eifueo/docs/1b/ece108.md

# 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.

- $h(gf)=(hg)f$
- $hgf\neq hfg$