# Lecture 012

Prime Number: A natural number n is called prime iff n > 1 and its only positive divisors are 1 and n.

## The Fundamental Theorem of Arithmetic

Claim: $(\forall n \in \mathbb{N})$ with n > 1 n is either prime or a product of primes.

Proof: We proceed by strong induction on $n\in \mathbb{Z} \land n \geq 2$. Define P(n) to be the variable proposition "n either prime or a product of primes"

• Base Case: when n=2, the only possible divisor is 1 and 2. n is prime

• Inductive Step: Let $k \in \mathbb{N}$ with $k \geq 2$, assume P(i) holds for all $i\in\mathbb{N}$ with $2 \leq i \leq k$. let $n=k+1$

• Case if $k+1$ is prime: $P(k+1)$ holds
• Case if $k+1$ is not prime: then let $p, q \in \mathbb{N}$, with $2\leq p \leq q \leq k$, and $k+1=pq$. Fix p and q. By IH, p and q are either prime or a product of primes. Then $(\exists a, b \in \mathbb{N})(p=ab)$ So we are good.

• (break k+1 to k+1=a*b, then assume a=product_of_prime_or_prime and b=product_of_prime_or_prime, k+1 must equal to product of prime) The Game of Chomp Claim: If Player I begins the game by choosing square (1, n) the top right corner, then Player 1 can always win the 2 x n game of Chomp. Proof: by strong induction. Let P(n) be the proposition.

• Base Case: for 2 x 1 game of Chomp. Player 1 will win because...

• Inductive Step

• Assume: for all 2 x i game, $i\in [k]$, $k \in \mathbb{Z}^+$, Player 1 will win.
• Consider: 2 x k+1 game.
• Case Player 2 take bottom: Then Player 1 will win by IH. (since it will be rectangle again)
• Case Player 2 take top: Then Player 1 will win by taking steps that again make the rectangle look like a winning strategy. (taking bottom right leave top right open one)

This is a strong induction because it require 2 x 1 is true to 2 x k.

## Well Ordering Property

Definition: A set X is called well-ordered iff every nonempty subset has a least element.

• Example: $\mathbb{N}$, $2^n|n\in \mathbb{N}$

• Non-Example: $\mathbb{Z}$, $[0, 1)$, $\mathbb{R}$

Claim: $\mathbb{N}$ is well-ordered Proof: by SPMI for all $n\in\mathbb{N}$. Let P(n) be variable proposition "every subsets of N containing n has a least element" (read: assume the subset containing n has the property, prove subset containing all number has property)

• Base: P(0) holds, cuz 0 is the least in N by definition (base case is crucial here)

• IS:

• Assume: $(\forall i \in [k])P(i)$ hold for $k\in\mathbb{N}$.
• Consider n = k+1. let $S\subseteq \mathbb{N}$ such that $k+1 \in S$
• Case: k+1 is the least element in S, done
• Case: k+1 is not the least element, then there is an element smaller than k+1, we call that i. apply IH, we know P(i) holds. (Then since the previous S containing i has a least element, this S containing the previous S has a least element)

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