifies an initial collection of elements. — The recursive step gives the rules for forming new elements in the set from those already known to be in the set. ¢ We will later develop a form of induction, called structural induction, to prove results about recursively defined sets. String Concatenation Definition: Two strings can be combined via the operation of concatenation. Let 2 be a set of symbols and 2* be the set of strings formed from the symbols in 2. We can define the concatenation of two strings, denoted by -, recursively as follows. BASIS STEP: Ifwex*,thenw-Qew neempty string RECURSIVE STEP: If w, € 2* and wz € 2* andx e€ 2, then w, ‘(w, x)= (W1" W2)x. Often w,* w, is written as w, Wp. If w, =abra and w, = cadabra, the concatenation W, W> = abracadabra. Length of a String Example: Give a recursive definition of /(w), the length of the string w. Solution: The length of a string can be recursively defined by: I(x) = 0; I(wx) = (w) + Lif wex* and xed. Well-Formed Formulae in Propositional Logic Definition: The set of well-formed formulae in propositional logic involving T, F, propositional variables, and operators from the set {,A,V,7,e}. BASIS STEP: T,F, and s, where s is a propositional variable, are well-formed formulae. RECURSIVE STEP: If E and Fare well formed formulae, then (4 £), (E AF), (EV F), (E> F), (E © F), are well-formed formulae. Examples: ((p Vq) — (q A F)) is a well-formed formula. pq is nota well formed formula. Full Binary Trees Definition: The set of fu// binary trees can be defined recursively by these steps. BASIS STEP: There is a full binary tree consisting of only a single vertex r. RECURSIVE STEP: If T, and T> are disjoint full binary trees, there is a full binary tree, denoted by 7,:T,, consisting of a root r together with edges connecting the root to each of the roots of the left subtree 7, and the right subtree T,. Building Up Full Binary Trees Basic step Apply recursive step 1 time Apply recursive step 2 times Induction and Recursively Defined Sets Example: Show that the set S defined by specifying that 3 €S and that if x eSand ye S,thenx+yisinS, is the set of all positive integers that are multiples of 3. Solution: Let A be the set of all positive integers divisible by 3. To prove that A=S, showthatA is a subset of S and S is a subset of A. — ACS: Let P(n) be the statement that 3n belongs to S. BASIS STEP: 3:1 = 3 €S, by the first part of recursive definition. INDUCTIVE STEP: Assume P(k) is true. By the second part of the recursive definition, if 3k ES, thensince 3 €S, 3k+ 3 = 3(K+ 1) €S. Hence, P(k + 1) is true. -— SCA: BASIS STEP: 3 € S by the first part of recursive definition, and 3 =3:1. INDUCTIVE STEP: The second part of the recursive definitionaddsx ty to S, if bothx and y are in S. If xandy are both in A, then both x andy are divisible by 3. By part (i) of Theorem 1 of Section 4.1, it follows that x + yis divisible by 3. * We used mathematical induction to prove a result about a recursively defined set. Next we study a more direct form induction for proving results about recursively defined sets. Structural Induction and Binary Trees Theorem: If Tis a full binary tree, then n(T) < 2'*1- 1, Proof: Use structural induction. — BASIS STEP: The result holds for a full binary tree consisting only of a root, n(T) = 1 and A(T) = 0. Hence, n(7)=1 <2%1!-1 =1. —RECURSH+INDUCTIVE STEP: Inductive hypothesis: Assume n(T,) < 2%7)+1—1 and also n(T) < 2°71 —1 whenever T, and T are full binary trees. Considering T=T,:T, n(T) = 1+n(T,) + n(T,) (by recursive formula of n(T)) $14 (2/71 4) + (24721 1) (by inductive hypothesis) < 2-max(2MTU+1 2AtT2)+1) — 4 = 2-2max(h(71),A(72)}+1 — 4 (max(2*, 2”)= 2™=*Xxy) ) = 2-2h7) —4 (by recursive definition of h(T)) = DAT _ 4 Generalized Induction Generalized induction is used to prove results about sets other than the integers that have the well-ordering property. (explored in more detail in Chapter 9) For example, consider an ordering on NXN, ordered pairs of nonnegative integers. Specify that (x, ,y;) is less than or equal to (x>,y2) if either x, < x», or x1 = xX, and y, <y2. This is called the /exicographic ordering. Strings are also commonly ordered by a lexicographic ordering. The next example uses generalized induction to prove a result about ordered pairs from NX N. Generalized Induction Example: Suppose that a,,,, is defined for (m,n) € N XN by ay)=0 and a — f Qm-tnt+1 ifm =Oandm > 0 mr") admn-1tn ifn>0 Show that a,,,,= m+ n(n + 1)/2 is defined for all (m,n)eN XN. Solution: Use generalized induction. BASIS STEP: dy g=0 = 0+ (0-1)/2 INDUCTIVE STEP: Inductive hypothesis: assume that O74,7= m+ n(n'4+ 1)/2 whenever(min) is less than (m,n) in the lexicographic ordering of N XN. * Ifn=0, by the inductive hypothesis we can conclude Onn = Im 1n+ 1=m—14+n(n+1)/2+1=m+n(n+1)/2. * Ifn> 0, by the inductivehypothesis we can conclude Onn = %mn1tn=M+n(n— 1)/2+n =m+n(n+1)/2.