lecture 3.1: mathematical induction*
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9/27/2011 Lecture 3.1 -- Mathematical Induction
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Lecture 3.1: Mathematical Induction*
CS 250, Discrete Structures, Fall 2011
Nitesh Saxena *Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag
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Course Admin Mid-Term 1
Hope it went well! Thanks cooperating with the TA as a proctor We are grading them now, and should have
the results by next weekend Solution will be posted today
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Course Admin HW1 graded
Scores have been posted To be distributed at the end of lecture Thanks for your patience waiting for the
results Any questions after taking a careful look,
please contact TA If that doesn’t help, please contact me
HW2 due Sep 30 (this Friday)
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Outline
Mathematical Induction Principle Examples Why it all works
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Mathematical InductionSuppose we have a sequence of propositions
which we would like to prove:P (0), P (1), P (2), P (3), P (4), … P (n), …EG: P (n) = “The sum of the first n positive odd numbers is
equal to n2”We can picture each proposition as a domino:
P (n)
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Mathematical InductionSo sequence of propositions is a sequence
of dominos.
…
P (n+1)P (n)P (2)P (1)P (0)
…
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Mathematical InductionWhen the domino falls, the corresponding
proposition is considered true:
P (n)
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Mathematical InductionWhen the domino falls (to right), the
corresponding proposition is considered true:
P (n)true
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Mathematical InductionSuppose that the dominos satisfy two
constraints.1) Well-positioned: If any domino falls (to
right), next domino (to right) must fall also.
2) First domino has fallen to right
P (0)true
P (n+1)P (n)
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Mathematical InductionSuppose that the dominos satisfy two
constraints.1) Well-positioned: If any domino falls to
right, the next domino to right must fall also.
2) First domino has fallen to right
P (0)true
P (n+1)P (n)
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Mathematical InductionSuppose that the dominos satisfy two
constraints.1) Well-positioned: If any domino falls to
right, the next domino to right must fall also.
2) First domino has fallen to right
P (0)true
P (n)true
P (n+1)true
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Mathematical InductionThen can conclude that all the dominos
fall!
…
P (n+1)P (n)P (2)P (1)P (0)
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Mathematical InductionThen can conclude that all the dominos
fall!
…
P (n+1)P (n)P (2)P (1)P (0)
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Mathematical InductionThen can conclude that all the dominos
fall!
…P (0)true
P (n+1)P (n)P (2)P (1)
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Mathematical InductionThen can conclude that all the dominos
fall!
…P (0)true
P (1)true
P (n+1)P (n)P (2)
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Mathematical InductionThen can conclude that all the dominos
fall!
P (2)true
…P (0)true
P (1)true
P (n+1)P (n)
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Mathematical InductionThen can conclude that all the dominos
fall!
P (2)true
…P (0)true
P (1)true
P (n+1)P (n)
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Mathematical InductionThen can conclude that all the dominos
fall!
P (2)true
…P (0)true
P (1)true
P (n)true
P (n+1)
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Mathematical InductionThen can conclude that all the dominos
fall!
P (2)true
…P (0)true
P (1)true
P (n)true
P (n+1)true
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Mathematical InductionPrinciple of Mathematical Induction: If:1) [basis] P (0) is true2) [induction] n P(n)P(n+1) is true
Then: n P(n) is trueThis formalizes what occurred to dominos.
P (2)true
…P (0)true
P (1)true
P (n)true
P (n+1)true
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Example 1Use induction to prove that the sum of the first
n odd integers is n2.Prove a base case (n=1)
Base case (n=1): the sum of the first 1 odd integer is 12. Yup, 1 = 12.
Prove P(k)P(k+1)
Assume P(k): the sum of the first k odd ints is k2. 1 + 3 + … + (2k - 1) = k2
Prove that 1 + 3 + … + (2k - 1) + (2k + 1) = (k+1)2
1 + 3 + … + (2k-1) + (2k+1) =k2 + (2k + 1)= (k+1)2 By arithmetic
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Example 2
Prove that 11! + 22! + … + nn! = (n+1)! - 1, nBase case (n=1): 11! = (1+1)! - 1?Yup, 11! = 1, 2! - 1 = 1
Assume P(k): 11! + 22! + … + kk! = (k+1)! - 1Prove that 11! + … + kk! + (k+1)(k+1)! = (k+2)! - 1
11! + … + kk! + (k+1)(k+1)! = (k+1)! - 1 + (k+1)(k+1)!
= (1 + (k+1))(k+1)! - 1
= (k+2)(k+1)! - 1
= (k+2)! - 1
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Example 3
Prove that if a set S has |S| = n, then |P(S)| = 2nBase case (n=0): S=ø, P(S) = {ø} and |P(S)| = 1 = 20
Assume P(k): If |S| = k, then |P(S)| = 2k
Prove that if |S’| = k+1, then |P(S’)| = 2k+1
S’ = S U {a} for some S S’ with |S| = k, and a S’.Partition the power set of S’ into the
sets containing a and those not.We count these sets separately.
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Example 3 (contd)Assume P(k): If |S| = k, then |P(S)| = 2k
Prove that if |S’| = k+1, then |P(S’)| = 2k+1
S’ = S U {a} for some S S’ with |S| = k, and a S’.Partition the power set of S’ into the
sets containing a and those not.
P(S’) = {X : a X} U {X : a X}
P(S’) = {X : a X} U P(S) Since these are all the subsets of elements in S.
Subsets containing a are made by taking any set from
P(S), and inserting an a.
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Example 3 (contd)Assume P(k): If |S| = k, then |P(S)| = 2k
Prove that if |S’| = k+1, then |P(S’)| = 2k+1
S’ = S U {a} for some S S’ with |S| = k, and a S’.P(S’) = {X : a X} U {X : a X}
P(S’) = {X : a X} U P(S) Subsets containing a are made by taking any set from
P(S), and inserting an a.
So |{X : a X}| = |P(S)|
|P(S’)| = |{X : a X}| + |P(S)|
= 2 |P(S)| = 22k = 2k+1
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Mathematical Induction - why does it work?
Proof of Mathematical Induction:
We prove that (P(0) (k P(k) P(k+1))) (n P(n))
Proof by contradiction.
Assume1. P(0)2. k P(k) P(k+1)3. n P(n) n P(n)
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Mathematical Induction - why does it work?
Assume1. P(0)2. k P(k) P(k+1)3. n P(n) n P(n)
Let S = { n : P(n) } Since N is well ordered, S has a least element. Call it k.
What do we know? P(k) is false because it’s in S. k 0 because P(0) is true. P(k-1) is true because P(k) is the least
element in S.
But by (2), P(k-1) P(k). Contradicts P(k-1) true, P(k)
false.
Done.
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More examples – prove by induction
1. Recall sum of arithmetic sequence:
2. Recall sum of geometric sequence:
2
)1(
1
nni
n
i
1
)1(...
12
0
r
raarararaar
nn
n
i
i
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Today’s Reading Rosen 5.1
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