Chapter 4, con't., Inductive Proofs
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Description

Inductive proofs must have:
– Base case:
• where you prove that what it is you are trying to prove is true
about the base case
– Inductive hypothesis:
• where you state what will be assumed in the proof
– Inductive step:
• show:
– where you state what will be proven below
• proof:
– where you prove what is stated in the show portion
– this proof must use the inductive hypothesis somewhere
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Example

n
n( n  1) 
Prove this statement: n  1 i 

2
 i 1

1



n(n  1) 1(1  1) 2

 1
2
2
2
i 1
Inductive hypothesis (assume the p
p ( p  1)
i  2
statement is true for n = p):
i 1
Base case (n = 1):
i  1
Inductive step (show the statement is true for n = p + 1),
i.e, show: p 1
( p  1)(( p  1)  1)
i 
2
i 1
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Variations


2 + 4 + 6 + 8 + … + 20 = ?
If you can, use the fact just proved, that:
n
n(n  1)
i 
i 1


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Can it be rearranged into a form that works?
If not, it must be proved from scratch
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Another example


k
n 1
n  0 2  2  1
 k 0

n
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Discrete Structures
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Lecture 23
March 26, 2008
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Another example- geometric progression

ar  a 
k
(r  R )(a  R)(n  Z ) ar 

r 1 
 k 0
1
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0
n
n1
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Another example- a divisibility property
0
(n  Z )[3 | ( n  n)]
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A sequence example

Assume the following definition of a sequence:
a1  1
(k  2)ak  ak 1  (2k  1)

Prove :
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1

(n  Z ) an  n
2

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Discrete Structures
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Lecture 24
March 28, 2008
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An example with an inequality
Prove this statement:
Base case (n = 3):
3

(n  Z ) 2n  1  2
n

LHS : 2(3)  1  6  1  7
RHS : 23  8
Inductive hypothesis (n = p): assume
Inductive step (n = p + 1):
LHS  RHS
2 p 1  2
p
2( p  1)  1  2
p 1
Show:
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Another example with an inequality
2


(n  Z )(x  R ) 1  nx  (1  x)
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n

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A less-mathematical example

If all we had was 2-cent coins and 5-cent coins, we could
form any value greater than 3 cents.
– Base case (n = 4):
– Inductive hypothesis (n = p):
– Inductive step (n = p + 1):
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Discrete Structures
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Lecture 25
March 31, 2008
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Recurrence relation example

Assume the following definition of a function:
a1  1
a0  1
a2  3
(k  Z )ak  ak 1  ak 2  ak 3 
3

Prove the following definition property:
0

(n  Z ) an  Z
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odd

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Strong induction

Regular induction:
P(n)  P(n+1)

With strong induction, the implication changes slightly:
– if the statement to be proven is true for all preceding elements,
then it's true for the current element
(n)[(i, a  i  n)[P(i)]  P(n+1)]

The strong induction principle:
P(0)  …  P(p)
(n)[P(0)  P(1)  P(2)  …  P(n)  P(n + 1)]
 (n ≥ 0)[P(n)]
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Now prove the recurrence relation property,
using strong induction

Here's the function definition again:
a0  1
a2  3
a1  1
(k  Z )ak  ak 1  ak 2  ak 3 
3

This is the property to be proven:
0

(n  Z ) an  Z
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odd

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Another recurrence relation example

Assume the following definition of a function:
a0  1
a1  2
(k  Z 2 )ak  ak 1  ak 2 

Prove the following definition property, using strong
induction:
0

( n  Z ) a n  2
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n

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Discrete Structures
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Lecture 26
April 2, 2008
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Another example- a divisibility property

Assume the following definition of a recurrence relation:
a0  0
a1  7
(i  2)ai  2ai 1  3ai  2 

Prove using strong induction that all elements in this
relation have this property:
(n  N )an  0 (mod 7)
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Another example

Assume the following definition of a recurrence relation:
a1  0
a2  2


(i  Z ) ai  3a i   2
 2 


3

Prove using strong induction that all elements in this
relation have this property:
1

(n  Z ) an  Z
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even
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Discrete Structures
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Lecture 27
April 4, 2008
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Another example

Theorem: for all n ≥ 2 there exist primes p1, p2,…,pk,
and exponents e1, e2,…ek, such that
n  p p2  pk
e1
1
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e2
ek
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Constructive induction
a1  7
a0  2
(k  Z )ak  12ak 1  3ak 2 
2

Show:
0

(n  Z ) an  A  B
n

(i.e., find integers A and B for which this is true)
In particular, we want to find the smallest A and B which will work
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