Strong Induction CMSC 250

Announcements HW5 up 5 minutes after lecture Online Quiz up 5 mins after lecture Midterm 1: 03-15, 05:30 – 07:30pm, various rooms (syllabus) Midterm 2:04-24, 05:30-07:30pm, various rooms (syllabus)

Strong Induction: The principle Let us recall the weak induction principle for a moment (Slides from last week)

Strong Induction: The principle Let us recall the weak induction principle for a moment (Slides from last week) The strong induction principle is different in only one thing: Instead of depending on just 𝑃(𝑘) to deduce 𝑃(𝑘+1), we will depend on many 𝑃 𝑖 , 0≤𝑖≤𝑘

Strong Induction: The principle Let us recall the weak induction principle for a moment (Slides from last week) The strong induction principle is fundamentally different in only one thing: Instead of depending on just 𝑃(𝑘) to deduce 𝑃(𝑘+1), we will depend on many 𝑃 𝑖 , 0≤𝑖≤𝑘 Visualization: Weak Induction Strong Induction

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true a … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true a … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 a … 𝑘 𝑘+1 𝑘+2 … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 a … 𝑘 𝑘+1 𝑘+2 … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 a … 𝑘 𝑘+1 𝑘+2 … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 𝑎=𝑘 𝑘+1 𝑘+2 … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 Then, 𝑃(𝑘+1) is also true a … 𝑘 𝑘+1 𝑘+2 … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 Then, 𝑃(𝑘+1) is also true a … 𝑘 𝑘+1 𝑘+2 … … −2 −1 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 Then, 𝑃(𝑘+1) is also true a … … 𝑘 𝑘+1 𝑘+2 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥ 0 0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ⇒𝑷 𝒌+𝟏 Then, 𝑃(𝑘+1) is also true But then, so is 𝑃 𝑘+2 , since we can simply expand the net to include 𝑃(𝑘+1)… a … … 𝑘 𝑘+1 𝑘+2 𝑘+3 𝑘+4 … 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌 ∧𝑷(𝒌+𝟏)⇒𝑷 𝒌+𝟐 Then, 𝑃(𝑘+1) is also true But then, so is 𝑃 𝑘+2 , since we can simply expand the net to include 𝑃(𝑘+1)… a … … 𝑘 𝑘+1 𝑘+2 𝑘+3 𝑘+4 … 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌+𝟏 ∧𝑷(𝒌+𝟐)⇒𝑷 𝒌+𝟑 Then, 𝑃(𝑘+1) is also true But then, so is 𝑃 𝑘+2 , since we can simply expand the net to include 𝑃(𝑘+1)… And 𝑃(𝑘+3)… a … … 𝑘 𝑘+1 𝑘+2 𝑘+3 𝑘+4 … 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌+𝟐 ∧𝑷(𝒌+𝟑)⇒𝑷 𝒌+𝟒 Then, 𝑃(𝑘+1) is also true But then, so is 𝑃 𝑘+2 , since we can simply expand the net to include 𝑃(𝑘+1)… And 𝑃(𝑘+4)… a … … 𝑘 𝑘+1 𝑘+2 𝑘+3 𝑘+4 … 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌+𝟐 ∧𝑷(𝒌+𝟑)⇒𝑷 𝒌+𝟒 Then, 𝑃(𝑘+1) is also true But then, so is 𝑃 𝑘+2 , since we can simply expand the net to include 𝑃(𝑘+1)… And 𝑃(𝑘+4)… and all 𝑛≥0 a … … 𝑘 𝑘+1 𝑘+2 𝑘+3 𝑘+4 … 1 2 …

Strong Induction: The principle The goal is the same: We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 The principle has, once again, two presuppositions. If: 𝑃 0 ,𝑃 1 , …,𝑃(𝑎) are true For a 𝑘≥𝑎, 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷 𝒂 ∧𝑷 𝒂+𝟏 ∧…∧𝑷 𝒌+𝟐 ∧𝑷(𝒌+𝟑)⇒𝑷 𝒌+𝟒 Then, 𝑃(𝑘+1) is also true But then, so is 𝑃 𝑘+2 , since we can simply expand the net to include 𝑃(𝑘+1)… And 𝑃(𝑘+4)… and all 𝑛≥0  a … … 𝑘 𝑘+1 𝑘+2 𝑘+3 𝑘+4 … 1 2 …

How we’ll make it work We want to prove a statement 𝑃 𝑛 ∀𝑛≥0 … … 3 2 … −2 −1 1 2

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 3 … a … … … −2 −1 1 2

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 3 … a … … … −2 −1 1 2 “Safety Net” Applied!

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 Inductive hypothesis: For some 𝑘≥𝑎 and for every 𝑖:0≤𝑖≤𝑘, we will assume that 𝑃(𝑖) holds 3 … a … … … −2 −1 1 2 “Safety Net” Applied!

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 Inductive hypothesis: For some 𝑘≥𝑎 and for every 𝑖:0≤𝑖≤𝑘, we will assume that 𝑃(𝑖) holds … a … k k+1 𝑘+2 … 3 … −2 −1 1 2 “Safety Net” Extended!

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 Inductive hypothesis: For some 𝑘≥𝑎 and for every 𝑖:0≤𝑖≤𝑘 We will assume that 𝑃(𝑖) holds. Inductive step: We attempt to prove 𝑃 𝑘+1 . … a … k k+1 𝑘+2 … 3 … −2 −1 1 2 “Safety Net” Extended!

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 Inductive hypothesis: For some 𝑘≥𝑎 and for every 𝑖:0≤𝑖≤𝑘 We will assume that 𝑃(𝑖) holds. Inductive step: We attempt to prove 𝑃 𝑘+1 . Note that we assume 𝑷 𝟎 ∧𝑷 𝟏 ∧…∧𝑷(𝒌) ! k k+1 𝑘+2 … 3 … a … … −2 −1 1 2 “Safety Net” Extended!

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 Inductive hypothesis: For some 𝑘≥𝑎 and for every 𝑖:0≤𝑖≤𝑘 We will assume that 𝑃(𝑖) holds. Inductive step: We attempt to prove 𝑃 𝑘+1 . But, by the inductive principle, this means that we can expand our net some more… … a … k k+1 𝑘+2 … 3 … −2 −1 1 2

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 Inductive hypothesis: For some 𝑘≥𝑎 and for every 𝑖:0≤𝑖≤𝑘 We will assume that 𝑃(𝑖) holds. Inductive step: We attempt to prove 𝑃 𝑘+1 . But, by the inductive principle, this means that we can expand our net some more… And prove the statement for 𝑛=𝑘+2 … a … k k+1 𝑘+2 𝑘+3 3 … −2 −1 1 2

How we’ll make it work Inductive base: We will explicitly prove (no matter how easy it might initially seem) that 𝑃 0 , 𝑃 1 , 𝑃 2 , …, 𝑃 𝑎 Inductive hypothesis: For some 𝑘≥𝑎 and for every 𝑖:0≤𝑖≤𝑘 We will assume that 𝑃(𝑖) holds. Inductive step: We attempt to prove 𝑃 𝑘+1 . But, by the inductive principle, this means that we can expand our net some more… And prove the statement for 𝑛=𝑘+2, 𝑛=𝑘+3, … … a … k k+1 𝑘+2 𝑘+3 3 … −2 −1 1 2

Utility of strong induction Enormous It is effectively a generalization of weak induction

Utility of strong induction Enormous It is effectively a generalization of weak induction Terrifically useful in sequences How many ways have we talked about that can be used to describe a sequence? 1 2 3 4

Utility of strong induction Enormous It is effectively a generalization of weak induction Terrifically useful in sequences How many ways have we talked about that can be used to describe a sequence? Also useful in the study of algorithm correctness and efficiency. Outlining terms Recursive definition Closed-form formula 1 2 3 4

A first example Let 𝑎 be a sequence such that: 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 Prove that 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 , 𝑛∈ ℕ

A first example Let 𝑎 be a sequence such that: 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 Prove that 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 , 𝑛∈ ℕ How many elements in my inductive base? 𝑃 𝑛 1 2 3 Something Else

A first example Let 𝑎 be a sequence such that: 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 Prove that 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 , 𝑛∈ ℕ How many elements in my inductive base? Make a note here about how we are given the closed-form formula and we are trying to prove that it exactly represents the sequence (i.e it’s not an upper bound) 𝑃 𝑛 1 2 3 Something Else

Inductive base 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 𝑃 𝑛 ⇔ 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 For 𝑛=0, 𝑎 0 =1 by the definition of 𝑎. 𝑃(0) says: 𝑎 0 =3⋅ 2 0 +2 −1 1 =3−2=1. So 𝑃(0) holds. For n = 1, 𝑎 1 =8 by the definition of 𝑎. 𝑃(1) says: 𝑎 1 =3⋅ 2 1 +2 −1 2 =6+2=8. So P(1) holds.

Inductive Base established! 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 𝑃 𝑛 ⇔ 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 For 𝑛=0, 𝑎 0 =1 by the definition of 𝑎. 𝑃(0) says: 𝑎 0 =3⋅ 2 0 +2 −1 1 =3−2=1. So 𝑃(0) holds. For n = 1, 𝑎 1 =8 by the definition of 𝑎. 𝑃(1) says: 𝑎 1 =3⋅ 2 1 +2 −1 2 =6+2=8. So P(1) holds. Inductive Base established!

Inductive Hypothesis 𝑃( 𝑃( ) ) 1 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 𝑃 𝑛 ⇔ 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 For 𝑘≥1, assume 𝑃 𝑖 , ∀𝑖=0, 1, …, 𝑘, i.e 𝑎 𝑖 =3⋅ 2 𝑖 +2 −1 𝑖+1 , 𝑖=0, 1, …, 𝑘

Inductive Hypothesis made! 𝑃( ) 𝑃( ) 1 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 𝑃 𝑛 ⇔ 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 For 𝑘≥1, assume 𝑃 𝑖 , ∀𝑖=0, 1, …, 𝑘, i.e 𝑎 𝑖 =3⋅ 2 𝑖 +2 −1 𝑖+1 , 𝑖=0, 1, …, 𝑘 Inductive Hypothesis made!

Inductive Step 𝑃( 𝑃( ) ) We will now prove 𝑃 𝑘+1 , i.e 1 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 𝑃 𝑛 ⇔ 𝑎 𝑛 =3⋅ 2 𝑛 +2 −1 𝑛+1 We will now prove 𝑃 𝑘+1 , i.e 𝑎 𝑘+1 =3⋅ 2 𝑘+1 +2 −1 𝑘+2 𝑃 𝑖 ∀𝑖=0,1, …𝑘

Inductive Step 𝑃( ) 𝑃( ) 1 𝑎 𝑘 = 1, 𝑘=0 8, 𝑘=1 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥2 𝑃 𝑛 ⇔ 𝑎 𝑛 =3⋅ 2 𝑛−1 +2 −1 𝑛 From the recursive definition of 𝑎 𝑘 , we obtain: 𝑎 𝑘+1 = 𝑎 𝑘 +2⋅ 𝑎 𝑘−1 = 𝐼.𝐻 3⋅ 2 𝑘 +2 −1 𝑘+1 +2⋅ 3⋅ 2 𝑘−1 +2 −1 𝑘 = =3⋅ 2 𝑘 +2⋅ 2 𝑘−1 +2⋅ −1 𝑘 [1 −2]= =3⋅ 2⋅ 2 𝑘−1 + 2⋅ −1 𝑘 −1 =3⋅ 2 𝑘 +2 −1 𝑘+1 𝑃 𝑖 ∀𝑖=0,1, …𝑘 My goal: Prove 𝑃 𝑘+1 : 𝑎 𝑘+1 =3⋅ 2 𝑘 +2 −1 𝑘+1

Inductive Step 𝑃( ) 𝑃( ) 1 2 𝑎 𝑘 = 1, 𝑘=1 8, 𝑘=2 𝑎 𝑘−1 +2⋅ 𝑎 𝑘−2 , 𝑘≥3 𝑃 𝑛 ⇔ 𝑎 𝑛 =3⋅ 2 𝑛−1 +2 −1 𝑛 From the recursive definition of 𝑎 𝑘 , we obtain: 𝑎 𝑘+1 = 𝑎 𝑘 +2⋅ 𝑎 𝑘−1 = 𝐼.𝐻 3⋅ 2 𝑘 +2 −1 𝑘+1 +2⋅ 3⋅ 2 𝑘−1 +2 −1 𝑘 = =3⋅ 2 𝑘 +2⋅ 2 𝑘−1 +2⋅ −1 𝑘 [1 −2]= =3⋅ 2⋅ 2 𝑘−1 + 2⋅ −1 𝑘 −1 =3⋅ 2 𝑘 +2 −1 𝑘+1 𝑃 𝑖 ∀𝑖=0,1, …𝑘 My goal: Prove 𝑃 𝑘+1 : 𝑎 𝑘+1 =3⋅ 2 𝑘 +2 −1 𝑘+1 Inductive step proven!

Here’s another Suppose that the sequence 𝑠 is as follows: 𝑠 𝑘 = 12, 𝑘=0 29, 𝑘=1 5 𝑠 𝑘−1 −6 𝑠 𝑘−2 , 𝑘≥2 Then, prove that 𝑠 𝑛 =5⋅ 3 𝑛 +7⋅ 2 𝑛 , ∀𝑛∈ℕ

Inductive base Let the statement to be proven be called 𝑃(𝑛). We proceed via strong induction on 𝑛. Inductive base: We want to prove 𝑃 0 , 𝑃(1). For 𝑛=0, 𝑃 0 is 𝑠 0 =5⋅ 3 0 +7⋅ 2 0 ⇔12=12 For 𝑛=1, 𝑃 1 is 𝑠 1 =5⋅ 3 1 +7⋅ 2 1 ⇔29=15+14 So the inductive base has been established!

Inductive hypothesis Inductive Hypothesis: Let 𝑘≥1 be a natural. Then, we assume that, for all 𝑖=0,1,…, 𝑘, 𝑃 𝑖 holds, i.e 𝑠 𝑖 = 5⋅ 3 𝑖 +7⋅ 2 𝑖 , 𝑖=0,1,…,𝑘

Inductive Step Inductive Step: We will attempt to prove 𝑃(𝑘+1), i.e 𝑠 𝑘+1 =5⋅ 3 𝑘+1 + 7⋅ 2 𝑘+1 From the recursive definition of 𝑠 we have: 𝑠 𝑘+1 =5 𝑠 𝑘 −6 𝑠 𝑘−1 = 𝐼.𝐻 5 5⋅ 3 𝑘 +7⋅ 2 𝑘 −6 5⋅ 3 𝑘−1 +7⋅ 2 𝑘−1 =25⋅ 3 𝑘 +35⋅ 2 𝑘 −30⋅ 3 𝑘−1 −42⋅ 2 𝑘−1 =5⋅ 5⋅ 3 𝑘 −2⋅ 3 𝑘 +7 5⋅ 2 𝑘 −3⋅ 2 𝑘 =5⋅ 3 𝑘+1 +7⋅ 2 𝑘+1 □ Since we need factors of 5 and 7 in our result, we force them to appear and our lives automatically become easier!

Any other classes of sequence problems? (solvable via strong induction) True False Strong induction can only help us recover closed-form formulae for sequences

Any other classes of sequence problems? (solvable via strong induction) Of course! Pretty much like weak induction. True False Prove that the Fibonacci sequence follows an odd-odd-even pattern, i.e ∀𝑚∈ ℕ ∗ , 𝐹 3𝑚−2 and 𝐹 3𝑚−1 are odd and 𝐹 3𝑚 are even. For the Tribonacci sequence, defined as: 𝑇 0 = 𝑇 1 = 𝑇 2 =1, and 𝑇 𝑛 = 𝑇 𝑛−1 + 𝑇 𝑛−2 + 𝑇 𝑛−3 , ∀𝑛≥3, prove that 𝑇 𝑛 < 2 𝑛+1

Any other classes of sequence problems? (solvable via strong induction) Of course! Pretty much like weak induction. True False Prove that the Fibonacci sequence follows an odd-odd-even pattern, i.e ∀𝑚∈ ℕ ∗ , 𝐹 3𝑚−2 and 𝐹 3𝑚−1 are odd and 𝐹 3𝑚 are even. For the Tribonacci sequence, defined as: 𝑇 0 = 𝑇 1 = 𝑇 2 =1, and 𝑇 𝑛 = 𝑇 𝑛−1 + 𝑇 𝑛−2 + 𝑇 𝑛−3 , ∀𝑛≥3, prove that 𝑇 𝑛 < 2 𝑛+1 Pick your favorite, and let’s see what you got!

And here’s a crucial strong induction proof not on recursive formulae! Any other classes of sequence problems? (solvable via strong induction) Of course! Pretty much like weak induction. True False Prove that the Fibonacci sequence follows an odd-odd-even pattern, i.e ∀𝑚∈ ℕ ∗ , 𝐹 3𝑚−2 and 𝐹 3𝑚−1 are odd and 𝐹 3𝑚 are even. For the Tribonacci sequence, defined as: 𝑇 0 = 𝑇 1 = 𝑇 2 =1, and 𝑇 𝑛 = 𝑇 𝑛−1 + 𝑇 𝑛−2 + 𝑇 𝑛−3 , ∀𝑛≥3, prove that 𝑇 𝑛 < 2 𝑛+1 Pick your favorite, and let’s see what you got! And here’s a crucial strong induction proof not on recursive formulae!

∀𝑛≥2 [(∃ 𝑝 𝑖 ∈𝐏, 𝑒 𝑖 ∈ ℕ ≥1 )[𝑛= 𝑖=0 𝑟 𝑝 𝑖 𝑒 𝑖 ]] Existence part of UFT Recall the statement of UFT: For any integer 𝑛≥2, there exist unique 𝑝 𝑖 ∈𝐏, 𝑒 𝑖 ∈ ℕ ≥1 such that: 𝑛= 𝑖=0 𝑟 𝑝 𝑖 𝑒 𝑖 We will prove the existence part, i.e that ∀𝑛≥2 [(∃ 𝑝 𝑖 ∈𝐏, 𝑒 𝑖 ∈ ℕ ≥1 )[𝑛= 𝑖=0 𝑟 𝑝 𝑖 𝑒 𝑖 ]] Maybe point out the difference between the book’s first proof and this elegant proof.

Proof IB: For 𝑛=2, 2= 2 1 , so ∃ 𝑝 1 ∈𝐏, 𝑒 1 ∈ ℕ ≥1 [2= 𝑝 1 𝑒 1 ]. Done. I.H: Let 𝑘≥2 be a natural. Then, assume that ∀𝑗∈ 2, 3, …,𝑘 ∃ 𝑝 𝑖 ∈𝐏, 𝑒 𝑖 ∈ ℕ ≥1 [𝑘= 𝑖=0 𝑟 𝑝 𝑖 𝑒 𝑖 ] I.S: We will now prove 𝑃(𝑘+1), i.e that ∃ 𝑝 𝑖 ′ ∈𝐏, 𝑒 𝑖 ′ ∈ ℕ ≥1 [𝑘+1= 𝑖=0 𝑟 𝑝 𝑖 ′ 𝑒 𝑖 ′ ]

Proof IB: For 𝑛=2, 2= 2 1 , so ∃ 𝑝 1 ∈𝐏, 𝑒 1 ∈ ℕ ≥1 [2= 𝑝 1 𝑒 1 ]. Done. I.H: Let 𝑘≥2 be a natural. Then, assume that ∀𝑗∈ 2, 3, …,𝑘 [ ∃ 𝑝 𝑖 ∈𝐏, 𝑒 𝑖 ∈ ℕ ≥1 [𝑘= 𝑖=0 𝑟 𝑝 𝑖 𝑒 𝑖 ] ] I.S: We will now prove 𝑃(𝑘+1), i.e that ∃ 𝑝 𝑖 ′ ∈𝐏, 𝑒 𝑖 ′ ∈ ℕ ≥1 [𝑘+1= 𝑖=0 𝑟 𝑝 𝑖 ′ 𝑒 𝑖 ′ ] Case #1: 𝑘+1∈𝐏. Then, 𝑝 1 ′ =𝑘+1, 𝑒 1 =1. Case #2: 𝑘+1∉𝐏. Then, ∃ ℓ 1 , ℓ 2 ∈{2, 3, …,𝑘} [𝑘+1= ℓ 1 ⋅ ℓ 2 ] Inductive step contd. in next slide for readability…

Inductive step (contd.) Case #2: 𝑘+1∉𝐏. Then, ∃ ℓ 1 , ℓ 2 ∈{2, 3, …,𝑘} [𝑘+1= ℓ 1 ⋅ ℓ 2 ]

Inductive step (contd.) Case #2: 𝑘+1∉𝐏. Then, ∃ ℓ 1 , ℓ 2 ∈{2, 3, …,𝑘} [𝑘+1= ℓ 1 ⋅ ℓ 2 ] By the Inductive Hypothesis, ∃ 𝑞 𝑖 ∈𝐏, 𝑠 𝑖 ∈ ℕ ≥1 [ ℓ 1 = 𝑖=0 𝑚 1 𝑞 𝑖 𝑒 𝑖 ] and ∃ 𝑞 𝑖 ′ ∈𝐏, 𝑠 𝑖 ′ ∈ ℕ ≥1 [ ℓ 2 = 𝑖=0 𝑚 2 𝑞 𝑖 ′ 𝑒 𝑖 ′ ]

Inductive step (contd.) Case #2: 𝑘+1∉𝐏. Then, ∃ ℓ 1 , ℓ 2 ∈{2, 3, …,𝑘} [𝑘+1= ℓ 1 ⋅ ℓ 2 ] By the Inductive Hypothesis, ∃ 𝑞 𝑖 ∈𝐏, 𝑠 𝑖 ∈ ℕ ≥1 [ ℓ 1 = 𝑖=0 𝑚 1 𝑞 𝑖 𝑒 𝑖 ] and ∃ 𝑞 𝑖 ′ ∈𝐏, 𝑠 𝑖 ′ ∈ ℕ ≥1 [ ℓ 2 = 𝑖=0 𝑚 2 𝑞 𝑖 ′ 𝑒 𝑖 ′ ] Therefore, 𝑘+1= 𝑖=0 max⁡{ 𝑚 1 , 𝑚 2 } 𝑞 𝑖 𝑒 𝑖 ⋅ 𝑞 𝑖 ′ 𝑒 𝑖 ′ which is a product of prime powers. Done.

Important note Recall case #2: 𝑘+1∉𝐏. Then, ∃ ℓ 1 , ℓ 2 ∈{2, 3, …,𝑘} [𝑘+1= ℓ 1 ⋅ ℓ 2 ] Note that 𝑃(𝑘) (the proposition for 𝑘+1) depends on falling back to the assumed 𝑃 ℓ 1 , 𝑃( ℓ 2 ).

Important note In our proof about recurrences, 𝑃(𝑘) dependent on stuff such as 𝑃 𝑘−1 , 𝑃 𝑘−2 , … It is possible (and common, e.g hw 5) for 𝑃(𝑘) to depend on 𝑃 𝑘 2 ,𝑃 𝑘 3 , 𝑃( 𝑘) , …

Important note In our proof about recurrences, 𝑃(𝑘) dependent on stuff such as 𝑃 𝑘−1 , 𝑃 𝑘−2 , … It is possible (and common, e.g hw 5) for 𝑃(𝑘) to depend on 𝑃 𝑘 2 ,𝑃 𝑘 3 , 𝑃( 𝑘) , … By contrast: case #2 in previous proof: 𝑘+1∉𝐏. Then, ∃ ℓ 1 , ℓ 2 ∈{2, 3, …,𝑘} [𝑘+1= ℓ 1 ⋅ ℓ 2 ] Note that 𝑃(𝑘) (the proposition for 𝑘+1) depends on falling back to the assumed 𝑃 ℓ 1 , 𝑃( ℓ 2 ). But where are ℓ 𝟏 , ℓ 𝟐 in reference to 𝒌+𝟏?

Important note In our proof about recurrences, 𝑃(𝑘) dependent on stuff such as 𝑃 𝑘−1 , 𝑃 𝑘−2 , … It is possible (and common, e.g hw 5) for 𝑃(𝑘) to depend on 𝑃 𝑘 2 ,𝑃 𝑘 3 , 𝑃( 𝑘) , … By contrast: case #2 in previous proof: 𝑘+1∉𝐏. Then, ∃ ℓ 1 , ℓ 2 ∈{2, 3, …,𝑘} [𝑘+1= ℓ 1 ⋅ ℓ 2 ] Note that 𝑃(𝑘) (the proposition for 𝑘+1) depends on falling back to the assumed 𝑃 ℓ 1 , 𝑃( ℓ 2 ). But where are ℓ 𝟏 , ℓ 𝟐 in reference to 𝒌+𝟏? We neither know nor care: the safety net applied by the I.H takes care of this! 2 3 4 … ℓ 1 … ℓ 2 … k k+1