From Privateer

Contents 1 Analysis 1.1 Logical Foundation 1.1.1 Quantifiers 1.1.2 Proof Types 1.1.2.1 Example direct proof 1.1.2.1.1 Algebraic Sketch 1.1.2.1.2 Proof 1.1.2.2 Example indirect proof 1.2 Field Axioms 1.2.1 Additive 1.2.2 Multiplicative 1.2.3 Ordering 1.2.4 Trichotomy 1.2.5 Transitive

Analysis

Logical Foundation

${\displaystyle p\Rightarrow q}$ means "p implies q"

${\displaystyle p\iff q}$ means "p implies q, and q implies p"

Quantifiers

${\displaystyle \forall }$: Universal quantifier; "for all". ${\displaystyle \exists }$: Existence quantifier; "there exists".

The negation of ${\displaystyle \forall }$ is ${\displaystyle \exists }$; "${\displaystyle \neg \forall =\exists }$". Similarly the negation of ${\displaystyle \exists }$ is ${\displaystyle \forall }$; "${\displaystyle \neg \exists =\forall }$".

Negation Example:
${\displaystyle \forall x,p\left(x\right)}$ means "for all ${\displaystyle x}$, the condition ${\displaystyle p\left(x\right)}$ is true."
${\displaystyle \neg (\forall x,p\left(x\right))=\exists x,\neg p\left(x\right)}$, which means "there exists an ${\displaystyle x}$ such that ${\displaystyle p\left(x\right)}$ is false."

Proof Types

Direct Proof: a series of logical statements that directly transform p into q.

Indirect Proof (proof by contradiction): Assume that the conclusion of a logical statement is false, then proceed with logical statements until a contradiction is found.

Example direct proof

Statement to prove: ${\displaystyle \forall \epsilon >0}$, ${\displaystyle \exists \delta >0}$ such that ${\displaystyle 1-\delta <x<1+\delta \Rightarrow 5-\epsilon <2x+3<5+\epsilon }$

In plain english, this means "for all epsilon greater than zero, there exists a delta that is greater than zero, such that the inequality x is less than 1 + delta and greater than 1 - delta, implies that ${\displaystyle 2x+3}$ less than 5 + epsilon and greater than 5 - epsilon."

Algebraic Sketch

Isolate the ${\displaystyle x}$, and find ${\displaystyle \delta }$
${\displaystyle 5-\epsilon <2x+3<5+\epsilon }$
${\displaystyle 2-\epsilon <2x<2+\epsilon }$
${\displaystyle \frac{2-\epsilon }{2}<x<\frac{2+\epsilon }{2}}$
${\displaystyle 1-\frac{\epsilon }{2}<x<1+\frac{\epsilon }{2}}$
${\displaystyle \Rightarrow \delta =\frac{\epsilon }{2}}$

Proof

let ${\displaystyle \epsilon >0}$

consider ${\displaystyle \delta =\frac{\epsilon }{2}>0}$

Since ${\displaystyle \epsilon >0}$

Assume ${\displaystyle 1-\delta <x<1+\delta }$

${\displaystyle 1-\frac{\epsilon }{2}<x<1+\frac{\epsilon }{2}}$, since ${\displaystyle \delta =\frac{\epsilon }{2}}$

and ${\displaystyle \frac{2-\epsilon }{2}<x<1+\frac{2+\epsilon }{2}}$

and ${\displaystyle 2-\epsilon <2x<2+\epsilon }$

So ${\displaystyle 5-\epsilon <2x+3<5+\epsilon }$

Example indirect proof

Theorem to prove: let ${\displaystyle x}$ be a real number.
if ${\displaystyle x>0}$, then ${\displaystyle \frac{1}{x}>0}$.

In this statement, ${\displaystyle p}$ is "${\displaystyle x>0}$," and ${\displaystyle q}$ is "$\frac{1}{x}>0$."

By way of contradiction, assume ${\displaystyle p}$ and ${\displaystyle \neg q}$, that is: if ${\displaystyle x>0}$, then ${\displaystyle \frac{1}{x}\le 0}$.

Field Axioms

Additive

if ${\displaystyle a<b}$, then ${\displaystyle a+\alpha <b+\alpha }$

Multiplicative

if ${\displaystyle a<b}$, and ${\displaystyle c>0}$, then ${\displaystyle ac<bc}$; if ${\displaystyle a<b}$, and ${\displaystyle c<0}$, then ${\displaystyle bc<ac}$.

Ordering

Trichotomy

Transitive