COMP 2711H: Lecture 4

Date: 2024-09-09 02:14:53

Reviewed:

Topic / Chapter: Infinite Descent and more PProofs

summary

❓Questions

Notes

Infinite Descent (cont.)

Relationship between three
- infinite descent: implies well-ordering
- well-ordering: implies induction
- 👨‍🏫 can we assume the reverse?
Well-ordering: implies infinite descent
- let $A$ be an infinite seq. of numbers
- then $B = {x \in N ∣\exists i a_{i} = x}$
- $B$ has a minimum $b$
  - as $B$ is non-empty
- and $\exists i, a_{i} = b$
- then $a_{i + 1}$ cannot be smaller than $a_{i} = b$
  - as $b$ is the minimum
- thus, there can't be an infinite sequence of $N$ that is always decreasing
Induction: implies well-ordering
- let $A \subseteq N \land A \neq = \emptyset$
- and does not have minimum (proof by contradiction)
- $B := A^{c}$
- $p (n) := \forall i \leq n, i \in B$
- to be proven by induction
  - $p (0)$ : for $i = 0$ , $i \in B$
    - unless: $0$ is the minimum element in $N$
    - and it cannot be in $A$
      - unless: 0 becomes the min. element of $A$
  - assuming $p (n)$ ,
    - suppose $\forall i \leq n + 1, i \neq \in B$
    - which means: $i \in A$
    - however, all elements $\leq n$ is not in $A$
    - $n + 1$ becomes the minimum element in $A$ !
      - thus: the inductive assumption is wrong: and $n + 1 \in B$
  - finally, $\forall n \in N, \forall i \leq n, i \in B$
    - $B = N$
  - i.e. $A = \emptyset$
    - contradicts our initial assumption: $A$ being non-empty
    - thus, every non-empty subset of $N$ has a minimum element

Fibonacci Numbers

Fibonacci numbers
$F_{n} = ⎩ ⎨ ⎧ n = 0 : 0 n = 1 : 1 n \geq 2 : F_{n - 1} + F_{n - 2}$
- e.g. $0, 1, 1, 2, 3, 5, 8, 13 \dots$

Problems

Problem 1
- $\forall F_{1} + F_{2} + \dots + F_{n} = F_{n + 2} - 1$
- base cases $⎩ ⎨ ⎧ n = 0 : 0 = 1 - 1 = 0 n = 1 : 1 = 2 - 1 = 1 n = 2 : 2 = 3 - 1 = 2$
- prove $\sum_{i = 1}^{n + 1} F_{i} = F_{n + 3} - 1$ $i = 1 \sum n F_{i} i = 1 \sum n + 1 F_{i} + F_{n + 1} i = 1 \sum n + 1 F_{i} i = 1 \sum n + 1 F_{i} = F_{n + 2} - 1 = F_{n + 2} - 1 + F_{n + 1} = F_{n + 2} - 1 + F_{n + 1} = F_{n + 3} - 1$
Problem 2
- $\forall n F_{1} + F_{3} + F_{5} + \dots F_{2 n - 1} = F_{2 n}$
- base case: $F_{1} = F_{2}$
- then: $\sum_{i = 1}^{n} F_{2 i - 1} = F_{2 n}$
  - $\sum_{i = 1}^{n} F_{2 i - 1} = F_{2 n}$
  - $\sum_{i = 1}^{n} F_{2 i - 1} + F_{2_{n} + 1} = F_{2 n} + F_{2_{n} + 1}$
  - $\sum_{i = 1}^{n + 1} F_{2 i - 1} = F_{2 n + 2}$
Problem 3
- $\forall n F_{2} + F_{4} + F_{6} + \dots F_{2 n} = F_{2 n + 1} - 1$
- 👨‍🏫 simply subtract the result of (2) from result of (1)
Problem 4
- $\forall n \in N, F_{n} ⊥ F_{n + 1}$
  - i.e. $g cd (F_{n}, F_{n + 1}) = 1$
  - or: they are relatively prime
- with well-ordering & infinite descent
- $A = {n \in N ∣ g cd (F_{n}, F_{n + 1}) \neq = 1}$
  - to be proven: $A = \emptyset$
- proof by contradiction: assume $A \neq = \emptyset$
  - let $m$ be the smallest element of $A$
  - $d := g cd (F_{m}, F_{m + 1}) \neq = 1$
  - $d ∣ F_{m} \land d ∣ F_{m + 1}$
    - i.e. $d$ divides both
    - and $d > 1$
  - if $F_{m + 1} = F_{m} + F_{m - 1}$
    - then also $d ∣ F_{m - 1}$
    - $⟹ m - 1 \in A$
    - however: $m$ is supposed to be the "smallest element" in $A$
    - thus: contradiction, and $A$ is empty
- or: using infinite descent
  - (continued from above) $m - 1 \in A ⟹ m - 2 \in A ⟹ m - 3 \in A$ ...
  - infinite sequence & decreasing -> impossible!
- also: using induction
  - base case holds: $g cd (F_{0}, F_{1}) = g cd (F_{1}, F_{2}) = 1$
  - induction: $g cd (F_{n}, F_{n + 1}) = 1 ⟹ g cd (F_{n + 1}, F_{n + 2} = 1)$
    - proof by contradiction
    - $d := g cd (F_{n + 1}, F_{n + 2})$ (hopefully $\neq = 1$ )
    - $d ∣ F_{n + 1} \land d ∣ F_{n + 2}$
    - i.e. $d$ divides both
    - then for $F_{n + 2} - F_{n + 1} = F_{n}$ , also $d ∣ F_{n}$
    - but, as $g cd (F_{n}, F_{n} + 1) = 1, d = 1$
- 👨‍🏫 any of three will work, but just that one will be more human-intuitive!
Problem 5:
- $\forall n \geq 1, \sum_{i = 1}^{n} i^{3} = (\sum_{i = 1}^{n})^{2}$
- by induction, so base case: $1^{3} = 1^{2}$
- assume $\sum_{i = 1}^{n} i^{3} = (\sum_{i = 1}^{n})^{2}$ , then: $i = 1 \sum n + 1 i^{3} (i = 1 \sum n + 1 i)^{2} = i = 1 \sum n i^{3} + (n + 1)^{3} = (i = 1 \sum n i)^{2} + (n + 1)^{3} = (\frac{n ( n + 1 )}{2})^{2} + (n + 1)^{3} = \frac{n ^{2} ( n + 1 ) ^{2}}{4} + (n + 1)^{3} = \frac{( n ^{2} + 4 n + 4 ) ( n + 1 ) ^{2}}{4} = (\frac{( n + 1 ) ( n + 2 )}{2})^{2} = \frac{( n + 1 ) ^{2} ( n + 2 ) ^{2}}{4} = \frac{( n ^{2} + 4 n + 4 ) ( n + 1 ) ^{2}}{4}$
Problem 6
- let $∣ A ∣ = n$ , then $A$ has $2^{n}$ subsets
- base case: $n = 0$ , then has only 1 subset: $\emptyset$
  - $1 = 2^{0}$ , so it holds
- proof by induction
  - let $A$ be a set of size $n + 1$
  - $A = {a_{1}, a_{2}, \dots, a_{n + 1}}$
  - any $B \subseteq A$ s.t. $B \neq ∋ a_{n + 1}$ is of size $n$ too
    - then, $A$ has two groups of subsets, one including $a_{n + 1}$ and those that are not
    - each of size $2^{n}$
    - together: of size $2^{n + 1}$
Problem 7
- let $A$ be a non-empty finite set
- let $E = {B \subseteq A ∣ ∣ B ∣ is even}$
- let $O = {B \subseteq A ∣ ∣ B ∣ is odd}$
- prove that $∣ E ∣ = ∣ O ∣$
- base case: $n = 1$
  - $A = {a_{1}}$
  - $\emptyset \in E$
  - $A \in O$
- let $A_{n + 1}$ be a set of size $n + 1$
  - let $A_{n} \subseteq A_{n + 1}$ with $∣ A_{n} ∣ = n$
  - $∣ E_{n} ∣ = ∣ O_{n} ∣$
  - let $x \in A_{n + 1} \land x \neq \in A_{n}$
  - subsets without $x$ :
    - odd: $∣ O_{n} ∣$
    - even: $∣ E_{n} ∣$
  - subsets with $x$ :
    - odd: $∣ E_{n} ∣$
    - even: $∣ O_{n} ∣$
  - and $∣ O_{n} ∣ + ∣ E_{n} ∣ + ∣ E_{n} ∣ + ∣ O_{n} ∣$
Problem 8
- suppose: we have a class $C$ of $n$ students
- let $A^{+} \subseteq A \subseteq C$
- in how many ways can we choose $A^{+}$ and $A$ ?
- 👨‍🏫
- if $n = 0$
  - $A^{+} = A = C = \emptyset$
- if $n = 1$
  - $C = {1}$
  - 3 choices
    - $A^{+} = A = \emptyset$
    - $A^{+} = A = {1}$
    - $A^{+} = \emptyset \land A = {1}$
- if $n = 2$
  - $C = {1, 2}$
  - 9 choices (as grade of a student is independent of another)
- assume: for $n$ , there are $3^{n}$ ways
- induction
  - $C = {1, 2, 3, \dots, n, n + 1}$
  - there are $3^{n}$ ways to allocate first $n$ students
  - and 3 ways to allocate the last student
  - as they are independent (in this course), there are $3^{n} \times 3 = 3^{n + 1}$ ways
Problem 9
- reducing game / game of Nim with two heaps
- given $a, b \in N$
- players, in turn, choose either $a, b$ and decreases it
  - if one can't make move: then the player loses
- assuming ideal play
  - if $a = b$ then player 2 wins
  - if $a \neq = b$ then player 1 wins
- and assume $a < b$ as the order doesn't matter
- base case: $a = b = 0$
  - player 2 wins
- using induction on $n$
  - with $a = b = n + 1$
  - if player 1 reduces $a$ to $n$
    - player 2 can also make $b$ to $n$
- strong induction
  - proving $\forall n \in N p (n)$
  - consists of:
    - $p (0)$ holds
    - $\forall n, (\forall i \leq n, p (i)) ⟹ p (n + 1)$
      - i.e. more stuff & requirements
- prove that player 2 wins $n + 1, n + 1$
  - $\forall i \leq n + 1$ , player 2 wins $i, i$
  - player 2: can copy player 1's move
    - i.e. player 2 makes $a, b$ equal
- can also be proven with infinite descent
  - game: ends at some point (i.e. when $0, 0$ )
  - and player 2 wins it

COMP 2711H: Honors Discrete Mathematical Tools for Computer Science