Definition of +

• definition will be recursive
• define some terms
  • $1:=s(0)$
  • $2:=s(s(0))$
  • ...
• rules
  1. $\forall n,n+0=n$
     • addition of 0 (identity)
  2. $\forall n,m,n+s(m)=s(n+m)$
     • addition of a number, that is successor of something else

Prove 1+1=2

• $s(0)+s(0)=s(s(0)+0)=s(s(0))$
• $2:=s(s(0))$

Prove 2+2=4

• 2+2=4 means
  • $ss(0)+ss(0)=ssss(0)$
• furthermore: $$\begin{array}{rl}ss(0)+ss(0)& =s(ss(0)+s(0))\\ & =s(s(ss(0)+0))\\ & =s(s(ss(0)))\\ & =ssss(0)\end{array}$$

Definition of $\times$

• rules
  1. $\forall n,n\times 0=0$
  2. $\forall n,m,n\times s(m)=s\times m+n$
• 👨‍🏫 can you define division?

Mathematical induction

• to show $\phi (n)$ holds $\forall n\in \mathbb{N}$
• it is sufficient to show
  • $\phi (0)$ holds
    • aka: induction base
  • $\forall n\in \mathbb{N},\phi (n)\to \phi (n+1)$
    • inductive step

Definition of $\le$

• $\forall n,m,n\le m⟺\exists x\in \mathbb{N},n+x=m$
• 👨‍🏫 other, better definitions exist!

Problems

• prove that ${\sum }_{i=1}^{n}i=\frac{n(n+1)}{2}$
• inductive base
  • $\phi (0)={\sum }_{i=1}^{0}i=\frac{0(1)}{2}=0$
• inductive step $$\begin{array}{rl}\phi (n)& \\ \sum _{i=1}^{n}i& =\frac{n(n+1)}{2}\\ \sum _{i=1}^{n}i+n+1& =\frac{n(n+1)}{2}+n+1\\ \sum _{i=1}^{n+1}i& =\frac{n(n+1)}{2}+n+1\\ \sum _{i=1}^{n+1}i& =\frac{n(n+1)}{2}+2\times \frac{n+1}{2}\\ \sum _{i=1}^{n+1}i& =\frac{(n+2)(n+1)}{2}=\frac{(n+1)(n+2)}{2}\\ \phi (n+1)& \end{array}$$

Well-ordering principle

• every non-empty subset A $\subseteq \mathbb{N}$ contains a least / smallest element
• alternatively: set of $\mathbb{N}$ is "well-ordered" by its natural / magnitude order
  • i.e. $x<y$ in order $⟺\exists n\in \mathbb{N},y=x+n$
    • $n$: can be 0
• principle: mostly driven from mathematical induction
  • taken as granted from Peano axioms
• $\mathbb{N}$: the smallest inductive set
  • one can show: set of all natural no. $n$ s.t. "$\{0,\dots ,n\}$ is well-ordered" is inductive
  • and therefore must contain all natural no.
• 👨‍🎓 ~= reverse direction of mathematical induction?

Proof by infinite descent

• aka Fermat's method of descent
• you can't have an infinite sequence of $n\in \mathbb{N}$ s.t. $\forall i\in \mathbb{N},n_i>n_{i+1}$
• particular kind of "proof by contradiction"
  • showing: if the statement was to hold a number
  • => it can hold a smaller number
  • => $\mathbb{N}$ has lower bound (0); thus it's wrong
• relies on the well-ordering principle
  • often used to show that no solution exists

Problem 2: proof on irrationality of $\sqrt{2}$

• theorem: $\exists a,b\in \mathbb{N},b\ne 0,\sqrt{2}=\frac{a}{b}$
• $x,y\in \mathbb{N}\wedge c_a,c_b\in \mathbb{Z}$ let $a=2^x{c}_a$ and $b=2^y{c}_b$
• $(x=0\vee y=0)\wedge (c_a\perp c_b)\wedge (2\perp c_a)\wedge (2\perp c_b)$ $$\begin{array}{rl}\sqrt{2}& =\frac{a}{b}\\ 2& =\frac{a^2}{b^2}\\ 2b^2& =a^2\end{array}$$
• i.e. $a^2$ is even => $a$ is also even
  • $a\%2=1\to a^2\%2=1$
  • thus $a^2\%2=0\to a\%2=0$
• $\exists c,a=2c$
• $2b^2=4c^2$
• $b^2=2c$
• $\exists d,b=2d$
• $\sqrt{2}=\frac{a}{b}=\frac{2c}{2d}=\frac{c}{d}=\dots$
• a ever-smaller fraction can be found, infinitely!
  • which is impossible, or is a "contradiction"
• i.e. no such integer $a,b$ exists!
• 👨‍🏫 also, keep in mind:

Proof in well-ordering principle

• let $A=\{a|\exists b,\sqrt{2}=a/b\}$
• proof by contradiction
• step
  • if $A$ is not empty, there must be smallest set from $A$
    • let $a$ be smallest number from $A$
  • however: we can obtain a even smaller member based on $a$
  • thus: it must be the case that $A$ is empty

Proving: "There are infinitely many prime numbers"

• let $p_1,p_2,\dots ,p_n$ be all the primes
• $k=\_1\times p_2\times \cdots \times p_n+1={\prod }_{i=1}^n{p}_i+1$
• if $k$ is prime: premise is wrong as $k$ is not in the list
• if $k=a\cdot b$
  • it must be the case that $a$ is a prime that's not on the list
    • (contradiction)
  • or: $a$ can be factorized to $a_1\cdot b_1$
• and so on, $a_2=a_2\cdot b_2$
• and $a>a_1>a_2>\dots$
• however: the sequence must be finite (infinite descent)
  • and the last element of the sequence, must be a prime
  • that is not on the list!
  • q.e.d.

Problem 4 (Pigeonhole principle)

• if we have $n$ holes and $n+1$ pigeons are put in them
  • $\exists$ a hole w/ at least two pigeons
  • (for $n\ge 1$)
• proof by induction
  • base case: starts from $n=1$
    • there is only 1 hole, with 2 pigeons
    • so there is a hole w/ 2 pigeons
  • induction step:
    • for $n+1$-th hole, you can either put
      • 2 pigeons: problem solved
      • 1 pigeon: $\phi (n)$, which is premise
      • 0 pigeon: $\phi (n)$ is sufficient as there are more than $n+1$ pigeons already

False proof: all cars are the same color

• base case: when $n=0,1$, $n$ cars have the same color!
• induction step $$\underset{n\text{ cars}}{\underbrace{\square \square \square }}\square$$ $$\square \stackrel{n\text{ cars}}{\overbrace{\square \square \square }}$$
  • first $n$ cars have the same color
  • and last $n$ cars have the same color too
  • 👨‍🏫 yeah, all crs have the same color!
• it's wrong when $n=1$!
  • and you can't prove that

False proof: all man are bald

• base case: if you have 0 threads of hair, you are bald!
• inductive case: if a man with $n$ threads of hair is bald, having $n+1$ threads doesn't really make you not bald either!
• thus, all man (unless they have negative or fractional hair) are bald!

Problem 5: $n$ players take part in a tournament

• every player plays against every other player
  • and every game has one winner
• prove that there exists a permutation $p_1,p_2,\dots ,p_n$
• s.t. $\forall i\le i\le n-1$ $p_i$ has lost to $p_{i+1}$
• proof
  • when $n=1$, the case holds
  • if we can sort for $n$ players
    • for $p_{n+1}$, has it won over $p_i{|}_{i=n}^1$?
      • if so, we can place it after $p_i$
    • if there was no place to put $p_{n+1}$, it shows that it has won against no one
      • in which, we can put $p_{n+1}$ in the first place

Proving induction is valid

• roadmap
  • show infinite descent $\to$ well ordering (1)
  • show induction holds from well-ordering

1. let $A\subseteq \mathbb{N}$ be a non-empty set w/o a smallest element
   • pick any $a_0\in A$
   • $A_0=\{a^{\prime }\in A|a^{\prime }<a_0\}$
   • and $A_0$: must be another non-empty set w/o a smallest element
     • and smaller than $A$
     • else: proof ends
   • pick any $a_1\in A_0$
   • $A_1=\{a^{\prime }\in A_0|a^{\prime }<a_1\}$
   • and $A_1$: must be another non-empty set w/o a smallest element
     • and smaller than $A_0$ // infinite descent!
   • thus: based on infinite descent principle, well ordering holds
2. let $k$ be a set of all of natural number s.t. $0\in K$
   • i.e. trying to prove the fifth Peano axiom
   • $0\in K$
   • $\forall n,n\in K\to n+1\in K$
   • proving: $K=N$; suppose $K\ne N$ (contradiction)
   • $A=\mathbb{N}-K$ // A: non-empty (by definition)
   • let $m$ be the smallest element of $A$
   • $m$ is not in $K$
     • thus, $m-1$ is also not in $K$
     • however, $m-1$ is smaller than $m$, and it also shouldn't be in $K$ either!