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

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

3. 1. Strong induction examples
DIVISIBILITY BY A PRIME

4. 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) …

5. 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!

6. 2. Strong induction examples
DIVISIBILITY BY A PRIME

7. 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).

8. 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

9. 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.

10. 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

11. 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

12. 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

13. 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.

14. 2. Strong induction examples
RECURSION SEQUENCE: PRODUCT OF FRACTIONS

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

16. 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.

17. 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

18. 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

19. 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

20. 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.

21. 3. Fibonacci numbers
Verifying a solution

22. Fibonacci numbers
1,1,2,3,5,8,…

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

24. Fibonacci in nature Many more examples:

25. 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!

26. 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

27. 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

28. 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

29. 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

30. 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

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

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

33. 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