# Lecture_12 - MA1100 Lecture 12 Mathematical Induction Using...

This preview shows pages 1–9. Sign up to view the full content.

1 MA1100 Lecture 12 Mathematical Induction Using PMI on other universal sets Strong PMI Variations of MI

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
MA1100 Lecture 12 2 Other Universal Sets Can we use PMI to prove statements of the form ( " x œ Q + ) P(x) ? ( " x œ R + ) P(x) ? Possible. No. ( " x œ Z ) P(x) ? Possible.
MA1100 Lecture 12 3 Other Universal Sets ( " n œ Z )P(n) 1. P(0) is true 2. For all k ¥ 0, if P(k) is true, then P(k+1) is true 3. For all k § 0, if P(k) is true, then P(k - 1) is true Then P(n) is true for all n œ Z P(0) Ø P(1) Ø P(2) Ø Ø P(n)… …P(-n) P(-2) P(-1) P(0)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
MA1100 Lecture 12 4 Other Universal Sets ( " q œ Q + )P(q) 1. P( 1 ) is true 2. For all k œ Z + ,ifP( k ) is true, then P( k+1 ) is true Then P(m/n) is true for all m/n œ Q + + where m, n m q n = Z Fix m œ Z + Then P(n) is true for all n œ Z + 3. P(m/ 1 ) is true (from 1 and 2 above) 4. For all k œ Z + ,ifP(m/ k ) is true, then P(m/ (k+1) ) is true
MA1100 Lecture 12 5 Other Universal Sets ( " q œ Q + )P(q) + where m, n m q n = Z P(1) Ø P(2) Ø P(3) Ø Ø P(m) … ) n 1 ( P ) 3 1 ( P ) 2 1 ( P # ) n 2 ( P ) 3 2 ( P ) 2 2 ( P # ) n 3 ( P ) 3 3 ( P ) 2 3 ( P # ) n m ( P ) 3 m ( P ) 2 m ( P #

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
MA1100 Lecture 12 6 Binary Sum Proposition Every positive integer can be written as the sum of distinct power of 2 . 4 = 2 2 10 = 2 3 + 2 1 45 = 2 5 + 2 3 + 2 2 + 2 0 Example Proof by Mathematical Induction Base case : prove P(1) is true P(n): n is a sum of distinct power of 2 . Since 1 = 2 0 , So P(1) is true. Binary Sum
MA1100 Lecture 12 7 Binary Sum Inductive step : P(k) Ø P(k+1) for all k ¥ 1 hypothesis P(k): want to get P(k+1): Is knowing P(k) enough to prove P(k+1) ? P(n): n is a sum of distinct power of 2 . k is a sum of distinct power of 2 k + 1 is a sum of distinct power of 2 Let’s first try the ordinary PMI : k = sum of distinct power of 2 k + 1 = (sum of distinct power of 2) + 2 0

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Lecture 12 8 Strong PMI Let P(n) be a predicate such that: 1. P(1) is true 2. For all k œ Z + , if P(1), P(2), …, P(k) are true, then P(k+1) is true. Then P(n) is true for all n
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 27

Lecture_12 - MA1100 Lecture 12 Mathematical Induction Using...

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online