Sequence: defined inductively (recursively)
sequence of real numbers
The Factorial Sequence
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}
By PMI, we conclude...
The Fibonacci Sequence
Assumption
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:
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}.
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}.
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.
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