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Math Induction

Math Induction. Section 2.2. First Principle of Induction. Relies on two assertions: Assertion of the basis step (1 st rung on the ladder) Assertion that, having gotten to the current step, one can always get to the next step (can go to the next rung on the ladder)

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Math Induction

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  1. Math Induction Section 2.2

  2. First Principle of Induction • Relies on two assertions: • Assertion of the basis step (1st rung on the ladder) • Assertion that, having gotten to the current step, one can always get to the next step (can go to the next rung on the ladder) • Using this logic, we can prove that we can climb an infinitely high ladder. CSC 333

  3. First Principle of Induction(stated in a more mathematical way) • Relies on two assertions: • Assertion of the basis step (or base case): P(1), or 1 has the property P. • Assertion of the inductive step: if P(k) is true, then P(k+1) is true, where k is an arbitrary positive integer. • Using this logic, we can prove that P(n) is true for any positive integer n. CSC 333

  4. First Principle of Induction(use of the inductive assumption) • The inductive step, if P(k) is true, then P(k+1) is true, relies on the inductive assumption that P(k) is true. • Also called the inductive hypothesis. • If we can show that P(k+1) follows from P(k), and we know that P(1) is true, then we can show that P(2) is true [let k = 1 . . . ]. • then we can show P(3) is true; then we can show P(4) is true; and so on, ad nauseum. CSC 333

  5. Example 1 • How many rows are required in a truth table? • We may observe that 2 rows are required for a table containing one statement letter: A T F We may notice that 4 rows are required for a table containing two statements: A v B T T T F F T F F CSC 333

  6. Example 1 (cont’d) • Pretty soon we decide that maybe for every n unique statement letters, 2n rows are required. • To prove this using math induction, we have a base case, i.e., when there is one statement letter, 2 rows are required, or stated in a more mathy way, P(1) = 21. • We can also assert the inductive assumption, i.e., P(k) = 2k • And the inductive step, if P(k) = 2k, then P(k+1) = 2k+1. CSC 333

  7. Example 1 (cont’d) • Now, having assumed P(k) = 2k , our job is to prove P(k+1) = 2k+1. • We can observe that adding a statement letter to a truth table requires that all of the existing combinations must be matched against a series of T’s and a new set of the existing combinations must be matched against a series of F’s. • This effectively multiplies the number of existing rows by 2. • So, going from k statement letters to k+1 statement letters requires going from 2k rows to 2k (2) rows, or 2k+1 rows. CSC 333

  8. Example 1 (cont’d) • We have used mathematical induction to confirm our conjecture that for n unique statement letters in a truth table, 2n rows are required. CSC 333

  9. Example 2 • Show that 2 + 4 + 6 + … + 2n = n(n+1) • P(1) = 2 = 1(1 + 1) [base step] • P(k)=2+4+6+…+2k= k(k+1) [inductive assumption] ? • P(k+1) = 2 + 4 + 6 + … + 2(k+1) =(k+1) (k+1 + 1) [inductive step]P (k + 1) = 2 + 4 + 6 + … + 2(k+1) = 2 + 4 + 6 + … + 2k + 2(k+1)= k (k + 1 ) + 2(k+1)= k2 + k + 2k + 2 = k2 + 3k + 2 = (k+1) (k+2) CSC 333

  10. Second Principle of Induction • Relies on two assertions: • Assertion of the basis step (or base case): P(1), or 1 has the property P. • Assertion of the inductive step: if P(r) is true, where 1 ≤ r ≤ k, then P(k+1) is true, where k is an arbitrary positive integer. • Using this logic, we can prove that P(n) is true for any positive integer n. CSC 333

  11. When to use the Second Principle of Induction • When the truth of P(k) is not enough to prove the truth of P(k+1), and • You can show that P(r) is true for one or more values of r that come before k. • See Example 24. CSC 333

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