MATH 120 - Day 1

2024-09-04 14:41:37 -0400 EDT

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Differentiation and Integration

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Some logic symbols to help explain

$ \forall $ = for all

$ \exists $ = exists

$ \in $ = in

$ \cup $ = includes

$ \Rightarrow $ = implies

$ \Leftrightarrow $ = equivalent

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Propositions (rules used to solve equations)

• $ \forall \text{ } a, \text{ } b \in \mathbb{R} \Rightarrow a \leq b \text{ } or \text{ } a \geq b$; there is always a relationship

• $ \forall \text{ } a \in \mathbb{R}, a \leq a $

• if $ a \leq b \text{ } and \text{ } b \leq c \Rightarrow a \leq c $

• $ a \leq b \text{ } and \text{ } a \geq b \Rightarrow a = b $

• $ a \leq b \text{ } and \text{ } c \in \mathbb{R} \Rightarrow a + c \leq b + c $

• $ a \leq b \text{ } and \text{ } c > 0 \Rightarrow ac \leq bc $

• $ a \leq b \Rightarrow -b \leq -a $

• $ a \leq b \Rightarrow \dfrac {1} {a} \leq \dfrac {1} {b} $

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Intervals and Set Notation

$ A = \lbrace x \in \mathbb{R} \text{ } | \text{ } P(x) \rbrace $; P(x) is a condition that x must meet in order to be in the set A.

• Closed Interval $ [a,b] = \lbrace x \in \mathbb{R} \text{ } | \text{ } a \leq x \leq b \rbrace $
• Open Interval $ (a,b) = \lbrace x \in \mathbb{R} \text{ } | \text{ } a < x < b \rbrace $
• Semi-closed Interval $ [a,b) = \lbrace x \in \mathbb{R} \text{ } | \text{ } a \leq x < \rbrace $

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Absolute Value

$P(x) = |x|$; returns the distance of the value from zero

this could be written more pragmatically with the peacewise definition:

$$ |x| = \begin{cases} x, & \text{if } x \geq 0 \\ -x, & \text{if } x < 0 \end{cases} $$

Absolute Value Propositions

• $ |ab| = |a| |b| $
• $ | \dfrac {a} {b} | = \dfrac {|a|} {|b|} $
• $ |a^n| = |a|^n, n \in \mathbb{R} $
• $ |b| = |b - 0| $
• $ |x| = a \Leftrightarrow x = \pm a $
• $ |x| < a \Leftrightarrow -a < x < a $
• $ |x| > a \Leftrightarrow -a > x > a $
• $ |a| = |b| \Leftrightarrow a = \pm b $

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Triangle Inequality

$ \forall a,b \in \mathbb{R}, |a + b| \leq |a| + |b| $

Proof: We will break the proposition into cases and bring each case to a easily solvable equality. this will also serve as an example of who to prove a proposition

$ (E): | a + b | \leq |a| + |b| $

Case 1. $ a \geq 0 \text{ and } b \geq 0 $

$$ \begin{gather} | a | = a \newline | b | = b \newline a + b \geq 0 \Leftrightarrow | a + b | = a + b \newline (E) \Leftrightarrow a + b \geq a + b \text{and } a + b \geq a + b \text{ TRUE!} \end{gather} $$

Case 2. $ a \geq 0 \text{ and } b \leqq 0 $

$$ \begin{gather} | a | = a \newline | b | = b \newline \end{gather} $$

$$ \begin{align} \text{1. either } a + b \geq 0 \Leftrightarrow a \geq -b \text{ then } | a + b | = a + b \text{ and } (E) & \Leftrightarrow a + b \leq a - b \newline & \Leftrightarrow b \leq -b \newline & \Leftrightarrow 2b \leq 0 \newline & \Leftrightarrow b \leq 0 \text{ TRUE!} \end{align} $$

$$ \begin{align} \text{2. or } a + b \leq 0 \Leftrightarrow a \leq -b \text{ then } | a + b | = -(a + b) \text{ and } (E) & \Leftrightarrow -(a + b) \leq a - b \newline & \Leftrightarrow -a - b \leq a - b \newline & \Leftrightarrow -a \leq a \newline & \Leftrightarrow a \leq 0 \text{ TRUE!} \end{align} $$

Case 3: $ a \leq 0 \text{ and } b \leq 0 \ $

$$ \begin{align} | a | = -a \newline | b | = -b \newline a + b \leq 0 \Rightarrow | a + b | = -(a + b) \newline (E) \Leftrightarrow -a - b \leq -a - b \text{ TRUE!} \end{align} $$

Case 4: a and be are the same (parallel) - therefore it must be TRUE!

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