mathematical induction - university of maryland...jason filippou (cmsc250 @ umcp) induction...
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Mathematical Induction
Jason Filippou
CMSC250 @ UMCP
06-27-2016
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 1 / 48
Outline
1 Sequences and seriesSequencesSeries and partial sums
2 Weak InductionIntro to InductionPractice
3 Strong Induction
4 Errors in proofs by mathematical induction
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 2 / 48
Sequences and series
Sequences and series
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 3 / 48
Sequences and series Sequences
Sequences
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 4 / 48
Sequences and series Sequences
Definitions
Definition (Sequence)
A function a : N 7→ R is called a sequence.
Examples:
2, 4, 6, . . .10, 20, 301, 1, 2, 3, 5, 8, 13, 21, . . .
So, sequences can be either finite or infinite.
We will mostly care about infinite sequences.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 5 / 48
Sequences and series Sequences
Definitions
Definition (Sequence)
A function a : N 7→ R is called a sequence.
Examples:
2, 4, 6, . . .10, 20, 301, 1, 2, 3, 5, 8, 13, 21, . . .
So, sequences can be either finite or infinite.
We will mostly care about infinite sequences.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 5 / 48
Sequences and series Sequences
Definitions
Definition (Sequence)
A function a : N 7→ R is called a sequence.
Examples:
2, 4, 6, . . .10, 20, 301, 1, 2, 3, 5, 8, 13, 21, . . .
So, sequences can be either finite or infinite.
We will mostly care about infinite sequences.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 5 / 48
Sequences and series Sequences
Denoting sequences
A sequence can be enumerated...
a : a1, a2, . . . (or just a1, a2, . . . )c0, c1, c2, . . . (notice the indices)
Described through an explicit formula...
bk = 2k
rn = (n + 1)!
Or a recursive formula...
Fn+1 = Fn + Fn−1 ∀n ≥ 1
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 6 / 48
Sequences and series Sequences
Denoting sequences
A sequence can be enumerated...
a : a1, a2, . . . (or just a1, a2, . . . )c0, c1, c2, . . . (notice the indices)
Described through an explicit formula...
bk = 2k
rn = (n + 1)!
Or a recursive formula...
Fn+1 = Fn + Fn−1 ∀n ≥ 1
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 6 / 48
Sequences and series Sequences
Denoting sequences
A sequence can be enumerated...
a : a1, a2, . . . (or just a1, a2, . . . )c0, c1, c2, . . . (notice the indices)
Described through an explicit formula...
bk = 2k
rn = (n + 1)!
Or a recursive formula...
Fn+1 = Fn + Fn−1 ∀n ≥ 1
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 6 / 48
Sequences and series Sequences
The arithmetic sequence
Definition
Let a : a0, a1, . . . be a sequence and ω ∈ R. If aj = aj−1 + ω ∀j ∈ N∗,a is an arithmetic sequence (or progression).
a0 and ω fully define the sequence.
So, how can I write ar?
a1 + r ∗ ω r ∗ a0 a0 + r ∗ ω r ∗ a0
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 7 / 48
Sequences and series Sequences
The arithmetic sequence
Definition
Let a : a0, a1, . . . be a sequence and ω ∈ R. If aj = aj−1 + ω ∀j ∈ N∗,a is an arithmetic sequence (or progression).
a0 and ω fully define the sequence.
So, how can I write ar?
a1 + r ∗ ω r ∗ a0 a0 + r ∗ ω r ∗ a0
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 7 / 48
Sequences and series Sequences
The geometric sequence
Definition
Let a : a0, a1, . . . be a sequence and k ∈ R∗. If aj = c ∗ aj−1 ∀j ∈ N∗,a is a geometric sequence (or progression).
a0 and c fully define the sequence.
ar. How can I write it?
cr ∗ a0 ar0 arc0 a0 + cr
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 8 / 48
Sequences and series Sequences
The geometric sequence
Definition
Let a : a0, a1, . . . be a sequence and k ∈ R∗. If aj = c ∗ aj−1 ∀j ∈ N∗,a is a geometric sequence (or progression).
a0 and c fully define the sequence.
ar. How can I write it?
cr ∗ a0 ar0 arc0 a0 + cr
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 8 / 48
Sequences and series Series and partial sums
Series and partial sums
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 9 / 48
Sequences and series Series and partial sums
Definitions
Definition (Series)
Let a0, a1, . . . be any sequence. Then, the sum
+∞∑i=0
ai is called a series.
Definition (Partial sum)
Let n ∈ N. Then, the n-th partial sum of the series+∞∑i=0
ai, denoted
Sn, is the sum
n∑i=0
ai.
The partial sums themselves also form a sequence!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 10 / 48
Sequences and series Series and partial sums
Definitions
Definition (Series)
Let a0, a1, . . . be any sequence. Then, the sum
+∞∑i=0
ai is called a series.
Definition (Partial sum)
Let n ∈ N. Then, the n-th partial sum of the series
+∞∑i=0
ai, denoted
Sn, is the sum
n∑i=0
ai.
The partial sums themselves also form a sequence!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 10 / 48
Sequences and series Series and partial sums
Definitions
Definition (Series)
Let a0, a1, . . . be any sequence. Then, the sum
+∞∑i=0
ai is called a series.
Definition (Partial sum)
Let n ∈ N. Then, the n-th partial sum of the series
+∞∑i=0
ai, denoted
Sn, is the sum
n∑i=0
ai.
The partial sums themselves also form a sequence!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 10 / 48
Sequences and series Series and partial sums
Statements to prove!
To kickstart the discussion on induction, here are two theoremsconcerning partial sums:
Theorem (Closed form of the arithmetic progression partial sum)
If a is an arithmetic progression, Sn = n(a1+an)2 .
Theorem (Closed form of the geometric progression partial sum)
If a is a geometric progression and c 6= 1, Sn = a1(cn−1)c−1 .
Both of those theorems can be proven via (weak) mathematicalinduction!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 11 / 48
Sequences and series Series and partial sums
Statements to prove!
To kickstart the discussion on induction, here are two theoremsconcerning partial sums:
Theorem (Closed form of the arithmetic progression partial sum)
If a is an arithmetic progression, Sn = n(a1+an)2 .
Theorem (Closed form of the geometric progression partial sum)
If a is a geometric progression and c 6= 1, Sn = a1(cn−1)c−1 .
Both of those theorems can be proven via (weak) mathematicalinduction!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 11 / 48
Sequences and series Series and partial sums
Statements to prove!
To kickstart the discussion on induction, here are two theoremsconcerning partial sums:
Theorem (Closed form of the arithmetic progression partial sum)
If a is an arithmetic progression, Sn = n(a1+an)2 .
Theorem (Closed form of the geometric progression partial sum)
If a is a geometric progression and c 6= 1, Sn = a1(cn−1)c−1 .
Both of those theorems can be proven via (weak) mathematicalinduction!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 11 / 48
Weak Induction
Weak Induction
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 12 / 48
Weak Induction Intro to Induction
Intro to Induction
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 13 / 48
Weak Induction Intro to Induction
Proof methods: The story so far...
S
Existential Stmt. ?
Existential Proof
Constructive Non-
constructive
Universal Proof
Direct Indirect
Contradiction ? ? ? Generic
Particular
+ + -
-
Universal Statement Contraposition Division into cases Exhaustion
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 14 / 48
Weak Induction Intro to Induction
Where induction fits
S
Existential Stmt. Universal Stmt
Existential Proof
Constructive Non-
constructive
Universal Proof
Direct Indirect
Contradiction Contraposition Exhaustion Cases
Generic Particular
+ + -
-
Induction
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 15 / 48
Weak Induction Intro to Induction
The penny proposition: Statement
Suppose I have at least 4¢ in my wallet.
Then, it turns out that all my money can be stacked as 2¢ and 5¢coins!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 16 / 48
Weak Induction Intro to Induction
The penny proposition: Direct (non-inductive) proof
The penny proposition
Every dollar amount greater than 3 ¢ s can be paid with only 2¢ and5¢ coins.
Proof (Direct, by cases).
Suppose we have a total amount of C cents in our wallet. If C is aneven number, then the statement is trivial by the definition of evennumbers: We can just stack k 2¢ coins for some positive integer k. If Cis an odd number greater than 3, then it is 5 or greater. If it is 5, theproblem is trivial: We only need one 5¢ coin. But every odd dollaramount after 5¢ can be retrieved by adding any number of 2¢ coins,because of the definition of parity. We are therefore done in bothcases.
What do you think of this proof?
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 17 / 48
Weak Induction Intro to Induction
The penny proposition: Direct (non-inductive) proof
The penny proposition
Every dollar amount greater than 3 ¢ s can be paid with only 2¢ and5¢ coins.
Proof (Direct, by cases).
Suppose we have a total amount of C cents in our wallet. If C is aneven number, then the statement is trivial by the definition of evennumbers: We can just stack k 2¢ coins for some positive integer k. If Cis an odd number greater than 3, then it is 5 or greater. If it is 5, theproblem is trivial: We only need one 5¢ coin. But every odd dollaramount after 5¢ can be retrieved by adding any number of 2¢ coins,because of the definition of parity. We are therefore done in bothcases.
What do you think of this proof?
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 17 / 48
Weak Induction Intro to Induction
The principle
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 18 / 48
Weak Induction Intro to Induction
The principle
Principle of Weak Mathematical Induction
Assume P (n) is a predicate applied on any natural number n, anda ∈ N. If:
P (a) is true
P (k + 1) is true when P (k) is true ∀k ≥ a
then, ∀n ≥ a, P (n) is true.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 19 / 48
Weak Induction Intro to Induction
The approach
Our task is to prove some proposition P (n), for all positiveintegers n ≥ n0.
Mathematical induction includes the following steps:
1 Inductive Base (IB): We prove P (n0). Most often, n0 will be0, 1, or 2.
2 Inductive hypothesis (IH): If k ∈ N is a generic particular suchthat k ≥ n0, we assume that P (k) is true.
3 Inductive Step (IS): We prove that P (k + 1) is true by makinguse of the Inductive Hypothesis where necessary.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 20 / 48
Weak Induction Intro to Induction
The approach
Our task is to prove some proposition P (n), for all positiveintegers n ≥ n0.
Mathematical induction includes the following steps:1 Inductive Base (IB): We prove P (n0). Most often, n0 will be
0, 1, or 2.
2 Inductive hypothesis (IH): If k ∈ N is a generic particular suchthat k ≥ n0, we assume that P (k) is true.
3 Inductive Step (IS): We prove that P (k + 1) is true by makinguse of the Inductive Hypothesis where necessary.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 20 / 48
Weak Induction Intro to Induction
The approach
Our task is to prove some proposition P (n), for all positiveintegers n ≥ n0.
Mathematical induction includes the following steps:1 Inductive Base (IB): We prove P (n0). Most often, n0 will be
0, 1, or 2.2 Inductive hypothesis (IH): If k ∈ N is a generic particular such
that k ≥ n0, we assume that P (k) is true.
3 Inductive Step (IS): We prove that P (k + 1) is true by makinguse of the Inductive Hypothesis where necessary.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 20 / 48
Weak Induction Intro to Induction
The approach
Our task is to prove some proposition P (n), for all positiveintegers n ≥ n0.
Mathematical induction includes the following steps:1 Inductive Base (IB): We prove P (n0). Most often, n0 will be
0, 1, or 2.2 Inductive hypothesis (IH): If k ∈ N is a generic particular such
that k ≥ n0, we assume that P (k) is true.3 Inductive Step (IS): We prove that P (k + 1) is true by making
use of the Inductive Hypothesis where necessary.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 20 / 48
Weak Induction Intro to Induction
The penny proposition, revisited
We want to prove that all amounts of pennies consisting of at least4 pennies can be built using only 2¢ and 4¢ coins.
The following proposition captures the essence of what we want toprove more generally:
The penny proposition, mathematical version
Let n be an integer equal to at least 4. Then, there exist integers m, ksuch that n = 2m + 5k.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 21 / 48
Weak Induction Intro to Induction
Inductive proof of the penny proposition
Proof (By weak mathematical induction).
Let P (n) be the proposition that we want to prove, where n ≥ 4. Let r ∈ Nbe a generic particular for all integers equal to at least 4. Then:
Inductive base: For r = 4, P (4) holds, since for m = 2 and k = 0,4 = 2m + 5k.
Inductive hypothesis: For a specific r ≥ 4, assume P (r): there existintegers m, k such that r = 2m + 5k.
Inductive step: We want to prove P (r + 1): there exist m′, k′ ∈ N suchthat r + 1 = 2m′ + 5k′. By the inductive hypothesis, we have thatr = 2m + 5k ⇔ r + 1 = 2m + 5k + 1⇔ r + 1 = 2m + 5k + (6− 5)⇔r + 1 = 2 (m + 3)︸ ︷︷ ︸
m′
+5 (k − 1)︸ ︷︷ ︸k′
⇔ r + 1 = 2m′ + 5k′. So P (r + 1) holds.
Since r is a generic particular for the set of integers above 3, the result holdsfor every n ∈ {4, 5, . . . }. We are done.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 22 / 48
Weak Induction Intro to Induction
Rules for authoring mathematical induction proofs
THOU SHALT NOT NEGLECT:
1 To state which variable you are inducing on (“Proof by induction onn, a, r, . . .
2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it
may seem.4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in
the inductive step.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 23 / 48
Weak Induction Intro to Induction
Rules for authoring mathematical induction proofs
THOU SHALT NOT NEGLECT:1 To state which variable you are inducing on (“Proof by induction on
n, a, r, . . .
2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it
may seem.4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in
the inductive step.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 23 / 48
Weak Induction Intro to Induction
Rules for authoring mathematical induction proofs
THOU SHALT NOT NEGLECT:1 To state which variable you are inducing on (“Proof by induction on
n, a, r, . . .2 Properly using your generic particulars.
3 Explicitly proving the inductive basis, no matter how obvious itmay seem.
4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in
the inductive step.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 23 / 48
Weak Induction Intro to Induction
Rules for authoring mathematical induction proofs
THOU SHALT NOT NEGLECT:1 To state which variable you are inducing on (“Proof by induction on
n, a, r, . . .2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it
may seem.
4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in
the inductive step.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 23 / 48
Weak Induction Intro to Induction
Rules for authoring mathematical induction proofs
THOU SHALT NOT NEGLECT:1 To state which variable you are inducing on (“Proof by induction on
n, a, r, . . .2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it
may seem.4 Clearly assuming the inductive hypothesis
5 Explicitly mentioning where the inductive hypothesis is used inthe inductive step.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 23 / 48
Weak Induction Intro to Induction
Rules for authoring mathematical induction proofs
THOU SHALT NOT NEGLECT:1 To state which variable you are inducing on (“Proof by induction on
n, a, r, . . .2 Properly using your generic particulars.3 Explicitly proving the inductive basis, no matter how obvious it
may seem.4 Clearly assuming the inductive hypothesis5 Explicitly mentioning where the inductive hypothesis is used in
the inductive step.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 23 / 48
Weak Induction Practice
Practice
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 24 / 48
Weak Induction Practice
Mini-quiz
Are the following partial sums (“Sn”s) of a series?
YES NO
1
100∑n=0
2 ∗ n
2
+∞∑n=0
2 ∗ n
3
+∞∑n=1
2 ∗ n
4
0∑n=0
2 ∗ i
5
100∑n=0
0
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 25 / 48
Weak Induction Practice
Mini-quiz
Are the following partial sums (“Sn”s) of a series?
YES NO
1
100∑n=0
2 ∗ n
2
+∞∑n=0
2 ∗ n
3
+∞∑n=1
2 ∗ n
4
0∑n=0
2 ∗ i
5
100∑n=0
0
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 25 / 48
Weak Induction Practice
Mini-quiz
Are the following partial sums (“Sn”s) of a series?
YES NO
1
100∑n=0
2 ∗ n
2
+∞∑n=0
2 ∗ n
3
+∞∑n=1
2 ∗ n
4
0∑n=0
2 ∗ i
5
100∑n=0
0
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 25 / 48
Weak Induction Practice
Mini-quiz
Are the following partial sums (“Sn”s) of a series?
YES NO
1
100∑n=0
2 ∗ n
2
+∞∑n=0
2 ∗ n
3
+∞∑n=1
2 ∗ n
4
0∑n=0
2 ∗ i
5
100∑n=0
0
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 25 / 48
Weak Induction Practice
Mini-quiz
Are the following partial sums (“Sn”s) of a series?
YES NO
1
100∑n=0
2 ∗ n
2
+∞∑n=0
2 ∗ n
3
+∞∑n=1
2 ∗ n
4
0∑n=0
2 ∗ i
5
100∑n=0
0
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 25 / 48
Weak Induction Practice
Three sums!
The following are some statements we will prove inductively inclass!
Note the different ways in which the theorems can be presented.
Theorem (Partial sum of a special-case arithmetic progression)
Let a be an arithmetic progression with a1 = ω = 1. Then, prove thatSn = n(n+1)
2 .
Theorem (Partial sum of a special-case geometric progression)
For r 6= 1 and n ∈ N, prove that∑n
i=0 ri = rn+1−1
r−1
Theorem (Some random theorem Jason found in an old book of his)
For n ≥ 1, 13 + 23 + · · ·+ n3 =[n(n+1)
2
]2.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 26 / 48
Weak Induction Practice
Problems beyond sums
Mathematical induction is useful for proving other properties ofintegers as well!
Let’s prove the following theorems inductively:
Theorem
For all integers n ≥ 1, 22n − 1 is divisible by 3.
Theorem
For all integers n ≥ 3, 2n + 1 < 2n.
Theorem
For all integers n ≥ 2,√n < 1√
1+ 1√
2+ · · ·+ 1√
n.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 27 / 48
Strong Induction
Strong Induction
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 28 / 48
Strong Induction
Nets instead of dominoes
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 29 / 48
Strong Induction
Strong induction principle
Principle of Strong Induction
Let n, a, b ∈ N : a ≤ b. If:
1 P (a), P (a + 1), . . . , P (b) are all true, and
2 ∀k > b, if P (i) is true ∀i : a ≤ i < k, then P (k) is true,
then, P (n) ∀n ≥ a.
Is there any relation to the principle of Weak Induction?
YES NO
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 30 / 48
Strong Induction
Strong induction principle
Principle of Strong Induction
Let n, a, b ∈ N : a ≤ b. If:
1 P (a), P (a + 1), . . . , P (b) are all true, and
2 ∀k > b, if P (i) is true ∀i : a ≤ i < k, then P (k) is true,
then, P (n) ∀n ≥ a.
Is there any relation to the principle of Weak Induction?
YES NO
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 30 / 48
Strong Induction
Strongly inductive proofs: Sketch
We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)
Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.
That’s our safety net!
Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)
Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.
Notice that the inductive hypothesis does not cover P (k)!
All 5 rules for cleanly authoring inductive proofs still apply.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 31 / 48
Strong Induction
Strongly inductive proofs: Sketch
We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)
Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.
That’s our safety net!
Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)
Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.
Notice that the inductive hypothesis does not cover P (k)!
All 5 rules for cleanly authoring inductive proofs still apply.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 31 / 48
Strong Induction
Strongly inductive proofs: Sketch
We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)
Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.
That’s our safety net!
Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)
Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.
Notice that the inductive hypothesis does not cover P (k)!
All 5 rules for cleanly authoring inductive proofs still apply.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 31 / 48
Strong Induction
Strongly inductive proofs: Sketch
We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)
Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.
That’s our safety net!
Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)
Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.
Notice that the inductive hypothesis does not cover P (k)!
All 5 rules for cleanly authoring inductive proofs still apply.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 31 / 48
Strong Induction
Strongly inductive proofs: Sketch
We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)
Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.
That’s our safety net!
Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)
Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.
Notice that the inductive hypothesis does not cover P (k)!
All 5 rules for cleanly authoring inductive proofs still apply.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 31 / 48
Strong Induction
Strongly inductive proofs: Sketch
We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)
Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.
That’s our safety net!
Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)
Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.
Notice that the inductive hypothesis does not cover P (k)!
All 5 rules for cleanly authoring inductive proofs still apply.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 31 / 48
Strong Induction
Strongly inductive proofs: Sketch
We want to prove a statement P (n) ∀n ≥ n0 (like in WeakInduction)
Inductive base: We prove that P (n0), P (n0 + 1), . . . , P (n0 + r)are true, for some r ∈ N.
That’s our safety net!
Inductive hypothesis: For some k > n0 + r, we assume thatP (i) is true for all i between n0 and k − 1. (Make sure you payextra attention to the inequalities here)
Inductive step: We prove that if the hypothesis holds, then P (k)should also hold.
Notice that the inductive hypothesis does not cover P (k)!
All 5 rules for cleanly authoring inductive proofs still apply.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 31 / 48
Strong Induction
An example proof
Theorem (Divisibility of an integer by a prime)
Prove that any integer n > 1 is divisible by a prime number.
We already proved this one through a very complicated directproof by cases.
The strong inductive hypothesis allows for a much more consiseproof!
Notice that, in this case, r = 0.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 32 / 48
Strong Induction
Reminder: recursively defined sequences
Strong induction is very popular for solving problems withrecursively defined sequences.
The Fibonacci sequence is perhaps the most famous such sequence.
Which one among those is a correct definition?1 F1 = 0, F2 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 32 F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 23 F0 = 1, F1 = 1, Fn+1 = Fn + Fn−1 ∀n ≥ 24 F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 2
1 2 3 4
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 33 / 48
Strong Induction
Reminder: recursively defined sequences
Strong induction is very popular for solving problems withrecursively defined sequences.
The Fibonacci sequence is perhaps the most famous such sequence.
Which one among those is a correct definition?1 F1 = 0, F2 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 32 F0 = 0, F1 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 23 F0 = 1, F1 = 1, Fn+1 = Fn + Fn−1 ∀n ≥ 24 F0 = 1, F1 = 1, Fn = Fn−1 + Fn−2 ∀n ≥ 2
1 2 3 4
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 33 / 48
Strong Induction
A proof on a recursive sequence
A statement on a recursive sequence
Let a : a1, a2, . . . be a sequence, recursively defined as follows:
a1 = 0
a2 = 0
ak =3abk/2c + 2 ∀k ≥ 3
Prove that an is even for each integer n ≥ 1.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 34 / 48
Strong Induction
Proof
Proof via strong induction on k.
Let r ≥ 1 be an integer generic particular and P (n) be the proposition that we wantto prove. We proceed inductively:
Inductive Base: For r = 1 and r = 2, a1 and a2 are even, so P (1) and P (2)hold.
(Strong) Inductive Hypothesis: For some r > 2 and for all 1 ≤ i < r,assume that P (i) is true: ai is even.
Inductive step: We want to prove that P (r) is true, i.e that ar = 3abr/2c + 2is even. From a known theorem, we know that the product of an even and anodd integer is even, while the sum of two even numbers is also even. Since wecovered the cases r = 1 and r = 2 in the inductive basis, it is the case thatr ≥ 3, and for all such choices of r, br/2c < r. But this means that abr/2c iseven, by the inductive hypothesis. Therefore, ar is even, since it consists ofa sum of an even number (2) added to another even number (3 times an evennumber).
Since r was chosen arbitrarily from the set N∗, the result holds ∀n ∈ N∗.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 35 / 48
Strong Induction
Another proof on a recursive sequence
The closed form for another recursive sequence
Let a be a sequence such that:
a1 = 1
a2 = 8
an = an−1 + 2an−2, ∀n ≥ 3
Prove that an = 3 · 2n−1 + 2(−1)n.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 36 / 48
Strong Induction
Proof
Proof (by strong induction on n).
Let k be a generic particular for N∗, and P (n) be the proposition that we want to show. Weproceed inductively:
Inductive base: P (1) is the proposition: a1 = 3 · 20 + 2 · (−1)1 ⇔ a1 = 1. Since, bythe definition of the sequence a, a1 = 1, P (1) is true. P (2) is the proposition:a2 = 3 · 21 + 2(−1)2 ⇔ a2 = 8. Once again, this is true by the definition of a, so P (2)also holds.
(Strong) inductive hypothesis: For some k > 2, we assume P (i) is true∀i ∈ {1, 2, . . . , k − 1}, i.e ai = 3 · 2i−1 + 2 · (−1)i.
Inductive step: We want to prove P (k), i.e ak = 3 · 2k−1 + 2(−1)k. By thedefinition of an for n ≥ 3, we have:
ak = ak−1 + 2ak−2
= 3 · 2k−2 + 2(−1)k−1 + 2[3 · 2k−3 + 2(−1)k−2] (By Inductive Hypothesis)
= 3 · 2k−2 + 2(−1)k−1 + 3 · 2k−2 + 2 · 2(−1)k−2 (By distributing the factor 2)
= 3 · 2k−1 + 2[(−1)k−1 + 2(−1)k−2] (By grouping terms)
= 3 · 2k−1 + 2(−1)k (By equality of the red terms)
⇔ P (k) is true
Since our result was reached for an arbitrarily selected k,it holds ∀n ∈ N∗.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 37 / 48
Strong Induction
Re-visiting an older theorem
Theorem (Divisibility by a prime)
Every integer n > 1 is divisible by a prime number.
Do you perhaps remember what the direct proof that we discussedpreviously looked like?
Strong induction will lead us to a much more consise proof!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 38 / 48
Strong Induction
Re-visiting an older theorem
Theorem (Divisibility by a prime)
Every integer n > 1 is divisible by a prime number.
Do you perhaps remember what the direct proof that we discussedpreviously looked like?
Strong induction will lead us to a much more consise proof!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 38 / 48
Strong Induction
Re-visiting an older theorem
Theorem (Divisibility by a prime)
Every integer n > 1 is divisible by a prime number.
Do you perhaps remember what the direct proof that we discussedpreviously looked like?
Strong induction will lead us to a much more consise proof!
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 38 / 48
Strong Induction
ProofProof (by strong induction on n).
Let q be a generic particular for the set {2, 3, . . . } and P (n) be the proposition we aretrying to prove. We proceed inductively:
Inductive base: For q = 2, we have that 2 is divisible by a prime number, namelyitself. Therefore, P (2) holds.
(Strong) Inductive hypothesis: For q > 2, we assume that P (i) holds∀i ∈ {2, 3, . . . , q − 1}, i.e i is divisible by a prime number.
Inductive step: We want to prove P (q), i.e that q > 2 is divisible by a prime
number. We distinguish between two cases:
1 q is prime. In this case, since q|q, we are done.2 q is not prime. Therefore, q is composite, which means that∃a, b ∈ N : q = a · b.a Since both a and b are above 1 (if one of them was 1, theother would have to be q, making q prime), we conclude that both of them haveto be smaller than q. By the inductive hypothesis, this means that thereexists some prime p that divides at least one of a or b, and by transitivity ofdivisibility we have that p|q. This is exactly P (q).
Since q was selected arbitrarily from {2, 3, . . . }, we have that the result holds for everyinteger in this set.
aWe only care about the positive divisors of q.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 39 / 48
Strong Induction
Practice
Let’s split into teams and solve these problems!
Yet another recursive sequence
Suppose that the sequence s is such that: s0 = 12, s1 = 29 andsk = 5sk−1 − 6sk−2 ∀k ≥ 2. Prove that sk = 5 · 3k + 7 · 2k, ∀k ∈ N.
More Fibonacci? Sure, why not.
Show that the Fibonacci sequence follows an “odd-odd-even” pattern.More formally, if by F we denote this sequence, show that∀m ∈ N∗, F3m−2 and F3m−1 are odd, while F3m is even.
The “Tribonacci” Sequence (yes, seriously)
The Tribonacci sequence is defined as: T0 = T1 = T2 = 1 andTn = Tn−1 + Tn−2 + Tn−3 for n ≥ 3. Prove that, for alln ∈ N, Tn < 2n+1
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 40 / 48
Errors in proofs by mathematical induction
Errors in proofs by mathematical induction
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 41 / 48
Errors in proofs by mathematical induction
This section
We will present some examples of non-sensical statements provenvia the principle of induction, or of correct statements withincorrect proofs!
The teaching staff assumes no responsibility for studentsbelieving those to be valid statements or proofs!
All examples courtesy of materials taken from MATH 347,Summer ’14, UIUC.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 42 / 48
Errors in proofs by mathematical induction
Non-sensical proof #1Sum of first n non-zero integers
∀n ∈ N∗,n∑
i=1
i =1
2(n +
1
2)2
Proof (By weak induction on n).
Let r ∈ N∗ be a generic particular and P (n) the statement we want to solve. We proceed inductively:
1 Inductive base: For r = 1, P (1) holds.
2 Inductive hypothesis: Assume that P (r) holds ∀r ≥ 1, i.er∑
i=1
i =1
2(r +
1
2)2
3 Inductive step: We want to prove P (r + 1), i.e
r+1∑i=1
i =1
2((r + 1) +
1
2)2. Beginning from the LHS
we have:
r+1∑i=1
i =r∑
i=1
i + (r + 1) =1
2(r +
1
2)2
+ (r + 1) (By breaking apart the sum and by I.H)
=1
2
[(r
2+ r +
1
4) + (2r + 2)
]=
1
2
[(r +
3
2)2 − 3r −
9
4+ r +
1
4+ (2r + 2)
](By algebra)
=1
2
[(r + 1) +
1
2
]2⇒ P (r + 1) holds. (By algebra)
Since r was chosen arbitrarily from N∗, we conclude that the result must hold for all positive integers.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 43 / 48
Errors in proofs by mathematical induction
Non-sensical proof #2
Raising 2 to a non-negative integer equals 1
∀n ∈ N, 2n = 1
Proof (by strong induction on n).
Let P (n) be the proposition that we’re trying to prove and r ∈ N be a generic particular.We proceed inductively:
1 Inductive base: For r = 0, 2r = 1. So P (0) = 1.
2 Inductive hypothesis: For r > 0, we assume that ∀i ∈ {0, 1, . . . , r}, P (i) holds, i.ethat 2i = 1.
3 Inductive step: We want to prove that P (r + 1) holds, i.e that 2r+1 = 1. We beginfrom the LHS:
2r+1 =21 · 2r (By properties of exponentiation)
=1 · 1 (By inductive hypothesis)
= 1
Therefore, P (r + 1) holds.
Since r was arbitrarily chosen within N, the result holds ∀n ∈ N.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 44 / 48
Errors in proofs by mathematical induction
Non-sensical proof #3
LOL WUT
All real numbers are equal.
Proof.
Equivalently, we want to prove that, ∀k ∈ N, a1 = a2 = · · · = ak, whereai ∈ R. We will proceed via strong induction on k. Let r ∈ N∗ be ageneric particular and P (k) be the proposition we’re trying to solve.We proceed inductively:
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 45 / 48
Errors in proofs by mathematical induction
“All reals are equal” proof, contd.
Proof.
1 Inductive base: For r = 1, the statement is trivially true: a1 = a1.
2 Inductive hypothesis: For r > 1, we will assume that ∀i ∈ {0, 1, . . . , r − 1},P (i) is true, i.e any i reals are true: a1 = a2 = · · · = ai.
3 Inductive step: We want to prove that P (r) is true, i.e thata1 = a2 = · · · = ar. Consider the first r − 1 reals. Since r − 1 < r, from the I.Hwe can deduce:
a1 = a2 = · · · = ar−1 (1)
The same can be deduced for the last r − 1 reals:
a2 = a3 = · · · = ar (2)
From (1) and (2) we conclude that a1 = a2 = · · · = ar−1 = ar, and P (r) is true.
Since r was arbitrarily chosen, the result holds ∀k ∈ N.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 46 / 48
Errors in proofs by mathematical induction
True statement, incorrect proof
Parity of the factorial
∀n > 1, n! is even
Proof (via strong induction on n).
Let r ∈ {2, 3, . . . } be a generic particular and P (n) be the proposition we’re tryingto prove. We proceed inductively:
1 Inductive base: For r = 2, 2! = 2 · 1 = 2. 2 is even, therefore P (2) holds.
2 Inductive hypothesis: For r > 2, we assume P (i) ∀i ∈ {2, 3, . . . , r − 1}, i.e i!is even.
3 Inductive step: We want to prove P (r + 1), i.e that (r + 1)! is even. By therecursive definition of the factorial, (r + 1)! = (r + 1) · r! From the inductivehypothesis, we know that r! is even. From a known theorem, we know that theproduct of two integers, where at least one is even, is also even. Therefore,(r + 1)! is even and P (r + 1) holds.
Since r was chosen arbitrarily, the result holds for every integer n ≥ 2.
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 47 / 48
Errors in proofs by mathematical induction
Vote!
Proof (via strong induction on n).
Let r ∈ {2, 3, . . . } be a generic particular and P (n) be the proposition we’re tryingto prove. We proceed inductively:
1 Inductive base: For r = 2, 2! = 2 · 1 = 2. 2 is even, therefore P (2) holds.
2 Inductive hypothesis: For r > 2, we assume P (i) ∀i ∈ {2, 3, . . . , r − 1}, i.e i!is even.
3 Inductive step: We want to prove P (r + 1), i.e that (r + 1)! is even. By therecursive definition of the factorial, (r + 1)! = (r + 1) · r! From the inductivehypothesis, we know that r! is even. From a known theorem, we know that theproduct of two integers, where at least one is even, is also even. Therefore,(r + 1)! is even and P (r + 1) holds.
Since r was chosen arbitrarily, the result holds for every integer n ≥ 2.
This proof has an error in:
The Ind. Base The Ind. Hypothesis The Ind. Step Somewhere else
Jason Filippou (CMSC250 @ UMCP) Induction 06-27-2016 48 / 48