CSE 20: Discrete Mathematics for Computer Science Prof. Shachar Lovett

Today’s Topics: Strong vs regular indiction Strong induction examples: Divisibility by a prime Recursion sequence: product of fractions

1. Strong induction examples DIVISIBILITY BY A PRIME

Strong vs regular induction Prove: n1 P(n) Base case: P(1) Regular induction: P(n)P(n+1) Strong induction: (P(1)…P(n))P(n+1) Can use more assumptions to prove P(n+1) P(1) P(2) P(3) … P(n) P(n+1) … P(1) P(2) P(3) … P(n) P(n+1) …

Example for the power of strong induction Theorem: For all prices p >= 8 cents, the price p can be paid using only 5-cent and 3-cent coins Proof: Base case: 8=3+5, 9=3+3+3, 10=5+5 Assume it holds for all prices 8..p-1, prove for price p when p11 Proof: since p-38 we can use the inductive hypothesis for p-3. To get price p simply add another 3-cent coin. Much easier than standard induction!

2. Strong induction examples DIVISIBILITY BY A PRIME

Definitions and properties for this proof n is prime if n is composite if n=ab for some 1<a,b<n Prime or Composite exclusivity: All integers greater than 1 are either prime or composite (exclusive or—can’t be both). Definition of divisible: n is divisible by d iff n = dk for some integer k. 2 is prime (you may assume this; it also follows from the definition).

Definitions and properties for this proof (cont.) Goes without saying at this point: The set of Integers is closed under addition and multiplication Use algebra as needed

Thm: For all integers n greater than 1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = ________. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof.

Thm: For all integers n greater than 1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = ________. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. 1 2 3 Other/none/more than one

Thm: For all integers n greater than 1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = _2______. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. For some integer n>1, n is divisible by a prime number. For some integer n>1, k is divisible by a prime number, for all integers k where 2kn. Other/none/more than one

Thm: For all integers n greater than 1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = _2______. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. n+1 is divisible by a prime number. k+1 is divisible by a prime number. Other/none/more than one

Thm: For all integers n greater than 1, n is divisible by a prime number. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n=2. Inductive step: Assume that for some n2, all integers 2kn are divisible by a prime. WTS that n+1 is divisible by a prime. Proof by cases: Case 1: n+1 is prime. n+1 divides itself so we are done. Case 2: n+1 is composite. Then n+1=ab with 1<a,b<n+1. By the induction hypothesis, since an there exists a prime p which divides a. So p|a and a|n+1. We’ve already seen that this Implies that p|n+1 (in exam – give full details!) So the inductive step holds, completing the proof.

2. Strong induction examples RECURSION SEQUENCE: PRODUCT OF FRACTIONS

Definitions and properties for this proof Product less than one: Algebra, etc., as usual

Definition of the sequence: d1 = 9/10 d2 = 10/11 dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = ________. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof.

Other/none/more than one Definition of the sequence: d1 = 9/10 d2 = 10/11 dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = ________. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. 1 2 3 Other/none/more than one

For some int n>0, 0<dn<1. Definition of the sequence: d1 = 9/10 d2 = 10/11 dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = _1,2_____. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. For some int n>0, 0<dn<1. For some int n>1, 0<dk<1, for all integers k where 1kn. Other/none/more than one

For some int n>0, 0<dn<1. Definition of the sequence: d1 = 9/10 d2 = 10/11 dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n = _1,2_____. Inductive step: Assume [or “Suppose”] that WTS that So the inductive step holds, completing the proof. For some int n>0, 0<dn<1. For some int n>1, 0<dk<1, for all integers k where 1kn. 0<dn+1<1 Other/none/more than one

Definition of the sequence: d1 = 9/10 d2 = 10/11 dk = (dk-1)(dk-2) for all integers k3 Thm: For all integers n>0, 0<dn<1. Proof (by strong mathematical induction): Basis step: Show the theorem holds for n=1,2. Inductive step: Assume [or “Suppose”] that the theorem holds for n2. WTS that 0<dn+1<1. By definition, dn+1=dn dn-1. By the inductive hypothesis, 0<dn-1<1 and 0<dn<1. Hence, 0<dn+1<1. So the inductive step holds, completing the proof.

3. Fibonacci numbers Verifying a solution

Fibonacci numbers 1,1,2,3,5,8,…

Fibonacci numbers 1,1,2,3,5,8,13,21,34,55,…

Fibonacci in nature Many more examples: http://jwilson.coe.uga.edu/emat6680/parveen/fib_nature.htm

Fibonacci numbers 1,1,2,3,5,8,13,21,… Rule: F1=1, F2=1, Fn=Fn-2+Fn-1. Question: can we derive an expression for the n-th term? YES!

Fibonacci numbers Rule: F1=1, F2=1, Fn=Fn-2+Fn-1. We will prove an upper bound: Proof by strong induction. Base case: n=1 n=2 n=1 and n=2 n=1 and n=2 and n=3 Other

Fibonacci numbers Rule: F1=1, F2=1, Fn=Fn-2+Fn-1. We will prove an upper bound: Proof by strong induction. Base case: n=1 n=2 n=1 and n=2 n=1 and n=2 and n=3 Other

Fibonacci numbers Rule: F1=1, F2=1, Fn=Fn-2+Fn-1. We will prove an upper bound: Proof by strong induction. Base case: n=1, n=2. Verify by direct calculation

Fibonacci numbers Rule: F1=1, F2=1, Fn=Fn-2+Fn-1. Theorem: Base cases: n=1,n=2 Inductive step: show… Fn=Fn-1+Fn-2 FnFn-1+Fn-2 Fn=rn Fn rn Other

Fibonacci numbers Rule: F1=1, F2=1, Fn=Fn-2+Fn-1. Theorem: Base cases: n=1,n=2 Inductive step: show… Fn=Fn-1+Fn-2 FnFn-1+Fn-2 Fn=rn Fn rn Other

Fibonacci numbers Inductive step: need to show , What can we use? Definition of Fn: Inductive hypothesis: That is, we need to show that

Fibonacci numbers Finishing the inductive step. Need to show: Simplifying, need to show: Choice of actually satisfied (this is why we chose it!) QED

Fibonacci numbers - recap Recursive definition of a sequence Base case: verify for n=1, n=2 Inductive step: Formulated what needed to be shown as an algebraic inequality, using the definition of Fn and the inductive hypothesis Simplified algebraic inequality Proved the simplified version