# Lecture 011

## Induction

### Sequences

Sequence: defined inductively (recursively)

sequence of real numbers

• definition: $S = \{(n, a_n) \in \mathbb{N} \times \mathbb{R}\}$ such that for all $n\in \mathbb{N}$, there exists a unique $a_n\in \mathbb{R}$ such that $(n,a_n)\in S$

The Factorial Sequence

• definition:
• "n factorial" = 1 if n = 0 (base case)
• = n(n-1)! if n > 0 (inductive step)

Conjecture: for all integer n, if n >=4, then n!>2^n Proof: proceed by induction on $n\in \mathbb{N}$ with $n \geq 4$

• Base Case:...

• Inductive Step: let $n\in \mathbb{N} \land n \geq 4$ assume $n! > 2^n$, WTS $(n+1)! > 2^{n+1}$

• $(n+1)!=(n+1)n!$ (definition)
• $>(n+1)$ (IH)
• $>2*2^n = 2^{n+1}$
• By PMI, we conclude...

The Fibonacci Sequence

• definition:
• fn = 0 if n = 0 (base case)
• fn = 1 if n = 1 (base case)
• fn = f_{n-1) + f_{n-2}} (inductive step)

### Strong Principle of Mathematical Induction

Assumption

1. P(0) holds
2. $(\forall k \in \mathbb{N})((\forall i \in [k] \cup \{0\})P(i) \implies P(k+1))$

We can use the PMI to prove Strong Principle of Mathematical Induction. Proof: Let P(n) be a variable proposition defined on $n \in \mathbb{N}$ satisfying conditions (1) and (2) of the theorem. Let Q(n) be variable proposition $(\forall i \in [n] \cup \{0\}P(i)$. We proceed by induction on $n\in \mathbb{N}$ to show Q(n) holds for all n.

• Base Case: when n=0, P(0) holds by assumption (1). $Q(0) \equiv P(0)$ Thus, Q(0) holds

• Inductive Step:

• let $k\in \mathbb{N}$ such that Q(k) holds. By definition of Q(k) and inductive hypothesis (IH), we have $(\forall i \in [k] \cup \{0\})P(i)$ holds.
• (2) implies P(k+1) holds. combine $(\forall i \in [k] \cup \{0\})P(i)$ and P(k+1), we have Q(k+1) holds.

Show: $\implies (\forall n \in \mathbb{N})P(n)$, fix n. $Q(n) \implies P(n)$ because $Q(n) = (\forall i \in [n] \cup {0})P(i) \land n\in[n]\cup {0}$.

#### Template for Strong Induction

Claim: $(\forall n \in \mathbb{N})P(n)$ Proof: We proceed by strong induction on $n\in \mathbb{N}$

• Base Case: P(0) holds because... (may be more base cases)

• Inductive Step: let $k\in \mathbb{N}$ and assume $(\forall i \in [k] \cup {0})P(i)$ holds. If there are M-many base cases ($k \geq M$), show $P(k+1)$ holds

• By SPMI, we conclude...

Note: we can use PMI for the number of base case is integer

Claim: $\forall n \in \mathbb{N}(f_n < 2^n)$ Proof:

• Base Case: verify when n=0 and n=1

• Inductive Step: Let $n \in \mathbb{N}$ with $n\geq 1$ such that $f_i < 2^i$ for all $i\in \mathbb{Z}$ with $0 \leq i \leq n$ WTS $f_{n+1}<2^{n+1}$.

• $f_{n+1}=f_n + f_{n+1}$ (by definition, since n+1>=2)
• $< 2^n + 2^{n+1}$ (by Inductive Hypothesis)
• $< 2^n + 2^n$
• $=2^{n+1}$
• Conclusion: therefore... By SPMI, we conclude...

Note: n=1 will be our last base case, n+1 to be our next term after base cases.

Table of Content