Lecture 010

Induction

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:

• $P(0)$ holds

• $(\forall k \in \mathbb{N}(P(k) \implies P(k+1)))$ holds

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 $S = \mathbb{N}$

• by definition $S \subseteq \mathbb{N}$

• $0 \in S$ and S is closed under successor function, thus $\mathbb{N} \subseteq S$.

• Therefore, $S = \mathbb{N}$

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.

• prove that the minimum point in the set hold

• prove that if every point in the set hold, every point + 1 also hold.

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 $(\forall n \in \mathbb{N})P(n).$

• be sure to state the variable on which you are inducting.

• be sure to state when you are invoking the inductive hypothesis (always)

Practice

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

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

• (Base Case): For n = 1, $\sum^1_{i=1}i = 1 = \frac{1(1+1)}{2}$

• (Inductive Step): Let $k \in \mathbb{Z}^{+}$ such that $\sum^k_{i=1}i = \frac{k(k+1)}{2}$ (inductive hypothesis). WTS $\sum^{k+1}_{i=1} = \frac{(k+1)(k+2)}{2}$

  • $\sum^{k+1}_{i = 1}i = (\sum^k_{i = 1}i)+k+1$
  • $= \sum^k_{i=1}i = \frac{k(k+1)}{2}$ (by Inductive Hypothesis)
  • $= \frac{(k+1)(k+2)}{2}$

• Thus we have $\sum^{k+1}_{i = 1} = \frac{(k+1)(k+2)}{2}$

• Therefore, by PMI, we conclude... Q.D.E.