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Review: Mathematical Induction

Review: Mathematical Induction. Prove a base case (n=1). Prove P(k)P(k+1). By inductive hypothesis. By arithmetic. Inductive hypothesis. Use induction to prove that the sum of the first n odd integers is n 2 . Base case (n=1): the sum of the first 1 odd integer is 1 2 . Yes, 1 = 1 2 .

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Review: Mathematical Induction

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  1. Review: Mathematical Induction Prove a base case (n=1) Prove P(k)P(k+1) By inductive hypothesis By arithmetic Inductive hypothesis Use induction to prove that the sum of the first n odd integers is n2. Base case (n=1): the sum of the first 1 odd integer is 12. Yes, 1 = 12. Assume P(k): the sum of the first k odd ints is k2. 1 + 3 + … + (2k - 1) = k2 Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2 1 + 3 + … + (2k-1) + (2k+1) = k2 + (2k + 1) = (k+1)2

  2. Mathematical Induction - a cool example A 2n x 2n sized grid is deficient if all but one cell is tiled. Deficient Tiling 2n 2n

  3. Mathematical Induction - a cool example • We want to show that all 2n x 2n sized deficient grids can be tiled with tiles, called triominoes, shaped like:

  4. Mathematical Induction - a cool example Yes! • Is it true for all 21 x 21 grids?

  5. Mathematical Induction - a cool example Inductive Hypothesis: We can tile any 2k x 2k deficient board using our fancy designer tiles. Use this to prove: We can tile any 2k+1 x 2k+1 deficient board using our fancy designer tiles.

  6. Mathematical Induction - a cool example 2k 2k ? ? 2k ? 2k 2k+1 OK!! (by IH)

  7. Mathematical Induction - a cool example 2k 2k 2k 2k OK!! (by IH) OK!! (by IH) 2k+1 OK!! (by IH) OK!! (by IH)

  8. Mathematical Induction - a cool example

  9. Mathematical Induction - why does it work? No. { x  Z : x < 0 } has no least element. Definition: A set S is “well-ordered” if every non-empty subset of S has a least element. Given (we take as an axiom): the set of natural numbers (N) is well-ordered. Is the set of integers (Z) well ordered?

  10. Mathematical Induction - why does it work? No. { x  R : x > 1 } has no least element. Is the set of non-negative reals (R) well ordered?

  11. Mathematical Induction - why does it work? Proof by contradiction. n P(n) Proof of Mathematical Induction: We prove that (P(0)  (k P(k)  P(k+1)))  (n P(n)) Assume • P(0) • k P(k)  P(k+1) • n P(n)

  12. Mathematical Induction - why does it work? But by (2), P(k-1)  P(k). Contradicts P(k-1) true, P(k) false. Done. n P(n) Since N is well ordered, S has a least element. Call it k. Assume • P(0) • n P(n)  P(n+1) • n P(n) Let S = { n : P(n) } What do we know? -P(k) is false because it’s in S. -k  0 because P(0) is true. -P(k-1) is true because P(k) is the least element in S.

  13. Strong Mathematical Induction In our proofs, to show P(k+1), our inductive hypothesis assumes that ALL of P(0), P(1), … P(k) are true, so we can use ANY of them to make the inference. If P(0) and n0 (P(0)  P(1)  …  P(n))  P(n+1) Then n0 P(n)

  14. Game with Matches • Two players take turns removing any number of matches from one of two piles of matches. The player who removes the last match wins • Show that if two piles contain the same number of matches initially, then the second player is guaranteed a win

  15. Strategy for Second Player • Let P(n) denote the statement “the second player wins when they are initially n matches in each pile” • Basis step: P(1) is true, because only 1 match in each pile, first player must remove one match from one pile. Second player removes other match and wins • Inductive step: suppose P(j) is True for all j 1<=j <= k. • Prove that P(k+1) is true, that is the second player wins when each piles contains k+1 matches

  16. Strategy for Second Player How is this different than regular induction? • Suppose that the first player removes r matches from one pile, leaving k+1 –r matches there • By removing the same number of matches from the other pile the second player creates the situation of two piles with k+1-r matches in each. Apply the inductive hypothesis and the second player wins each time.

  17. Postage Stamp Example • Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps • P(n) : Postage of n cents can be formed using 4-cent and 5-cent stamps • All n >= 12, P(n) is true

  18. Postage Stamp Proof Why does this work? • Base Case: n = 12, n = 13, n = 14, n = 15 • We can form postage of 12 cents using 3, 4-cent stamps • We can form postage of 13 cents using 2, 4- cent stamps and 1 5-cent stamp • We can form postage of 14 cents using 1, 4-cent stamp and 2 5-cent stamps • We can form postage of 15 cents using 3, 5-cent stamps • Induction Step • Let n >= 15 • Assume P(k) is true for 12 <= k <= n, that is postage of k cents can be formed with 4-cent and 5-cent stamps (Inductive Hypothesis) • Prove P(n+1) • To form postage of n +1 cents, use the stamps that form postage of n-3 cents (from I.H) with a 4-cent stamp

  19. Recursive Definitions Inductive (Recursive) Definition Recursive Case Base Case We completely understand the function f(n) = n!, right? As a reminder, here’s the definition: n! = 1 · 2 · 3 · … · (n-1) · n, n  1 But equivalently, we could define it like this:

  20. Recursive Definitions Recursive Case Base Cases Is there a non-recursive definition for the Fibonacci Numbers? Another VERY common example: Fibonacci Numbers

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