Download 11.5 Mathematical Induction

Chapter 11: Further Topics in Algebra
11.1Sequences and Series
11.2Arithmetic Sequences and Series
11.3Geometric Sequences and Series
11.4The Binomial Theorem
11.5Mathematical Induction
11.6Counting Theory
11.7Probability
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Slide 11-2
11.5 Mathematical Induction
• Mathematical induction is used to prove
statements claimed true for every positive
integer n.
For example, the summation rule
n(n  1)
1  2  3  ...  n 
2
is true for each integer n > 1.
Copyright © 2007 Pearson Education, Inc.
Slide 11-3
11.5 Mathematical Induction
Label the statement Sn.
n(n  1)
Sn : 1  2  3  ...  n 
2
For any one value of n, the statement can be
verified to be true.
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Slide 11-4
11.5 Mathematical Induction
n  1,
n  2,
n  3,
1(1  1)
S1 : 1 
true since 1=1
2
2(2  1)
S2 : 1  2 
true since 3=3
2
3(3  1)
S3 : 1  2  3 
true since 6=6
2
To show Sn is true for every n requires
mathematical induction.
Copyright © 2007 Pearson Education, Inc.
Slide 11-5
11.5 Mathematical Induction
Principle of Mathematical Induction
Let Sn be a statement concerning the positive integer n.
Suppose that
1. S1 is true;
2. For any positive integer k, k < n, if Sk is true,
then Sk+1 is also true.
Then, Sn is true for every positive integer n.
Copyright © 2007 Pearson Education, Inc.
Slide 11-6
11.5 Mathematical Induction
Proof by Mathematical Induction
Step 1 Prove that the statement is true for n = 1.
Step 2 Show that for any positive integer k, if Sk is
true, then Sk+1 is also true.
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Slide 11-7
11.5 Proving an Equality Statement
Example Let Sn be the statement
n(n  1)
Sn : 1  2  3  ...  n 
2
Prove that Sn is true for every positive integer n.
Solution The proof uses mathematical induction.
Copyright © 2007 Pearson Education, Inc.
Slide 11-8
11.5 Proving an Equality Statement
Solution Step 1 Show that the statement is true when
n = 1. S1 is the statement
1(1  1)
1
2
which is true since both sides equal 1.
Copyright © 2007 Pearson Education, Inc.
Slide 11-9
11.5 Proving an Equality Statement
Solution Step 2 Show that if Sk is true then Sk+1 is
also true. Start with Sk
k (k  1)
1  2  3  ...  k 
2
and assume it is a true statement. Add k + 1 to
each side
k (k  1)
1  2  3  ...  k  (k  1) 
 (k  1)
2
Copyright © 2007 Pearson Education, Inc.
Slide 11-10
11.5 Proving an Equality Statement
Solution Step 2
k (k  1)
1  2  3  ...  k  (k  1) 
 (k  1)
2
k 
 (k  1)   1
2 
 k 2
 (k  1) 

 2 
Copyright © 2007 Pearson Education, Inc.
Slide 11-11
11.5 Proving an Equality Statement
Solution Step 2
 k 2
1  2  3  ...  k  ( k  1)  ( k  1) 

 2 
 (k  1)  1 
 (k  1) 

2


This is the statement for n = k + 1. It has been shown
that if Sk is true then Sk+1 is also true. By
mathematical induction Sn is true for all positive
integers n.
Copyright © 2007 Pearson Education, Inc.
Slide 11-12
11.5 Mathematical Induction
Generalized Principle of Mathematical Induction
Let Sn be a statement concerning the positive integer n.
Suppose that
Step 1 Sj is true;
Step 2 For any positive integer k, k > j, if Sk implies
Sk+1.
Then, Sn is true for all positive integers n > j.
Copyright © 2007 Pearson Education, Inc.
Slide 11-13
11.5 Using the Generalized Principle
Example Let Sn represent the statement
S n : 2 n  2n  1
Show that Sn is true for all values of n > 3.
Solution Since the statement is claimed to be true for
values of n beginning with 3 and not 1, the proof uses
the generalized principle of mathematical induction.
Copyright © 2007 Pearson Education, Inc.
Slide 11-14
11.5 Using the Generalized Principle
Solution Step 1 Show that Sn is true when n = 3. S3
is the statement
2  2  3 1
3
which is true since 8 > 7.
Copyright © 2007 Pearson Education, Inc.
Slide 11-15
11.5 Using the Generalized Principle
Solution Step 2 Show that Sk implies Sk+1 for k > 3.
Assume Sk
2 k  2k  1
is true. Multiply each side by 2, giving
2  2k  2(2k  1)
or
2k 1  4k  2
or, equivalently 2
Copyright © 2007 Pearson Education, Inc.
k 1
 2(k  1)  2k .
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11.5 Using the Generalized Principle
Solution Step 2 Since k > 3, then 2k >1 and it follows
that
2
k 1
 2(k  1)  2k  2(k  1)  1
or
2
k 1
 2(k  1)  1
which is the statement Sk+1. Thus Sk implies Sk+1 and,
by the generalized principle, Sn is true for all n > 3.
Copyright © 2007 Pearson Education, Inc.
Slide 11-17