{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

lect14_1 - Induction proofs To prove that some proposition...

This preview shows pages 1–8. Sign up to view the full content.

1 Induction proofs To prove that some proposition P ( n ) is true for all n n 0 by induction on n n 0 1. Basis : Check that P ( n 0 ) is true. 2. Inductive step . Assume that P ( n ) is true for n = k , where k is some integer, k n 0 Prove that P ( n ) is true for n = k +1 So, in inductive step we prove that for any n n 0 P ( n ) P ( n +1) Usually this is simpler than prove directly that for any n n 0 P ( n ).

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

View Full Document
2 Example. Let P ( n ) : n 3 - n is divisible by 6 To prove that P ( n ) is true for all n 0: 1. Basis. P (0): “0 3 - 0=0 is divisible by 6”. P (0) is true 2. Inductive step P ( n = k ) P ( n = k +1), k 0. Assume P ( n = k ) is true, where k is some integer, k 0. So, we assume that k 3 - k is divisible by 6 i.e. k 3 - k = 6 s ( s is any integer) Goal: to prove that P ( n = k +1) is true, i. e. that ( k +1) 3 - ( k +1) is divisible by 6. ( k +1) 3 - ( k +1) = k 3 +3 k 2 + 3 k +1 - k - 1 = = ( k 3 - k ) + 3 k 2 + 3 k = = 6 s + 3 k ( k + 1)
3 Suppose we want to prove the formula for the sum of ‘geometric progression’ q q q ... q q q n n - - = + + + + + + 1 1 1 1 3 2 where 1 0 < < q , and n is any integer, n 0 How can we prove that the formula is correct for any n 0 ? Notation: = = + + + + = n i i n n q q ... q q S 0 2 1 q q S n n - - = + 1 1 1 Show it for arbitrary fixed n 0 (prove directly, without induction) Use induction on n 0 Two alternatives:

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

View Full Document
4 Show it for some n 0 (prove P ( n ) for any n directly) 1 2 1 2 1 ) (1 + + + - + + + + = + + + = n n n n q q ... q q q ... q q S q n S 1 1 - + = + n n n q S S q 1 1 ) (1 + - = - n n q q S q q S n n - - = + 1 1 1
5 1) Base: n =0 1; 0 0 0 0 = = = = q q S i i 1 1 1 1 1 1 0 = - - = - - + q q q q 2) Induction Hypothesis : Assume the formula is correct for n = k , where k is some integer, k 0 q q S k k - - = + 1 1 1 Induction Step : prove the formula for n = k +1, i. e. q q S k k - - = + + + 1 1 1 1) ( 1 1 1 2 1 1 + + + + = + + + + + = k k k k k q S q q ... q q S q q q q q q q q q k k k k k k - - = - - + - = + - - = + + + + + + + + 1 1 1 1 1 1 1 1) ( 1 1) ( 1 1 1 1 Proof by induction on n 0 that q q S n n - - = + 1 1 1

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

View Full Document
6 ( 2200 n n 0 ) P ( n ) P ( n 0 ) ( 2200 n n 0 ) [ P ( n ) P ( n +1)] ? Induction Principle . If A is a subset of N that satisfies two properties: 1) n 0 A 2) 2200 n n 0 [ n A ( n +1) A ], then A = N We just can think about A as a set of integers, that satisfy P : A ={ n N | P ( n )} Then, 1) n 0 A P ( n 0 ) 2) 2200 n n 0 [ n A ( n +1) A ] 2200 n n 0 ( P ( n ) P ( n +1))
7 Although it sounds very obvious, the induction

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 25

lect14_1 - Induction proofs To prove that some proposition...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online