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.
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 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.
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)
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}
Thus we have \sum^{k+1}_{i = 1} = \frac{(k+1)(k+2)}{2}
Therefore, by PMI, we conclude... Q.D.E.
Table of Content