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Lecture 3.1: Mathematical Induction
CS 250, Discrete Structures, Fall 2014
Nitesh Saxena
Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag
04/21/23Lecture 3.1 -- Mathematical
Induction
Course Admin Mid-Term 1
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04/21/23Lecture 3.1 -- Mathematical
Induction
Course Admin HW2 (don’t forget)
Due Oct 14 (Tues)
04/21/23Lecture 3.1 -- Mathematical
Induction
Outline
Mathematical Induction Principle Examples Why it all works
04/21/23Lecture 3.1 -- Mathematical
Induction
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)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionSo sequence of propositions is a sequence
of dominos.
…
P (n+1)P (n)P (2)P (1)P (0)
…
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionWhen the domino falls, the corresponding
proposition is considered true:
P (n)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionWhen the domino falls (to right), the
corresponding proposition is considered true:
P (n)true
04/21/23Lecture 3.1 -- Mathematical
Induction
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)
04/21/23Lecture 3.1 -- Mathematical
Induction
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)
04/21/23Lecture 3.1 -- Mathematical
Induction
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
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionThen can conclude that all the dominos
fall!
…
P (n+1)P (n)P (2)P (1)P (0)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionThen can conclude that all the dominos
fall!
…
P (n+1)P (n)P (2)P (1)P (0)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionThen can conclude that all the dominos
fall!
…P (0)true
P (n+1)P (n)P (2)P (1)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionThen can conclude that all the dominos
fall!
…P (0)true
P (1)true
P (n+1)P (n)P (2)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionThen can conclude that all the dominos
fall!
P (2)true
…P (0)true
P (1)true
P (n+1)P (n)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionThen can conclude that all the dominos
fall!
P (2)true
…P (0)true
P (1)true
P (n+1)P (n)
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionThen can conclude that all the dominos
fall!
P (2)true
…P (0)true
P (1)true
P (n)true
P (n+1)
04/21/23Lecture 3.1 -- Mathematical
Induction
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
04/21/23Lecture 3.1 -- Mathematical
Induction
Mathematical InductionPrinciple of Mathematical Induction: If:1) [basis] P (0) is true2) [induction] k P(k)P(k+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
04/21/23Lecture 3.1 -- Mathematical
Induction
Exercise 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
04/21/23Lecture 3.1 -- Mathematical
Induction
Exercise 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)! - 111! + … + 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
04/21/23Lecture 3.1 -- Mathematical
Induction
Exercises 3 and 4 (have seen before?)
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
04/21/23Lecture 3.1 -- Mathematical
Induction
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)
04/21/23Lecture 3.1 -- Mathematical
Induction
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.
04/21/23Lecture 3.1 -- Mathematical
Induction
Today’s Reading Rosen 5.1