# induction - diamondmath Page 1 of 6 Notes for MATH/COSC...

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Unformatted text preview: diamondmath Page 1 of 6 Notes for MATH/COSC 1056 August 5, 2005 Mathematical Induction B.G.Adams An important proof technique is mathematical induction . A proof by induction begins with a propositional function S ( n ) defined for all integers n m for some smallest integer m . The goal is to show that S ( n ) is true for all integers n m . First we try to show that the statement S ( n ) is true for the first case n = m . Then we use mathematical induction to show that S ( n ) is true for all n m . In numerical examples it is usually necessary to guess a formula for the first few values of n and and then use induction to verify that the formula is correct for all vaules of n . Standard Example (sum of the first n integers) Consider the problem of calculating the sum of the first n integers. Define S n = 1 + 2 + 3 + + n for n 1. Then we have the following results for the first few cases. S 1 = 1 S 2 = 1 + 2 = 3 S 3 = 1 + 2 + 3 = 6 S 4 = 1 + 2 + 3 + 4 = 10 At this stage we might be clever enough to notice that the values of the sums are n ( n + 1 ) / 2 for n = 1 , 2 , 3 , 4. For example, 4 ( 4 + 1 ) / 2 = 10. Later we will show how to justify this guess. We would like to show that this formula is true for all values of n 1. This means we must show that the propositional function S defined by the statements S ( n ) : S n = n ( n + 1 ) 2 is true for all n 1. But how do we prove an infinite number of statements? We can do this using mathematical induction. The first statement S ( 1 ) is true since 1 = 1 ( 1 + 1 ) / 2. Suppose we can show for an arbitrary n that S ( n ) S ( n + 1 ) . Then we can reason as follows S ( 1 ) is true , S ( 2 ) is true since S ( 1 ) S ( 2 ) and S ( 1 ) is true (modus ponens) , S ( 3 ) is true since S ( 2 ) S ( 3 ) and S ( 2 ) is true (modus ponens) , S ( 4 ) is true since S ( 3 ) S ( 4 ) and S ( 3 ) is true (modus ponens) , Thus, since S ( 1 ) is true, we can prove an infinite number of statements if we can show that for an arbitrary n &gt; 1 S ( n + 1 ) : S n + 1 = ( n + 1 )( n + 2 ) 2 follows from S ( n ) : S n = n ( n + 1 ) 2 Notes for MATH/COSC 1056 Page 2 of 6 Here the formula for S n + 1 is obtained by substituting...
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## This note was uploaded on 08/01/2008 for the course PHIL 140A taught by Professor Fitelson during the Spring '07 term at University of California, Berkeley.

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induction - diamondmath Page 1 of 6 Notes for MATH/COSC...

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