eggy/eifueo
1
1
Fork 1
Code Issues 2 Pull Requests Releases Wiki Activity
master
eifueo/docs/1b/ece108.md
eggy 57e8d08e96 Revert "chore: add discontinuation notices" 
This reverts commit 8c9d83851d.
2023-10-30 15:47:39 -04:00

959 lines
31 KiB
Markdown
Raw Permalink Blame History

# ECE 108: Discrete Math 1
An **axiom** is a defined core assumption of the mathematical system held to be true without proof.
!!! example
True is not false.
A **theorem** is a true statement derived from axioms via logic or other theorems.
!!! example
True or false is true.
A **proposition/statement** must be able to have the property that it is exclusively true or false.
!!! example
The square root of 2 is a rational number.
An **open sentence** becomes a proposition if a value is assigned to the variable.
!!! example
$x^2-x\geq 0$
## Truth tables
A truth table lists all possible **truth values** of a proposition, containing independent **statement variables**.
!!! example
| p | q | p and q |
| --- | --- | --- |
| T | T | T |
| T | F | F |
| F | T | F |
| F | F | F |
## Logical operators
!!! definition
- A **compound statement** is composed of **component statements** joined by logical operators AND and OR.
The **negation** operator is equivalent to logical **NOT**.
$$\neg p$$
The **conjunction** operaetor is equivalent to logical **AND**.
$$p\wedge q$$
The **disjunction** operator is equivalent to logical **OR**.
$$p\vee q$$
### Proposation relations
!!! definition
A **tautology** is a statement that is always true, regardless of its statement variables.
The **implication** sign requires that if $p$ is true, $q$ is true, such that *$p$ implies $q$*. The first symbol is the **hypothesis** and the second symbol is the **conclusion**.
$$p\implies q$$
| $p$ | $q$ | $p\implies q$ |
| --- | --- | --- |
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | F |
The **inference** sign represents the inverse of the implication sign, such that $p$ **is implied by** $q$. It is equivalent to $q\implies p$.
$$p\impliedby q$$
The **if and only if** sign requires that the two propositions imply each other — i.e., that the state of $p$ is the same as the state of $q$. It is equivalent to $(p\implies q)\wedge (p\impliedby q)$.
$$p\iff q$$
The **logical equivalence** sign represents if the truth values for both statements are **the same for all possible variables**, such that the two are **equivalent statements**.
$$p\equiv q$$
$p\equiv q$ can also be defined as true when $p\iff q$ is a tautology.
!!! warning
$p\equiv q$ is *not a proposition* itself but instead *describes* propositions. $p\iff q$ is the propositional equivalent.
## Common theorems
The **double negation rule** states that if $p$ is a proposition:
$$\neg(\neg p)\equiv p$$
!!! tip "Proof"
Note that:
| $p$ | $\neg p$ | $\neg(\neg p)$ |
| --- | --- | --- |
| T | F | T |
| F | T | F |
Because the truth values of $p$ and $\neg(\neg p)$ for all possible truth values are equal, by definition, it follows that $p\equiv\neg(\neg p)$.
!!! warning
Proofs must include the definition of what is being proven, and any relevant evidence must be used to describe why.
The two **De Morgan's Laws** allow distributing the negation operator in a dis/conjunction if the junction is inverted.
$$
\neg(p\vee q)\equiv(\neg p)\wedge(\neg q) \\
\neg(p\wedge q)\equiv(\neg p)\vee(\neg q)
$$
An implication can be expressed as a disjunction. As long as it is stated, it can used as its definition.
$$p\implies \equiv (\neg p)\vee q$$
Two **converse** propositions imply each other:
$$p\implies q\text{ is the converse of }q\implies p$$
A **contrapositive** is the negatated converse, and is **logically equivalent to the original implication**. This allows proof by contrapositive.
$$\neg p\implies\neg q\text{ is the contrapositive of }q\implies p$$
### Operator laws
Both **AND** and **OR** are commutative.
$$
p\wedge q\equiv q\wedge p \\
p\vee q\equiv q\vee p
$$
Both **AND** and **OR** are associative.
$$
(p\wedge q)\wedge r\equiv p\wedge(q\wedge r) \\
(p\vee q)\vee r\equiv p\vee(q\vee r)
$$
Both **AND** and **OR** are distributive with one another.
$$
p\wedge(q\vee r)\equiv(p\wedge q)\vee(p\wedge r) \\
p\vee(q\wedge r)\equiv(p\vee q)\wedge(p\vee r)
$$
!!! tip "Proof"
$$
\begin{align*}
(\neg p\vee\neg r)\wedge s\wedge\neg t&\equiv\neg(p\wedge r\vee s\implies t) \\
\tag*{definition of implication} &\equiv \neg (p\wedge r\vee[\neg s\vee t]) \\
\tag*{DML} &\equiv\neg(p\wedge r)\wedge\neg[(\neg s)\vee t)] \\
\tag*{DML} &\equiv(\neg p\vee\neg r)\wedge\neg[(\neg t)\vee t] \\
\tag*{DML} &\equiv(\neg p\vee\neg r)\wedge\neg(\neg s)\wedge\neg t \\
\tag*{double negation} &\equiv(\neg p\vee\neg r)\wedge s\wedge\neg t
\end{align*}
$$
### Quantifiers
A **quantified statement** includes a **quantifier**, **variable**, **domain**, and **open sentence**.
$$
\underbrace{\text{for all}}_\text{quantifier}\ \underbrace{\text{real numbers}\overbrace{x}^\text{variable}\geq 5}_\text{domain}, \underbrace{x^2-x\geq 0}_\text{open sentence}
$$
The **universal quantifier** $\forall$ indicates "for all".
$$\forall x\in S,P(x)$$
!!! example
All real numbers greater than or equal to 5, defined as $x$, satisfy the condition $x^2-x\geq 0$.
$$\forall x\in\mathbb R\geq 5,x^2-x\geq 0$$
The **existential quantifier** $\exists$ indicates "there exists at least one".
$$\exists x\in S, P(x)$$
!!! example
There exists at least one real number greater than or equal to 5, defined as $x$, satisfies the condition $x^2-x\geq 0$.
$$\exists x\in\mathbb R\geq 5,x^2-x\geq 0$$
Quantifiers can also be negated and nested. The opposite of "for each ... that satisfies $P(x)$" is "there exists ... that does **not** satisfy $P(x)$".
$$\neg(\forall x\in S,P(x))\equiv\exists x\in S,\neg P(x)$$
Nested quantifiers are **evaluated in sequence**. If the quantifiers are the same, they can be grouped together per the commutative and/or associative laws.
$$\forall x\in\mathbb R,\forall y\in\mathbb R\equiv \forall x,y\in\mathbb R$$
!!! warning
This means that the order of the quantifiers is relevant if the quantifiers are different:
$\forall x\in\mathbb R,\exists y\in\mathbb R,x-y=1$ is **true** as setting $y$ to $x-1$ always fulfills the condition.
$\exists y\in\mathbb R,\forall x\in\mathbb R, x-y=1$ is **false** as when $x$ is selected first, it is impossible for every value of $y$ to satisfy the open sentence.
## Proof techniques
There are a variety of methods to prove or disprove statements.
- **Deduction**: a chain of logical inferences from a starting assumption to a conclusion
- **Case analysis**: exhausting all possible cases (e.g., truth table)
- **Contradiction**: assuming the conclusion is false, which follows that a core assumption is false, therefore the conclusion must be true
- **Contrapositive**: is equivalent to the original statement
- **Counterexample**: disproves things
- **Induction**: Prove for a small case, then prove that that applies for all cases
Implications can be proven in two simple steps:
1. It is assumed that the hypothesis is true (the implication is always true when it is false)
2. Proving that it follows that the conclusion is true
!!! example "Proving implications"
Prove that if $n+7$ is even, $n+2$ is odd.
$\text{Proof:}$
$\text{Assume }n+7\text{ is an even number. It follows that for some }k\in\mathbb Z$
$$
\begin{align*}
n+7&=2k \\
\text{s.t.} n+2&=2k-5 \\
&=2(k-3)+1
\end{align*}
$$
$\text{which is of the form }2z+1,z\in\mathbb Z,\text{ thus } n+2\text{ is odd.}$
!!! example "Proof by contradiction"
Prove that there is no greatest integer.
$\text{Proof:}$
$\text{ Let }n\in\mathbb Z\text{ be given and assume }\overbrace{\text{for the sake of contradiction}^\text{FTSOC}}\text{ that }n\text{ is the largest integer. Note that }n+1\in\mathbb Z\text{ and }n+1>n.\text{ This contradicts the initial assumption that }n\text{ is the largest integer, therefore there is no largest integer.}$
### Formal theorems
An **even number** is a multiple of two.
$$\boxed{n\ \text{is even}\iff\exists k\in\mathbb Z,n=2k}$$
An **odd number** is a multiple of two plus one.
$$\boxed{n\text{ is odd}\iff\exists k\in\mathbb Z,n=2k+1}$$
A number is **divisible** by another $m|n$ if it can be part of its product.
$$\boxed{n\text{ is divisible by } m\iff\exists k\in\mathbb Z,n=mk}$$
A number is a **perfect square** if it is the square of an integer.
$$n\text{ is a perfect square}\iff \exists k\in\mathbb Z,n=k^2$$
### Induction
!!! definition
- A proof **without loss of generality** (WLOG) indicates that the roles of variables do not matter — so long as the symbols CTRL-H'd, the proof remains exactly the same. For example, "WLOG, let $x,y\in\mathbb Z$ st. $x<y$."
Induction is a proof technique that can be used if the open sentence $P(n)$ depends on the parameter $n\in\mathbb N$. Because induction works in discrete steps, it generally cannot be applied domains of all real numbers.
To do so, the following must be proven:
- $P(1)$ must be true (the base case)
- $P(k+1)$ must be true for all $P(k)$, assuming $P(k)$ is true (the inductive case)
!!! warning
The statement **cannot** be assumed to be true, so one side must be derived into the other side.
!!! tip "Proof"
This should more or less be exactly followed. For the statement $\forall n\in\mathbb Z,n!>2^n$:
> We use mathematical induction on $n$, where $P(n)$ is the statement $n!>2^n$.
>
> **Base case**: Our base case is $P(4)$. Note that $4!=24>16=2^4$, so the base case holds.
>
> **Inductive step**: Let $k\geq 4$ for an arbitrary natural number and assume that $k!>2^k$. Multiplying by $k+1$ gives
>
> $$(k+1)k^2>(k+1)2^k$$
>
> By definition $(K=1)k!=(k+1)!$. Since $k\geq 4$, $k+1>2$ and thus $(k+1)2^k>2\cdot 2^k=2^{k+1}$. Putting this together gives
>
> $$(k+1)!>2^{k+1}$$
>
> Thus $P(k+1)$ is true and by the Principle of Mathematical Induction (POMI), $P(n)$ is true for all $n\geq 4$.
Induction can be applied to the whole set of integers by proving the following:
- $P(0)$
- if $i\geq 0, P(i)\implies P(i+1)$
- if $i\leq 0, P(i)\implies P(i-1)$
Alternatively, some steps can be skipped in **strong induction** by proving that if for $k\in\mathbb N$, $P(i)$ holds for all $i\leq k$, so $P(k+1)$ holds. In other words, by assuming that the statement is true for all values before $k$. If strong induction is true, regular induction must also be true, but not vice versa.
## Sets
!!! definition
- A **set** is an unordered collection of distinct objects.
- An **element/member** of a set is an object in that set.
- A **multiset** is an unordered collection of objects.
Sets are expressed with curly brackets:
$$\{s_1, s_2,\dots\}$$
Numbers are defined as sets of recursively empty sets:
$$
\begin{align*}
0&:=\empty \\
1&:=\{\empty\} \\
2&:=\{\empty,\{\empty\}\}
\end{align*}
$$
### Special sets
- $\mathbb N$ is the set of **natural numbers** $\{1, 2, 3,\dots\}$
- $\mathbb W$ is the set of **whole numbers** $\{0, 1, 2,\dots\}$
- $\mathbb Z$ is the set of **integers** $\{\dots, -1, 0, 1, \dots\}$
- $\mathbb Z^+_0$ is the set of **positive integers, including zero** — these modifiers can be applied to the set of negative integers and real numbers as well
- $2\mathbb Z$ is the set of **even integers**
- $2\mathbb Z + 1$ is the set of **odd integers**
- $\mathbb Q$ is the set of **rational numbers**
- $\mathbb R$ is the set of **real numbers**
- $\empty$ or $\{\}$ is the **empty set** with no elements
### Set builder notation
!!! definition
- The **domain of discourse** is the context of the current problem, which may limit the universal set (e.g., if only integers are discussed, the domain is integers only)
$x$ is an element if $x$ is in $\mathcal U$ and $P(x)$ is true.
$$\{x\in\mathcal U|P(x)\}$$
!!! example
All even numbers: $A=\{n\in\mathbb Z,\exists k\in\mathbb Z,n=2k\}$
$f(x)$ is an element if $x$ is in $\mathcal U$, and $P(x)$ is true:
$$\{f(x)|\underbrace{x\in\mathcal U, P(x)}_\text{swappable, omittable}\}$$
!!! example
- All even numbers: $A=\{2k|k\in\mathbb Z\}$
- All rational numbers: $\mathbb Q=\{\frac a b | a,b\in\mathbb Z,b\neq 0\}$
The **complement** of a set is the set containing every element **not** in the set.
$$\overline S$$
The **universal set** is the set containing everything, and is the complement of the empty set.
$$\mathcal U=\overline\empty$$
Two sets are **disjoint** if they do not have any elements in common.
$$S\cup T=\empty$$
### Set operations
A **subset** is inside another that is a **superset**.
$$
S\subseteq T \\
S\subseteq T\iff \forall x\in\mathcal U,(x\in S\implies x\in T)
$$
A **strict or proper subset** is a subset that is not equal to its **strict or proper superset**.
$$S\subset T$$
Two sets are equal if they are subsets of each other.
$$S=T\equiv (S\subseteq T)\wedge (T\subseteq S)$$
The **union** of two sets is the set that contains any element in either set.
$$S\cup T=\{x\in\mathcal U|(x\in S)\vee(x\in T)\}$$
The **intersection** of two sets is the set that only contains elements in both sets.
$$S\cap T=\{x\in\mathcal U|(x\in S)\wedge(x\in T)\}$$
The **difference** of two sets is the set that contains elements in the first but not the second. The remainder is dropped.
$$S-T=S\backslash T$$
The **complementary** set is every element not in that set.
$$
\overline S=\{x:x\not\in S\} \\
\overline S=\mathcal U-S
$$
The intersection and union operators have the same properties as **AND** and **OR** and so are equally commutative / associative.
**De Morgan's laws** still hold with sets.
### Intervals
An interval can be represented as a bounded set.
$$[a,b)=\{x\in\mathcal U|a\leq x\wedge x<b\}$$
$\empty$ is any impossible interval.
### Ordered pairs
!!! definition
- A **binary relation** on two sets $A, B$ is a subset of their Cartesian product.
- An ***n*-ary relation** between $n$ sets is a subset of their *n*-Cartesian product.
Also known as **tuples**, ordered pairs are represented by angle brackets.
$$\left<a,b\right> = \left<c,d\right>\iff (a=c)\wedge(b=d)$$
The **Cartesian product** of two sets is the set of all ordered pair combinations within the two sets.
$$A\times B=\{\left<a,b\right> | (a\in A)\wedge (b\in B)\}$$
It is effectively the cross product, so is not commutative, although distributing unions, intersections, and differences works as expected.
The **n-Cartesian product** of $n$ sets expands the Cartesian product.
$$A\times B\times\dots\times Z=\{\left<a, b,\dots z\right>|a\in A, b\in B,\dots,z\in Z\}$$
### Powersets
!!! definition
- An **index set** $I$ is the set containing all relevant indices.
A **partition** of a set $S$ is a set of **disjoint** sets that create the original set when unioned.
$$S=\bigcup_{i\in I}A_i$$
!!! example
$\{\{1\},\{2,3\},\{4,\dots\}\}$ is a partition of $\mathbb N$.
A **powerset** of a set $A$ is the set of all possible subsets of that set.
$$\mathcal P(A)=\{X|X\subseteq A\}$$
The empty set is the subset of every set so is part of each powerset. The number of elements in a subset is equal to the the number of elements in the original set as a power of two.
$$\dim(\mathcal P(A))=2^{\dim(A)}$$
!!! example
- $\mathcal P(\empty)=\empty$
- $\mathcal P(\{1,2\})=\{\empty, \{1\}, \{2\}, \{1, 2\}\}$
By definition, any subset is an element in the powerset.
$$A\subseteq B\equiv A\in\mathcal P(B)$$
- $\empty\in\mathcal P(A)$
- $A\in\mathcal P(A)$
- $A\subseteq B\implies (\mathcal P(A)\subseteq \mathcal P(B))$
- $A\in C\implies (C-A\subseteq C)$
!!! example
To prove $A\subseteq B\implies \mathcalP(A)\subseteq \mathcal P(B)$:
**Proof:** Let $A\subseteq B$ and $X\in\mathcal P(A)$. By definition, since $X\in\mathcal P(A), X\subseteq A$. Since $A\subseteq B$, it follows that $X\subseteq B$. Thus by the definition of the powerset, $X\in\mathcal P(B)$.
## Functions
!!! definition
- A **surjective** function has an equal codomain and range.
A **function** a relation between two sets $f:X\to Y$ such that each $x\in X$ **maps to** a unique $f(x)\in Y$.
$$
\begin{align*}
f:\ &X\to Y \\
&x\longmapsto f(x)
\end{align*}
$$
!!! example
Sample function with multiple cases and indices:
$$
\begin{align*}
f:\ &X\to Y \\
&x_i\longmapsto \begin{cases}
y_1 & i\in\{1,2\} \\
y_3 & i\in\{3,4,5\}
\end{cases}
\end{align*}
$$
The **domain** $\text{dom}(f)$ is the input set.
$$X=\text{dom}(f)$$
The **codomain** $\text{cod}(f)$ is the output set.
$$Y=\text{cod}(f)$$
The **range** $\text{rang}(f)$ is the subset of $Y$ that is actually mapped to by the domain.
$$
\begin{align*}
\text{rang}(f)&=\{y\in Y|\exists x\in X,y=f(x)\} \\
&=\{f(x)|x\in X\}
\end{align*}
$$
The **pre-image** is the subset of the domain that maps to a specific subset $B$ of the codomain.
$$\text{preimage}(f)=\{x\in X|\exists y\in B,y=f(x)\}$$
The **image** is the subset of the codomain that is mapped by a specific subset $A$ of the domain.
$$\text{image}(f)=\{f(x)|\exists x\in A\}$$
!!! example
For the function $f: \mathbb R^+_0\to \mathbb R$ defined by $x\longmapsto x^2$:
- the domain is $\mathbb R^+_0$
- the codomain is $\mathbb R$
- the range is $\mathbb R^+_0$
- the preimage for $\{1\}$ is $\{1,-1\}$
- the image for $0$ is $\{0\}$
Two functions $f=g$ are equal if and only if:
- their domains are equal
- their codomains are equal
- $f(x)=g(x)$ for all $x\in \text{dom}(f)$
### Function types
An **injective function**, **injection**, or **one-to-one function** is a function that maps only one $y$-value to each $x$.
$$\forall x_1,x_2\in\text{dom}(f), \text{ if } f(x_1)=f(x_2),x_1=x_2$$
A **surjective function**, **surjection**, or **onto** is a function that has its codomain equal to its range. A surjection $g:Y\to X$ exists if and only if an injection $f:X\to Y$ exists.
$$
\forall y\in\text{cod}(f),\exists x\in\text{dom}(f), f(x)=y \\
\text{rang}(f)=\text{cod}(f)
$$
A **bijective function** is both injective and surjective.
An **inverse relation** swaps the domain, codomain, and ordered pairs.
$$
\begin{align*{
R^{-1}:Y&\to X \\
R(x)&\mapsto x
$$
A function is **invective** or **invertible** if and only if it is bijective. All inversions are also bijective.
$$f^{-1^{-1}}=f$$
A **composition** maps the codomain of one to the domain of another function only if the first is a subset ($Y_1\subseteq Y_2$).
$$
\begin{align*}
f&:X\to Y_1,x\mapsto f(x) \\
g&:Y_2\to Z,y\mapsto g(y) \\
gf&: X\to Z,x\mapsto g(f(x))
\end{align*}
$$
Compositions are commutative but not associative.
- $h(gf)=(hg)f$
- $hgf\neq hfg$
- $f, g$ are injective $\implies$ $gf$ is injective
- $f, g$ are surjective $\implies$ $gf$ is surjective
- $gf$ is injective $\implies$ $f$ is injective
- $gf$ is surjective $\implies$ $g$ is surjective
The **identity function** is the function that returns its argument. Generally, a function composed with its inverse is the identity function.
$$
\begin{align*}
I:X&\to X \\
x&\mapsto x
\end{align*}
$$
If $f: X\to Y$ is bijective:
- the identity on $Y$ is $f(f^{-1}(y))$
- the identity on $X$ is $f^{-1}(f(x))$
If $f: X\to Y$ and $g: Y\to Z$ are bijective:
- $gf$ exists and is invertible
- $f^{-1}g^{-1}=(gf)^{-1}$ and exists
## Cardinality
!!! definition
- A **countably infinite** set is such that there exists a **bijective** function that maps the set to the set of natural numbers.
- A **countable** set is a finite set or a countably infinite set.
- An **uncountable** or **uncountably infinite** set is not countable.
The **cardinality** of a set is the number of elements in that set.
$$|S|$$
If two sets have a finite number of elements, their Cartesian product will have the same number of elements as the product of their elements.
$$|A|,|B|\in\mathbb N\implies|A\times B|=|A||B|$$
If two sets $X$ and $Y$ have finite cardinality and $f:X\to Y$:
- An injective function must have $|X|\leq |Y|$.
- A surjective function must have $|X|\geq |Y|$.
- A bijective function occurs if and only if $|X|=|Y|$.
A set is **finite** if it is empty or it is mappable to a subset of the natural numbers. By definition, the set of natural numbers is infinite.
$$\exists n\in\mathbb N,\exists f\text{ is bijective}, f:S\to \mathbb N_n,|s|=n$$
### Uncountable sets
The cardinality of countable sets is relative to the cardinality of the set of **natural numbers**.
$$|\mathbb N|=\aleph_0$$
By Contor's theorem, the powerset of the natural numbers must have a larger cardinality than the set of natural numbers.
$$|X|=\aleph_0\implies|\mathcal P(X)|=2^{\aleph_0}>\aleph_0$$
The following can be taken for granted:
- $|\mathbb R|>|\mathbb N|$
- $|\mathcal P(\mathbb N)|>|\mathbb N|$
- $|\mathcal P(\mathbb N)|=|\mathbb R|$
## Relations
A **binary relation** $R$ from sets $A$ to $B$ must be a subset of the two. A relation from $A$ to $A$ can be written as $R\subseteq A^2$.
$$R\subseteq A\times B$$.
!!! example
- $\forall x,y\in A,B,x<y$ is a subset. $<$ is a binary relation.
For $R\subseteq X\times Y$:
- $\text{dom}(R)=\{x\in X|\exists y\in Y,xRy\}$
- $\text{cod}(R)=Y$
- $\text{rang}(R)=\{y\in Y|\exists x\in X,xRy\}$
- The **image** of $X_1\subseteq X$ under $R$: $R(X_1)=\{y\in Y|\exists x\in X_1xRy\}$
- The **pre-image** is: $R^{-1}(Y_1)=\{x\in X|\exists y\in Y_1,xRy\}$
Relations are trivially proven to be relations through subset analysis.
!!! example
For the relation $L$\subseteq R^2=\{\left<x,y\right>\in\mathbb R^2|x<y\}$:
Clearly it is a subset of $R^2$, so it is a relation.
- The domain is $\mathbb R$.
- The range is $\mathbb R$.
- $L(\{1,4\})=\{y>4|y\in\mathbb R\}$ (1 OR 4)
- $L^{-1}(\{-1,2\})=\{x\in\mathbb R|x<2\}$ (-1 OR 2)
The **empty relation** $\empty$ is a relation on all sets.
The **identity relation** on all sets returns itself.
$$E=\{\left<a,a\right>|a\in A\}$$
The **universal relation** relates each element in the first set to every element to the second set.
$$U=A^2$$
The **restriction** of relation $R$ to set $B$ limits a previous relation on a superset $A$ such that $B\subseteq A$.
$$R\big|_B=R\cap B^2$$
Graphs are often used to represent relations. A node from $4\to3$ can be represented as $\left<3,4\right>$, much like an adjacency list.
### Reflexivity
A **reflexive** relation $R\subseteq X^2$ is such that every element in $X$ is related to itself by the relation.
$$\forall x\in X,\left<x,x\right>\in R$$
An **irreflexive** relation is such that each element is *not* related to itself.
$$\forall x\in X,\left<x,x\right>\not\in R$$
Reflexivity is determined graphically by checking if the main diagonal of a truth table is true.
!!! example
For the reflexive relation $R$, $A=\{1,2\},R=\{\left<1,1\right>,\left<2,2\right>\}$:
|$A\times A$ | 1 | 2 |
| --- | --- | --- |
| 1 | T | F |
| 2 | F | T |
!!! warning
$\empty$ is often vacuously true for most conditions.
If $R$ is a **non-empty** relation on a **non-empty** set $X$, $R$ cannot be both reflexive and irreflexive.
### Symmetry
A **symmetric** relation $R\subseteq X^2$ is such that every relation goes both ways.
$$\forall x,y\in X^2,\left<x,y\right>\in R\iff\left<y,x\right>\in R$$
An **asymmetric** relation is such that **no** relation goes both ways.
$$\forall x,y\in X^2,\left<x,y\right>\in R\implies\left<y,x\right>\not\in R$$
An **antisymmetric** relation is such that **no** relation goes both ways, *except* if compared to itself, and that the relation relates identical items.
$$\forall x,y\in X^2,\left<x,y\right>\in R\wedge\left<y,x\right>\in R\implies x=y$$
Where $x,y,z$ are elements in $X$, and $p,q,r$ are arbitrary proposition results (true/false):
- Symmetric relations must be symmetrical across the main diagonal of a truth table.
| $X^2$ | $x$ | $y$ | $z$ |
| --- | --- | --- | --- |
| $x$ | ? | $p$ | $q$ |
| $y$ | $\neg p$ | ? | $r$ |
| $z$ | $\neg q$ | $\neg r$ | ? |
- Asymmetric relations must be oppositely symmetrical across the main diagonal. The main diagonal also must be false.
- Antisymmetric relations must be false only if there is a true.
### Transitivity
A **transitive** relation links related terms. For example, $a<b$ and $b<c$ implies $a<c$.
$$\forall x,y,z\in X,\left<x,y\right>\in R\wedge\left<y,z\right>\in R\implies\left<x,z\right>\in R$$
## Orders
!!! definition
- A **partial order** is reflextive, antisymmetric, and transitive.
A **partially ordered set (poset)** is a set $S$ partially ordered with relation $R$.
$$\left<S,R\right>\text{ on } P=R_{S,P}$$
!!! example
$R_{\mathbb Z,\geq}$ is a poset. $\left<\mathcal P(A),\subseteq\right>$ on $A$ is also a poset.
A **strict poset** is irreflexive, asymmetric, and transitive.
A **total order** is a strict poset such that the relation is defined between every possible pair on the set.
$$\forall x,y\in S,xPy\wedge yPx\in\left<S,P\right>$$
### Equivalence relations
An **equivalence class** is a criterion that determines whether two objects are equivalent. The original set must be the union of all equivalence classes.
!!! example
The following are all in the equivalence class $=_1$: $\{1,\frac 2 2,\frac 3 3,\frac 4 4,...\right}$
## Combinatorics
!!! definition
- **and** usually requires you to multiply sets together.
- **or** usually requires you to add then subtract unions.
The number of ways to choose exactly one element from finite sets is the product of their dimensions.
$$|A_1||A_2|...|A_n|$$
!!! example
The number of unique combinations (including order) from four dice is $|6|^4$.
### Ordered with replacement
These problems count order as separate permutations and replace an item after it is taken for the future. If there are $n$ outcomes, and $m$ events that take one of those outcomes:
$$P=n^m$$
To pick $m$ items out of $n$ elements:
$$P(n,m)=\frac{n!}{(n-m!)}$$
If there are duplicates that would otherwise result in an identical string, divide the result by $m!$, where $m$ is the number of repetitions for each duplicate $n_1,n_2,...$.
$${n\choose n_1!n_2!n_k!}=\frac{n!}{n_1!n_2!...n_k!}$$
!!! example
The number of permutations of "ECE119" has two characters that have duplicates. Therefore, the number of possibilities is:
$$\frac{6!}{2!2!}$$
### Unordered with replacement
To rearrange $n$ unique items, the number of possibilities is:
$$n!$$
To choose $n$ items $m$ times, regardless of order, the number of possibilities is:
$${n\choose m}=\frac{n!}{(n-m)!m!}={n\choose(n-m),m}$$
Clearly ${n\choose m}=0$ if $m>n$ or $m<0$.
To choose $k$ out of $n$ items one time, multichoose can be used:
$$\left({n\choose k}\right)={n+k-1\choose k}={n-1+k\choose n-1,k}$$
### Binomial coefficients
A **slack variable** is used to change inequalities into equalities.
!!! example
If solving $x+y\leq 7$, setting $z=7-(x+y)$ to make everything the same domain ($\mathbb Z^+_0$) to use choose.
**Pascal's identity** defines the choose operator recursively.
$${n\choose m}={n-1\choose m-1}+{n-1\choose m}$$
The **binomial theorem** expands a binomial.
$$\forall a,b\in\mathbb R,(a+b)^n=\sum^n_{i=0}{n\choose i}a^{n-i}b^i$$
The sum of choosing integers is its power to 2. Therefore, a finite set with dimension $n$ must have exactly $2^n$ possible subsets.
$$\forall n\in\mathbb Z^+_0,\sum^n_{k=0}{n\choose k}=2^n$$
### Inclusion-exclusion
The inclusion-exclusion principle removes duplicate counting.
$$|A\cup B|=|A|+|B|-|A\cap B|$$
This can be extended to 3+ sets, proven by a bijection to $\mathbb N_{|A| + |B|+|A\cap B|}$:
$$|A\cup B\cup C|=|A| + |B| + |C| - (|A\cap B| + |A\cap C| + |B\cap C|)-|A\cap B\cap C|$$
If $B$ is a subset of $A$, the dimension of $B$ is related to that of $A$.
$$B\subseteq A\implies|B|=|A|-|\overline B|$$
## Probability
!!! definition
- An **experiment** is an event that has a number of outcomes.
- **Elementary events** are the outcomes of an experiment compose the set of all events.
- An **event** $E$ is a subset of the sample space $S$, which is the **certain event**.
- The **null event** is the empty set.
- Sets of events are **mutually exclusive** if they are disjoint.
- Elementary events are **equiprobable** if they are equally probable.
- A **uniform probability distribution** on $S$ is such that all elementary events are equiprobable.
A **probability distribution function (PDF)** $Pr$ converts the elements of the powerset of all outcomes to a real number — its probability.
$$Pr:\mathcal P(S)\to\mathbb R,0\leq P(A)\leq 1$$
A PDF must have, if $S$ is the sample space:
- $\forall A\subseteq S,Pr\{A\}\geq 0$
- $Pr\{S\}=1$
- The union of all mutually exclusive sets is the sample space
A **discrete probability distribution** is such that the sample space is a countable set.
For all $A\subseteq S$, the probability of event $A$ is the sum of the probabilities of all elementary events in $A$.
- $Pr\{A\}=\sum_{e\in A}Pr\{\{e\}\}$
- $Pr\{\empty\}=0$
- $Pr\{A'\}=1-Pr\{A\}$
Adding events together can never decrease their probability, and the sum of all probabilities must equal $1$ such that $\text{rang}(Pr)\subseteq[0,1]$.
$$A\subseteq B\subseteq S\implies Pr\{A\}\leq Pr\{B\}$$
The **inclusion-exclusion principle** also applies.
$$Pr\{A\cup B\}=Pr\{A\}+Pr\{B\}-Pr\{A\cup B\}$$
### Named PDFs
!!! definition
- An **emperical PDF** is collected from empirical data.
A **Bernouilli trial** is an event with exactly two options, pass $P$ with probability $p$, or fail $F$ with probability $q=1-p$. For the event $X$:
$$
Pr\{X\}=\begin{cases}
p &\text{if }X=\{P\} \\
1-p&\text{if }X=\{F\}
\end{cases}
$$
For exactly two options for $x$ (1 or 0):
$$Pr\{X=x\}=p^x(1-p)^{1-x}$$
Please see [SL Math - Analysis and Approaches 2#Binomial distribution](/g11/mcv4u7/#binomial-distribution) for more information.
A **random variable** is a function that assigns a real number to every item in the sample space. A **discrete random variable** is used if the sample space is discrete. The probability of all events that lead to a possible discrete random variable $x\in\mathbb R$, where $X$ is the function to transform those variables:
$$Pr\{X^{-1}(\{x\})\}$$
Thus the **binomial distribution** for $r$ successes of $n$ total tries, if they are independent, is:
$$Pr\{X=r\}{n\choose r}p^rq^{n-r}$$
### Independence
Please see [SL Math - Analysis and Approaches 2#Conditional probability](/g11/mcv4u7/#conditional-probability) for more information.
Two events are independent if they can be treated separately.
$$\text{independent}\iff Pr\{A\cap B\}=Pr\{A\}Pr\{B\}$$
Or, via the inclusion-exclusion theorem:
$$\text{independent}\iff Pr\{A\cup B\}=Pr\{A\}+Pr\{B\}-Pr\{A\}Pr\{B\}$$
**Bayes' theorem** provides a general formula for conditional probability:
$$Pr\{A|B\}=\frac{Pr\{B|A\}}{Pr\{B\}}$$
Formally, this can be solved without $Pr\{B\}$:
$$Pr\{A|B\}=\frac{Pr\{A\}Pr\{B|A\}}{Pr\{A\}Pr\{B|A\}+Pr\{\overline A\}Pr\{B|\overline A\}}$$
### Expected value
The **expected value**, **mean**, or **expectation of $X$** is:
$$E[X]=\sum_{x\in\mathbb R}x\cdot Pr\{X=x\}=\sum_{s\in S}X(s)\cdot Pr\{\{s\}\}$$
This operation is **linear**, but multiplies using AND:
$$
E[X+Y]=E[X}+E[Y] \\
E[XY]=\sum_{x\in X,y\in Y}xy\cdotPr\{X=x\wedge y\=y\}
$$
Thus if $X$ and $Y$ are independent:
$$E[XY]=E[X]E[Y]$$
An **indicator random variable** only has two possible outcomes: zero or one. Thus an indicator random variable $X$ has an expected value equal to its probability:
$$E[X]=Pr\{X=1\}$$
The **covariance** of $X$ and $Y$ represents the direction of difference of $X$ and $Y$ from their means.
$$Cov[X,Y]=E[XY]-E[X]E[Y]$$
Reference in New Issue View Git Blame Copy Permalink