Lecture 010


For n\in \mathbb{Z}, WTS (\forall n \in \mathbb{N}P(n)), we show generality by showing n is arbitrary.

However, sometimes the truth value of P(n) depends on the truth value of P(k) for some k<n. This should be dealt with induction.

The Unsolvable Problem

The Principle of Mathematical Induction (PMI)

Let P(n) be variable proposition (n\in \mathbb{N}) if:

then we can conclude (\forall n \in \mathbb{N})P(n) holds.

Claim: (\forall n \in \mathbb{N})P(n) Proof: let S = \{n \in \mathbb{N} | P(n)\}, we want to show S = \mathbb{N}

Corollary to PMI

Corollary to PM: Let M\in \mathbb{Z} \land S = \{z \in \mathbb{Z} | z \geq M\}. P(n) is a variable proposition defined on S such that 1. P(M) holds 2. (\forall k \in S)(P(k) \implies P(k+1)) holds. Then (\forall n \in S)P(n) holds.

Template for Induction Proofs using PMI

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

- (Base Case): P(0) holds because ____.

- (Inductive Step): Let $n \in \mathbb{N}$ and assume P(n) holds. (This is the Inductive Hypothesis) Use this assumption to show P(n+1) holds also

- (Conclusion) By PMI, we have $(\forall n \in \mathbb{N})P(n).$


Conjecture: (\forall n \in \mathbb{Z}^{+})(\sum^n_{i=1}i = \frac{n(n+1)}{2}).

\sum^{n+1}_{i=1}f(i) = (\sum^{n}_{i=1}f(i))+f(i+1)

Proof: We process by induction on n\in \mathbb{Z}^{+}. WTS P(1) \land P(k) \implies P(k+1) (inductive step)

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