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1.2E: Domain and Range
https://math.libretexts.org/Courses/Northeast_Wisconsin_Technical_College/College_Algebra_(NWTC)/01%3A_Functions/1.02%3A_Domain_and_Range/1.2E%3A_Domain_and_Range
f\left(x\right)=\left\{ $\begin{array}{ccc} {5x} & {if} & {x<0} \\ {3} & {if} & {0\le x\le 3} \\ {x^{2} } & {if} & {x>3} \end{array}$ \right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ 24. f\left(x\right)=\left\{\begin{a...f\left(x\right)=\left\{ $\begin{array}{ccc} {5x} & {if} & {x<0} \\ {3} & {if} & {0\le x\le 3} \\ {x^{2} } & {if} & {x>3} \end{array}$ \right.\ \ \ \ \ \ \ \ \ \ \ \ \ \ 24. f\left(x\right)=\left\{ $\begin{array}{ccc} {x^{3} +1} & {if} & {x<0} \\ {4} & {if} & {0\le x\le 3} \\ {3x+1} & {if} & {x>3} \end{array}$ \right.\) f\left(x\right)=\left\{ $\begin{array}{ccc} {3} & {if} & {x\le -2} \\ {-x+1} & {if} & {-2<x\le 1} \\ {3} & {if} & {x>1} \end{array}$ \right.\ \ \ \ \ \ \ \ 36.
6.10: Summary
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/06%3A_Functions/6.10%3A_Summary
Important definitions: Cartesian product function domain codomain, range one-to-one onto bijection identity map inverse function composition image pre-image Notation: $A \times B$ \(f : A \rightarro...Important definitions: Cartesian product function domain codomain, range one-to-one onto bijection identity map inverse function composition image pre-image Notation: $A \times B$ $f : A \rightarrow B$ $f(a)$ $\forall a_{1}, a_{2} \in A$ $I_{A}$ $f^{−1}$ $g \circ f$ $\{f(a) \mid a \in A\}$ $f(A_{1})$ $f^{−1} (B_{1})$ A function $f : A \rightarrow B$ has an inverse $f^{−1} : B \righarrow A \text { iff} f$ is a bijection. The inverse of a bijection is a bijection
2.1: First Example of a Two-Column Proof
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/02%3A_Two-Column_Proofs/2.01%3A_First_example_of_a_two-column_proof
Now, we can translate the deduction into English:For clarity, we draw a dark horizontal line to separate the hypotheses from the rest of the proof. (In addition, we will number each row of the proof, ...Now, we can translate the deduction into English:For clarity, we draw a dark horizontal line to separate the hypotheses from the rest of the proof. (In addition, we will number each row of the proof, for ease of reference, and we will make the left border of the figure a dark line.) For example, here is a two-column proof that justifies the deduction above.
8.4: The Natural Numbers are Well-Ordered
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/08%3A_Proof_by_Induction/8.04%3A_The_natural_numbers_are_well-ordered
Namely, if $P(n)$ can be proved for all $n \in \mathbb{N}^{+}$ by using any one of the many forms of Mathematical Induction, then it can also be proved by applying Theorem $8.4.4$ to the set \[S...Namely, if $P(n)$ can be proved for all $n \in \mathbb{N}^{+}$ by using any one of the many forms of Mathematical Induction, then it can also be proved by applying Theorem $8.4.4$ to the set $S=\left\{n \in \mathbb{N}^{+} \mid \neg P(n)\right\} .$ Suppose it is not true that $F_{n} < 2^{n}$ for all $n \in \mathbb{N}^{+}$ . (This will lead to a contradiction.) Then, since $\mathbb{N}$ is well-ordered, there is a smallest $n$ , such that $F_{n} \geq 2^{n}$ .
6.1: Cartesian Product
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/06%3A_Functions/6.01%3A_Cartesian_product
The Cartesian product is another important set operation. Before introducing it, let us recall the notation for an ordered pair.
1.10: Summary
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/01%3A_Propositional_Logic/1.10%3A_Summary
$\lnot$ (not; means “It is not the case that”) $\&$ (and; means “Both ______ and ______”) $\lor$ (or; means “Either ______ or ______”) $\Rightarrow$ (implies; means “If ______ then ______”) \(... $\lnot$ (not; means “It is not the case that”) $\&$ (and; means “Both ______ and ______”) $\lor$ (or; means “Either ______ or ______”) $\Rightarrow$ (implies; means “If ______ then ______”) $\Leftrightarrow$ (iff; means “______ if and only if ______”) Determining whether an assertion is true (for particular values of its variables) commutativity of $\&$ , $\lor$ , and $\Leftrightarrow$ introduction and elimination rules for $\&$ , $\lor$ , and $\Leftrightarrow$
6: Functions
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/06%3A_Functions
It is the pervading law of all things . . . that form ever follows function. This is the law. Louis Sullivan (1856–1924), American architect The tall office building artistically considered
8.2: Three cases for λ
https://math.libretexts.org/Bookshelves/Differential_Equations/Partial_Differential_Equations_(Walet)/08%3A_Separation_of_Variables_in_Polar_Coordinates/8.02%3A_Three_cases_for_%CE%BB
Now let me look at the solution of the $R$ equation for each of the two cases (they can be treated as one), $\rho^2 R''(\rho) + \rho R'(\rho) -n^2R(\rho)=0. \nonumber$ Let us attempt a power-seri...Now let me look at the solution of the $R$ equation for each of the two cases (they can be treated as one), $\rho^2 R''(\rho) + \rho R'(\rho) -n^2R(\rho)=0. \nonumber$ Let us attempt a power-series solution (this method will be discussed in great detail in a future lecture) $R(\rho) =\rho^\alpha. \nonumber$ We find the equation $\rho^\alpha[\alpha(\alpha-1)+\alpha^2-n^2]= \rho^\alpha[\alpha^2-n^2]=0 \nonumber$ If $n\neq 0$ we thus have two independent solutions (as should be) \[R_n…
4: First-Order Logic
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/04%3A_First-Order_Logic
Three Logicians walk into a bar, and the barkeeper asks “Would you all like something to drink?” The 1st Logician says “I don’t know,” and the 2nd Logician says “I don’t know.” Then the 3rd Logician s...Three Logicians walk into a bar, and the barkeeper asks “Would you all like something to drink?” The 1st Logician says “I don’t know,” and the 2nd Logician says “I don’t know.” Then the 3rd Logician says “yes.” author unknown
TitlePage
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/00%3A_Front_Matter/01%3A_TitlePage
University of Lethbridge Proofs and Concepts - The Fundamentals of Abstract Mathematics Dave Witte Morris & Joy Morris
1.5: Determining whether an assertion is true
https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/Proofs_and_Concepts_-_The_Fundamentals_of_Abstract_Mathematics_(Morris_and_Morris)/01%3A_Propositional_Logic/1.05%3A_Determining_whether_an_assertion_is_true
We have \[\begin{aligned} (A \lor C) \Rightarrow \lnot (A \Rightarrow B) &\quad= \quad (\mathsf{T} \lor \mathsf{T}) \Rightarrow \lnot (\mathsf{T} \Rightarrow \mathsf{F}) \\& \quad= \quad \mathsf{T} \R...We have $\begin{aligned} (A \lor C) \Rightarrow \lnot (A \Rightarrow B) &\quad= \quad (\mathsf{T} \lor \mathsf{T}) \Rightarrow \lnot (\mathsf{T} \Rightarrow \mathsf{F}) \\& \quad= \quad \mathsf{T} \Rightarrow \lnot \mathsf{F} \\& \quad= \quad \mathsf{T} \Rightarrow \mathsf{T} \\& \quad= \quad \mathsf{T} . \end{aligned}$ The assertion is true.

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