COMP 2711H: Lecture 30

Date: 2024-11-11 02:56:43

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Rational Numbers

Rational numbers
- for $a, r \in Q$
- a rational number: all numbers that can be written as a fraction
  - can be expressed as ("expansion") $i = 1 \sum n a_{i} r^{i}$
- decimal expansion of rationals $0. a_{1} a_{2} a_{2} \dots a_{n} := i = 1 \sum n \frac{a _{i}}{1 0 ^{i}}$
- similarly: an infinitely repeating one: e.g. $a = 0.248 \overline{97}$ $0. a_{1} a_{2} a_{2} \dots a_{n} := i = 1 \sum \infty \frac{a _{i}}{1 0 ^{i}}$
  - aka repeating decimal expansion
- 👨‍🏫 terminating / repeating decimal expansion: can be converted to a rational number
  - and vice versa
- theorem: suppose $x$ : a terminating / repeating decimal expansion
  - then $x \in Q$
  - case 1: $x = 0. a_{1} a_{2} \dots a_{n}$ $\sum_{i = 1}^{n} \frac{a _{i}}{1 0 ^{i}} \times \frac{1 0 ^{n}}{1 0 ^{n}} = \frac{\sum _{i = 1}^{n} a _{i} 1 0 ^{n - i}}{1 0 ^{n}}$
  - case 2: $x = 0. \overline{a_{1} a_{2} \dots a_{n}}$ $10 ? x (1 0^{n} - 1) x x = \underline{a_{1} a_{2} \dots a_{n}} \cdot \overline{a_{1} a_{2} \dots a_{n}} = a_{2} a_{2} \dots a_{n} = a_{1} a_{2} \dots a_{n} / (1 0^{n} - 1)$
  - 👨‍🏫 combination of two: also trivial
- theorem: let $0 < p < q$ and $p ⊥ q$
  - then $p / q$ has a decimal representation: either terminating or repeating
  - case 1: $g cd (q, 10) = 1$
    - 👨‍🏫 10: our base
    - w/ Euler's theorem: $1 0^{φ (q)} = 1 mod q$
      - let $q := φ (q)$
      - $1 0^{t} = 1 mod q$
      - $q ∣1 0^{t} - 1$
      - $\exists a 1 0^{t} - 1 = q \cdot a$
      - $\frac{p}{q} = \frac{p a}{q a} = \frac{p \cdot a}{1 0 ^{t} - 1} = \frac{p a}{1 0 ^{t}} + \frac{p a}{1 0 ^{2 t}} + \frac{p a}{1 0 ^{3 t}} + \dots$
    - 👨‍🏫 also: easy to show that only $t$ digits are needed
      - $= 0. \overline{p a}$
      - $p a$ : of $t$ digits
  - case 2: $g cd (q, 10) \neq = 1$
    - $q = 2^{α} \cdot 5^{β} \dots$
    - $γ = lcm (α, β)$
    - $x = \frac{p}{q} = \frac{p}{2 ^{α} 5 ^{β}} \times \frac{1 0 ^{γ}}{1 0 ^{γ}} = \frac{p \cdot a}{1 0 ^{γ}}$
Order
- yet: cannot express all number
  - e.g. $x = 0.1234567891011 \dots$
  - 👨‍🏫 real numbers will be discussed later
- order: property of set: relation $< \subseteq A \times A$ , s.t.
  1. $\forall a, b \in A$ , we have exactly one of
    - $a < b$ , $a = b$ , or $a > b$
  2. $\forall a, b, c \in A, a < b \land b < c ⟹ a < c$
  3. $a \leq b ⟺ a < b \lor a = b$
  4. d
- precise definition on minimum element
  - let $U$ : an ordered set, $A \subseteq U$
  - element: $b \in U$ : an upper bound of $A$ if:
    - $\forall a \in A, a \leq b$
  - let $B$ : set of all upper-bounds of $A$
    - and let minimum element of $B$ : an supremum of $A$ ( $sup A$ )
  - if there exists $s$ s.t.
    - $s \in B, \forall a \in A a \leq s$
    - $\forall s^{'} s^{'} \in B s^{'} \geq s$
    - then: $s$ an supremum of $A$ : $s = sup (A)$
  - let $U = N$
    - $A = {1}$ , then $sup A = 1$
      - i..e maximum
    - any infinite set: $sup A \neq \in A$
  - let $U = Q$
    - $A = {x \in Q ∣1 < x}$ , then $sup A = 1$
- ordered set $U$ : has supremum / least-upper-bound property if:
  - every non-empty subset of $U$ : w/ an upper bound: also has a supremum within it
- does $N$ : have the least-upper-bound property?
  - 👨‍🏫 yes
  - assume: $A$ be bounded above
  - $B :=$ upper bounds of $A$
    - by assumption (of bound): $B \neq = \emptyset$
  - and every non-empty subset: has minimum element
    - which is reachable
    - thus: $N$ must also have the least upper bound property
- does $Q$ have one?
  - $A = {x \in Q ∣ x \geq 0 \land x^{2} \leq 2}$
  - let $ε$ : an upper-bound for $A$ ( $= 2 \neq \in A$ )
    - 👨‍🏫 assume $ε \in A$ , and it will lead to infinite descent, etc.
  - $Q$ : doesn't have least upper-bound for $A$
- theorem: $A \subseteq U$ , then $l$ : a lower bound for $A$ if
  - $\forall a \in A$ , $l \leq a$
  - and $\exists l \in L$ s.t. $\forall l^{'} \in L l \geq l^{'}$
  - then: $l$ : infimum of $A$
  - 👨‍🏫 i.e. largest lower bound property
- theorem: if $U$ an ordered set w/ least upper bound property
  - then $U$ : also satisfies largest lower bound property
- proof: let $A \subseteq A, A \neq = \emptyset$ and $A$ : bounded below
  - let $L$ : set of all lower bounds of $A$
    - $L = {x \in U ∣\forall a \in A x \leq a}$
    - 👨‍🏫 must not be empty: as it is "bounded below" and lower bound exists
    - $\forall l \in L l \leq a_{0}$ (supremum)
    - let $α = sup L$
  - every $a \in A$ : an upper bound of $L$
    - $α$ : $sup L$
    - $⟹ \forall a \in A, α \leq a$
    - 👨‍🏫 by definition: $α \in L$
      - $sup L = in f A = α \in A$
- ⭐👨‍🏫 thus: having infimum property implies supremum, and vice versa
Real number
- let set $2 = {x \in Q ∣ x \leq 2}$

COMP 2711H: Honors Discrete Mathematical Tools for Computer Science