Mathematical Induction - Discrete Mathematical Structures - Lecture Slides, Slides of Discrete Mathematics

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Chapter 4

Sequences and Mathematical

Induction

Mathematical Induction

Principle of Mathematical Induction

• Let P(n) be a property that is defined for

integers n, and let a be a fixed integer. Suppose the following two statements are true:

1. P(a) is true.
2. For all integers k ≥ a, if P(k) is true then

P(k+1) is true.

• Then the statement, for all integers n ≥ a,

P(a) is true

Example

• For all integers n ≥ 8, n cents can be obtained

using 3 cents and 5 cents.

• or, for all integers n ≥ 8, P(n) is true, where

P(n) is the sentence “n cents can be obtained

using 3 cents and 5 cents.”

Method of Proof Mathematical

Induction

• Statement: “For all integers n≥a, a property P(n)

is true.”

• (basis step) Show that the property is true for n =

a.

• (inductive step) Show that for all integers k≥a, if

the property is true for n=k then it is true for

n=k+1.

• (inductive hypothesis) suppose that the property is true for n=k, where k is any particular but arbitrarily chosen integer with k≥a.
• then, show that the property is true for n = k+

Example Coins Revisited

• Proposition 4.2.1: Let P(n) be the property “n

cents can be obtained using 3 and 5 cent

coins.” Then P(n) is true for all integers n≥8.

• Proof:
• Show that the property is true for n=8:
  • The property is true b/c 8=3+5.

Example Formula

• Prove with mathematical induction

• Identify P(n)

• Basis step

1 + 2 + ...+ n =

n(n +1) 2

,n ≥ 1

P(n) = 1 + 2 + ...+ n =

n(n +1) 2

,n ≥ 1

P(1) = 1 =

Example Formula cont.

• Inductive step
  • assume P(k) is true, k>=
  • show that P(k+1) is true by subing k+1 for n
  • show that left side 1+2+…k+1 = right side (k+1)(k+2)/
  • 1+2+…+k+1 = (1+2+…+k) + k+
  • sub from inductive hypothesis:

P(k) = 1 + 2 + ...+ k =

k(k +1) 2

P(k +1) = 1 + 2 + ...+ k + 1 =

(k +1)(k + 1 +1) 2

(k +1)(k + 2) 2

k(k +1) 2 +^ (k^ +1)^ =^

k(k +1) 2 +^

2(k +1) 2 = (k^ +1)(k^ +^ 2)

Example

• Sum of the First n Integers
  • Find 2+4+6+…+
    • Get in form of Theorem 4.2.2 (1+2+…+n)
    • factor out 2: 2(1+2+3+…+250)
    • sum = 2( n(n+1)/2 ), n = 250
    • sum = 2( 250(250+1)/2 ) = 62,
  • Find 5+6+7+8+…+
    • add first 4 terms 1+2+3+4 to problem then subtract back out after computation with 4.2.
    • 1+2+3+4+5+6+7+…+50 – (1+2+3+4)
    • (50 (50+1)/ 2) – 10
    • =1,

Sum of Geometric Sequence

• Prove that , for all integers n≥0 and all

real numbers r except 1.

• P(n):

• Basis:

• Inductive: (n=k)

• Inductive Hypothesis: (n=k+1)

r i i= 0

n ∑ =^

r n^ +^1 − 1 r − 1

r i i= 0

n ∑ =^

r n^ +^1 − 1 r − 1

r i i= 0

0 ∑ =^

r^0 +^1 − 1 r − 1 =^1

r i i= 0

k ∑ =^

r k^ +^1 − 1 r − 1

r i i= 0

k + 1 ∑ =^

r k^ +^1 +^1 − 1 r − 1

= r^

k + (^2) − 1 r − 1

r i i= 0

k + 1

∑ =^ r^ i

i= 0

k

∑ +^ r^ k^ +^1 =^

r k^ +^1 − 1 r − 1

• r k^ +^1 =

r k^ +^1 − 1 r − 1

r k^ +^1 (r −1) r − 1

=

r k^ +^1 − 1 + r k^ +^2 − r k^ +^1 r − 1

=

r k^ +^2 − 1 r − 1

Examples

• Sum of Geometric Sequences

• Find 1 + 3 + 3^2 + … + 3m-
  • Sequence is in geometric series, apply 4.2.3 directly
• Find 3^2 + 3^3 +… + 3m
  • rearrange into proper geometric sequence by factoring out 3^2 from sequence
  • 32 (1 + 3 + 3^2 + … + 3m-2) =

r i i= 0

n ∑ =^

r n^ +^1 − 1 r − 1

30 + 31 + 32 + ...+ 3 m^ −^2 =

3 m^ −^2 +^1 − 1 3 − 1

=

3 m^ −^1 − 1 2

32 3 i i= 0

m − 2 ∑ =^

3 m^ −^1 − 1 3 − 1