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Discrete Maths

Discrete Maths. 242-213 , Semester 2, 2013 - 2014. Objective to introduce mathematical induction through examples . 8. Mathematical Induction. Overview. 1 . Motivation 2 . Induction Defined 3 . Maths Notation Reminder 4 . Four Examples 5 . More General Induction Proofs

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Discrete Maths

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  1. Discrete Maths 242-213, Semester 2,2013-2014 • Objective • to introduce mathematical induction through examples 8. Mathematical Induction

  2. Overview 1. Motivation 2. Induction Defined 3. Maths Notation Reminder 4. Four Examples 5. More General Induction Proofs 6. Further Information

  3. 1. Motivation • Induction is used in mathematical proofs of recursive algorithms • e.g. quicksort, binary search • Induction is used to mathematically define recursive data structures • e.g. lists, trees, graphs Part 9 continued

  4. Part 10 • Induction is often used to derive mathematical estimates of program running time • timings based on the size of input data • e.g. time increases linearly with the number of data items processed • timings based on the number of times a loop executes

  5. 2. Induction Defined • Induction is used to solve problems such as: • is S(n) correct/true for all n values? • usually for all n >= 0 or all n >=1 • Example: • let S(n) be "n2 + 1 > 0" • is S(n) true for all n >= 1? S(n) can be much more complicated, such as a program that reads in a n value. continued

  6. How do we prove (disprove) S(n)? • One approach is to try every value of n: • is S(1) true? • is S(2) true? • ... • is S(10,000) true? • ... forever!!! Not very practical

  7. Induction to the Rescue • Induction is a technique for quickly proving S(n) true or false for all n • we only have to do two things • First show that S(1) is true • do that by calculation as before continued

  8. Second, assume that S(n) is true, and use it to show that S(n+1) is true • mathematically, we show thatS(n)  S(n+1) • Now we know S(n) is true for all n>=1. • Why? "stands for "implies" continued

  9. With S(1) and S(n)  S(n+1)then S(2) is true • S(1)  S(2) when n == 1 • With S(2) and S(n)  S(n+1)then S(3) is true • S(2)  S(3) when n == 2 • With S(3) and S(n)  S(n+1)then S(4) is true • S(3)  S(4) when n == 3 • and so on, for all n

  10. Let’s do it • Prove S(n): "n2 + 1 > 0" for all n >= 1. • First task: show S(1) is true • S(1) == 12 + 1 == 2, which is > 0 • so S(1) is true • Second task: show S(n+1) is true by assuming S(n) is true continued

  11. Assume S(n) is true, so n2+1 > 0 • Prove S(n+1) • S(n+1) == (n+1)2 + 1 == n2 + 2n + 1 +1 == (n2 + 1) + 2n + 1 • since n2 + 1 > 0, then (n2 + 1) + 2n + 1 > 0 • so S(n+1) is true, by assuming S(n) is true • so S(n)  S(n+1) continued

  12. We have used induction to show two things: • S(1) is true • S(n)  S(n+1) is true • From these it follows that S(n) is true for all n >= 1

  13. Induction More Formally • Three pieces: • 1. A statement S(n) to be proved • the statement must be about an integer n • 2. A basis for the proof. This is the statement S(b) for some integer. Often b = 0 or b = 1. continued

  14. 3. Aninductive step for the proof. We prove the statement “S(n)  S(n+1)” for all n. • The statement S(n), used in this proof, is called the inductive hypothesis • We conclude that S(n) is true for all n >= b • S(n) might not be true for some n < b

  15. 3. Maths Notation Reminder • Summation: means 1+2+3+4+…+n • e.g. means 4+9+16+…+m2 • Product: means 1*2*3*…*n

  16. 4. Example 1 • Prove the statement S(n):for all n >= 1 • e.g. 1+2+3+4 = (4*5)/2 = 10 • Basis. S(1), n = 1so 1 = (1*2)/2 (1) continued

  17. Induction. Assume S(n) to be true.Prove S(n+1), which is: • Notice that: (2) (3) continued

  18. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = (n2 + n + 2n + 2) /2 = (n2+3n+2)/2 which is (2)

  19. Example 2 • Prove the statement S(n):for all n >= 0 • e.g. 1+2+4+8 = 16-1 • Basis. S(0), n = 0so 20 = 21 -1 (1) continued

  20. Induction. Assume S(n) to be true.Prove S(n+1), which is: • Notice that: (2) (3) continued

  21. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = 2*(2n+1) - 1 which is (2) +1

  22. Example 3 • Prove the statement S(n): n! >= 2n-1for all n >= 1 • e.g. 5! >= 24, which is 120 >= 16 • Basis. S(1), n = 1: 1! >= 20so 1 >= 1 (1) continued

  23. Induction. Assume S(n) to be true.Prove S(n+1), which is: (n+1)! >= 2(n+1)-1 >= 2n (2)Notice that: (n+1)! = n! * (n+1) (3) continued

  24. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: (n+1)! >= 2n-1 * (n+1) >= 2n-1 * 2 since (n+1) >= 2 (n+1)! >= 2n which is (2) why?

  25. Example 4 (1) • Prove the statement S(n):for all n >= 1 • This proof can be used to show that. the limit of the sum:is 1 • Basis. S(1), n = 1so 1/2 = 1/2 continued

  26. Induction. Assume S(n) to be true.Prove S(n+1), which is:Notice that: (2) (3) continued

  27. Substitute the right hand side (rhs) of (1) for the first operand of (3), to give: = which is (2)

  28. 5. More General Inductive Proofs • There can be more than one basis case. • We can do a complete induction (or “strong” induction) where the proof of S(n+1) may use any of S(b), S(b+1), …, S(n) • b is the lowest basis value

  29. Example • We claim that every integer >= 24 can be written as 5a+7b for non-negative integers a and b. • note that some integers < 24 cannot be expressed this way (e.g. 16, 23). • Let S(n) be the statement (for n >= 24) “n = 5a + 7b, for a >= 0 and b >= 0” continued

  30. Basis. The 5 basis cases are 24 through 28. • 24 = (5*2) + (7*2) • 25 = (5*5) + (7*0) • 26 = (5*1) + (7*3) • 27 = (5*4) + (7*1) • 28 = (5*0) + (7*4) continued

  31. Induction uses S(n-4)  S(n+1) • Induction: Let n+1 >= 29. Then n-4 >= 24, the lowest basis case. • Thus S(n-4) is true, and we can write n - 4 = 5a + 7b • Thus, n+1 = 5(a+1) + 7b, proving S(n+1)

  32. 6. Further Information • Discrete Mathematics and its ApplicationsKenneth H. RosenMcGraw Hill, 2007, 7th edition • chapter 5, section 5.1

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