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Discrete Mathematics Tutorial 11

Discrete Mathematics Tutorial 11. Chin chlee@cse.cuhk.edu.hk. Induction. Suppose P(1) is true, and If P(n) is true, P(n+1) is also true Then P(n) is true for every n ≥ 1. 6. 5. 4. 3. 2. 1. Induction. Suppose I proved that P(1) and P(2) are true.

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Discrete Mathematics Tutorial 11

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  1. Discrete MathematicsTutorial 11 Chin chlee@cse.cuhk.edu.hk

  2. Induction • Suppose • P(1) is true, and • If P(n) is true, P(n+1) is also true • Then P(n) is true for every n ≥ 1 6 5 4 3 2 1

  3. Induction • Suppose I proved that • P(1) and P(2) are true. • If P(n) is true, then P(n+2) is true. • Can I conclude that • P(n) is true for every n?

  4. Induction • Suppose I proved that • P(1) and P(2) are true. • If P(n) is true, then P(n+2) is true. • Two stack of dominoes • P(1)→P(3)→P(5)→… • P(2)→P(4)→P(6) →… .. .. . . 7 8 5 6 3 4 1 2

  5. Induction • Suppose I proved that • P(1, 1) is true. • If P(n, m) is true, then P(n+1, m+1) is true. • Can I conclude that • P(n, m) is true for every n, m?

  6. Induction • Suppose I proved that • P(1, 1) is true. • If P(n, m) is true, then P(n+1, m+1) is true. • What about P(2,1)?

  7. Induction • Suppose I proved that • P(1, 1) is true. • If P(n, m) is true, then P(n+1, m) is true. • If P(n, m) is true, then P(n-1, m+1) is true. • Can I conclude that • P(n, m) is true for every n, m?

  8. Induction • Suppose I proved that • P(1, 1) is true. • If P(n, m) is true, then P(n+1, m) is true. • If P(n, m) is true, then P(n-1, m+1) is true. 2,3 1,3 3,2 4,1 2,2 1,2 3,1 2,1 1,1

  9. Induction • Show that ∑ i2 = n(2n+1)(n+1)/6 n i=1

  10. Induction • Show that ∑ i2 = n(2n+1)(n+1)/6 i.e. 12 + 22 + 32 + … + n2 = n(2n+1)(n+1)/6 • Base case: • when n = 1 • L.H.S. = 1 • R.H.S. = 1 x 3 x 2 / 6 = 1 n i=1

  11. Induction • Assume P(n) is true: ∑ i2 = n(2n+1)(n+1)/6 i.e. 12 + 22 + 32 + … + n2 = n(2n+1)(n+1)/6 • Then we need to show P(n+1) is also true: ∑ i2 = (n+1)(2(n+1)+1)((n+1)+1)/6 i.e. 12 + 22 + … + n2 + (n+1)2= (n+1)(2n+3)(n+2)/6 • How to use the induction assumption? n i=1 n+1 i=1

  12. Induction • Then we need to show P(n+1) is also true: ∑ i2=(n+1)(2n+3)(n+2)/6 i.e. 12 + 22 + … + n2 + (n+1)2 = (n+1)(2n+3)(n+2)/6 • How to use the induction assumption? ∑ i2 = ∑ i2 + (n+1)2 ∑ i2 = n(2n+1)(n+1)/6 + (n+1)2 n+1 i=1 n n+1 i=1 i=1 n+1 i=1

  13. Induction • Now we have ∑ i2 = n(2n+1)(n+1)/6 + (n+1)2 • Our goal is to make R.H.S. equal (n+1)(2n+3)(n+2)/6 • Just expand everything out and factorize n+1 i=1

  14. Induction n+1 ∑ i2 = n(2n + 1)(n + 1)/6 + (n + 1)2 = (2n3 + 3n2 + n)/6 + (n2 + 2n + 1) =((2n3 + 3n2 + n) + (6n2 + 12n + 6))/6 = (2n3 + 9n2 + 13n + 6)/6 How to factor? i=1

  15. Induction n+1 ∑ i2 = (2n3 + 9n2 + 13n + 6)/6 = (n + 1)(2n2 + 7n + 6)/6 = (n + 1)(2n + 3)(n + 2)/6 i=1 • How to factor? • Remember, our goal is to make R.H.S. equal (n+1)(2n+3)(n+2)/6 • So P(n+1) is also true. Done.

  16. Strong induction • Suppose • P(1) is true, and • If P(s) is true for every s < n, P(n+1) is also true • Then P(n) is true for every n ≥ 1 6 4 5 3 2 1

  17. Strong induction • Prove that To divide up a chocolate bar with m x n squares, we need at least mn - 1 splits 5 = 2x3 - 1 splits

  18. Strong induction • Prove that To divide up a chocolate bar with m x n squares, we need at least mn - 1 splits • We prove a stronger statement P(s) := To divide up a chocolate bar with s squares, we need at least s - 1 splits

  19. Strong Induction • Prove that P(s) := To divide up a chocolate bar with s squares, we need at least s-1 splits • Base caseP(1): • No need to split.

  20. Strong Induction • Prove that P(s) := To divide up a chocolate bar with s squares, we need at least s-1 splits • Induction step: • (If P(k) is true for every k < s) • Assume for any chocolate bar with 1 ≤ k < s squares,we need at least k – 1 splits.

  21. Strong induction • Induction step (If P(k) is true for every k < s) : • Assume for any chocolate bar with 1 ≤ k < s squares, we need at least k – 1 splits. • Given a chocolate bar with s squares Split it into two smaller bars with i and j squares,i + j = s i squares s squares j squares

  22. Strong induction • By induction assumption (P(k) is true for any k < s) • Small chocolate bar with i < s squares • needs at least i-1 splits for one of the two bars • Small chocolate bar with j < s squares • needs at least j-1 splits for one of the two bars • Therefore we need (i-1) + (j-1) + 1 = s-1 splits • because i + j = s • P(s) is true. Done. i-1 splits 1 split j-1 splits

  23. Well-ordering principle • Every nonempty set of nonnegative integers has a least element • This is equivalent to Mathematical Induction: • If MI is true, then we can prove WOP is also true. • If WOP is true, then we can prove MI is also true.

  24. Well-ordering principle • Prove by well-ordering principle that ∑ i = n(n+1)/2 n i=1

  25. Well-ordering principle • Prove by well-ordering principle that ∑ i = n(n+1)/2 • Prove by contradiction • Suppose there exist some m, ∑ i ≠ m(m+1)/2 • Let S be the set containing all such m • S is nonempty (exists at least one) • S is a set of nonnegative integers n i=1 m i=1

  26. Well-ordering principle • Let S be the set containing all the m ∑ i ≠ m(m+1)/2 • S is nonempty (exists at least one) • S is a set of nonnegative integers • By well-ordering principle, • there exists a least element m’ in S • i.e. m’ is the smallest number such that ∑ i ≠ m’(m’+1)/2 m i=1 m’ i=1

  27. Well-ordering principle • m’ is the smallest number such that ∑ i ≠ m’(m’+1)/2 • This means for any 0 ≤ n < m’ ∑ i = n(n+1)/2 m’ i=1 n i=1

  28. Well-ordering principle • This means for any 0 ≤ n < m’ ∑ i = n(n+1)/2 • This is true when n = 0, so • m’ > 0 • m’-1 is non-negative and m’-1 < m’, so ∑ i = (m’-1)((m’-1)+1)/2 n i=1 m’-1 i=1

  29. Well-ordering principle m’-1 ∑ i = (m’-1)((m’-1)+1)/2 ∑ i = ∑ i + m’ = (m’-1)m’/2 + m’ = m’(m’+1)/2 • But ∑ i ≠ m’(m’+1)/2 • Contradiction i=1 m’ m’-1 i=1 i=1 m’ i=1

  30. Well-ordering principle • Show that a2 + b2 = 3(s2 + t2) has no non-zero integer solutions.

  31. Well-ordering principle • Suppose it has non-zero integer solutions. • Let S be the collection of (a, b, s, t) such that • (a, b, s, t) ≠ (0, 0, 0, 0), and • a2 + b2 = 3(s2 + t2) • S is nonempty, by well-ordering principle • there is a least element (a1, b1, s1, t1) such that a12 + b12 = 3(s12 + t12)

  32. Well-ordering principle • S is nonempty, by well-ordering principle • there is a least element (a1, b1, s1, t1) such that a12 + b12 = 3(s12 + t12) • Right idea, but… • S is a set containing 4-tuples (a, b, s, t) • not a set of non-negative integers • Cannot apply the well-ordering principle

  33. Well-ordering principle • Suppose there exists (a, b, s, t) ≠ (0, 0, 0, 0), a2+ b2 = 3(s2 + t2) • Let S be the collection of |a| such that there exist b, s and t a2 + b2 = 3(s2 + t2) • S is a nonempty set of non-negative integers • By the well-ordering principle • There exists an a1 in S such that |a1| is smallest

  34. Well-ordering principle • Let S be the collection of |a| such that there exist b, s and t a2 + b2 = 3(s2 + t2) • By the well-ordering principle • There exists an a1 in S such that |a1| is smallest • And for this smallest |a1|, there exist b1, s1 and t1, a12+ b12= 3(s12+ t12)

  35. Well-ordering principle • And for this smallest |a1|, there exist b1, s1 and t1, a12+ b12= 3(s12+ t12) • This means • a12 + b12is a multiple of 3 • a1 and b1are both multiples of 3 • proof by contrapositive • a1= 3a2 and b1 = 3b2 for some a2 and b2

  36. Well-ordering principle • And for this smallest |a1|, there exist b1, s1 and t1, a12+ b12= 3(s12+ t12) • a1 = 3a2 and b1 = 3b2 for some a2 and b2 • (3a2)2 + (3b2)2 = 3(s12 + t12) • 9a22 + 9b22 = 3(s12 + t12) • 3(a22 + b22) = s12 + t12 • s12 + t12 = 3(a22 + b22) • therefore (s1, t1, a2, b2) is also a solution

  37. Well-ordering principle • There is a least element (a1, b1, s1, t1) a12 + b12 = 3(s12 + t12) • We have just showed • (s1, t1, a2= a1/3, b2 = b1/3) is also a solution • Repeat the argument • (a2=a1/3, b2=b1/3, s2=s1/3, t2=t1/3) is also solution • But |a2| < |a1|, contradiction.

  38. Invariant Method • The numbers 1, 2, 3, 4 and 5 are written on a board • Repeat the following until there is only one number left: • pick any two of the numbers • erase them • write the absolute value of their difference on board • Can the remaining number be 2?

  39. Invariant Method • Example • 1 2 3 4 5 → 2 3 4 4 • 2 3 4 4 → 2 4 1 • 2 41→ 2 3 • 2 3 → 1

  40. Invariant Method • Observe • 1 2 3 4 5 (1 + 2 + 3 + 4 + 5 = 15) • 2 3 4 4 (2 + 3 + 4 + 4 = 13) • 2 41(2 + 4 + 1 = 7) • 2 3 (2 + 3 = 5) • 1 (1) • What is the pattern?

  41. Invariant Method • Observation • can only get an odd number in the final answer • total sum of the numbers on the board is always odd • Suppose I pick m and n on the board • Change in total sum = m + n - (m - n) = 2n • the change in total sum must be even • Therefore the total sum is always odd

  42. Invariant Method • At the beginning the total sum is 15 • which is odd • Therefore the total sum is always odd • Impossible to get 2 as final answer • In fact, impossible to get any even number as final answer

  43. End • Questions?

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