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% Document Info
\author{j24}
\pdfinfo{
  /Title (a.pdf)
  /Creator (Cheatography)
  /Author (j24)
  /Subject (A Cheat Sheet)
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{A Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{j24} via \textcolor{DarkBackground}{\uline{cheatography.com/195607/cs/41005/}}}
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\noindent
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}j24 \\
  \uline{cheatography.com/j24} \\
  \end{tabulary}
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 26th October, 2023.\\
   Page {\thepage} of \pageref{LastPage}.
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
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  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
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\begin{multicols*}{4}

\begin{tabularx}{3.833cm}{x{1.61351 cm} x{1.81949 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Propositions}}  \tn
% Row 0
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Different Ways of Expressing p → q & q unless ¬p, q if p, q whenever p, q follows from p, p only if q, q when p, p is sufficient for q, q is necessary for p. \tn 
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Proposition & True/False, with no variables. Ex) The sky is blue = Prop. n+1 is even Not prop bc n is unknown. \tn 
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Tautology & a proposition which is always true. Ex) p ∨¬ p \tn 
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contradiction & a proposition which is always false. Ex) p ∧¬ p \tn 
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contingency & a proposition which is neither a tautology nor a contradiction, such as p \tn 
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satisfiable & at least one truth table is true. \tn 
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p -\textgreater{} q & Only false when p = T q = F. everthing else true. \tn 
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converse & q -\textgreater{} p \tn 
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inverse & -p -\textgreater{} -q \tn 
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contrapositive & -q -\textgreater{} -p \tn 
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p \textless{}-\textgreater{} q & if and only if. true if and only if p and q have the same truth value ex) p = t q = t or p = f q = f \tn 
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\begin{tabularx}{3.833cm}{x{1.61351 cm} x{1.81949 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Propositions (cont)}}  \tn
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Logically equivalent & p≡q. all truth values have to be the equal aka same results. \tn 
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% Row 12
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Negate Quantifiers & ¬∀xP (x) ≡ ∃x ¬P (x). ¬∃xQ(x) ≡ ∀x ¬Q(x) \tn 
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\hhline{>{\arrayrulecolor{DarkBackground}}--}
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\begin{tabularx}{3.833cm}{x{1.7165 cm} x{1.7165 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Functions}}  \tn
% Row 0
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function from A to B. f: A -\textgreater{} B & an assignment of exactly one element of B to each element of A \tn 
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domain of f & the set A, where f is a function from A to B. ans is A \tn 
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codomain of f & the set B, where f is a function from A to B. ans is B \tn 
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b is the image of a under f & b = f (a). "what does this map to" \tn 
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a is a pre-image of b under f & f (a) = b. "what values map to this". \tn 
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range & values of codomain that were mapped to by domain. \tn 
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Injective Function (one to one) & a function f is one-to-one if and only if f (a) =/= f (b) whenever a =/= b. each value in the range is mapped to exactly one element of domain. (each range value is mapped once). ∀a∀b(f (a) = f (b) → a = b) or equivalently ∀a∀b(a =/= b → f (a) =/= f (b)) \tn 
% Row Count 31 (+ 14)
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\begin{tabularx}{3.833cm}{x{1.7165 cm} x{1.7165 cm} }
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Functions (cont)}}  \tn
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Surjective function (onto) & every element in codomain maps to at least one element in domain. (each element in codomain is mapped). if and only if for every element b ∈ B there is an element a ∈ A with f (a) = b \tn 
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To show that f is injective & f(x1)=f(x2) =\textgreater{} x1=x2. x1=/=x2 =\textgreater{} f(x1)=/=f(x2).  ex) f(a) = f(b) =\textgreater{} a=b. ex) f(x) = x+3. f(a) = 7, a+3=7, a=4. f(b)=7. b=4. f(a)=f(b) a=b, 1to1 \tn 
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To show that f is not injective & Find particular elements x, y ∈ A such that x /= y and f (x) = f (y) \tn 
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% Row 10
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To show that f is surjective & solve in terms of x. pick 2 random ys, if x eqns comes back in domain, surjective. domain matters, Z, R, N has to map x and y in same.  ex) f(x)=x+3. f(4)=7, f(5)=8. always mapped, onto. \tn 
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\begin{tabularx}{3.833cm}{x{1.7165 cm} x{1.7165 cm} }
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Functions (cont)}}  \tn
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To show that f is not surjective & Find a particular y ∈ B such that f (x)  = y for all x ∈ A \tn 
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Bijective function & all range is mapped to and mapped to once (injective and surjective) \tn 
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Inverse & has to be bijective. f\textasciicircum{}-1(y) = x if and only if f(x) = y. because this is both 1to1 and onto, its a bijection, therefore invertible. \tn 
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composition of fns & f(g(a)) or f o g(a) \tn 
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floor/ceiling & bracketwithlow only/highonly. round down/up to nearest integer. ex) -2.2 floor = -3. 5.5 ceil = 6. \tn 
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% Row 16
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properties & x floor = n if and only if n ≤ x \textless{} n + 1. x ciel = n if and only if n − 1 \textless{} x ≤ n. x floor = n if and only if x − 1 \textless{} n ≤ x. x ciel = n if and only if x ≤ n \textless{} x + 1. x − 1 \textless{} floorx ≤ x ≤ cielx \textless{} x + 1. floorx = −cielx. cielx = -floorx. ciel(x + n) = cielx + n. opp of last floor \tn 
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Functions (cont)}}  \tn
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ex) let f(x) = floor((x\textasciicircum{}2)/2). find f(S) if S=\{0,1,2,3\} & f(0) = 0, f(1) = 0. f(2) = 2, f(3) = 4 \tn 
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equal functions & Two functions are equal when they have the same domain, the same codomain and map each element of the domain to the same element of the codomain \tn 
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\begin{tabularx}{3.833cm}{x{1.51052 cm} x{1.92248 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Proofs}}  \tn
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Direct Proof & assume p is true, prove q. p =\textgreater{} q. Always start with this then try contraposition. \tn 
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Proof by contraposition & assume \textasciitilde{}q is true, prove \textasciitilde{}p. (\textasciitilde{}q =\textgreater{} \textasciitilde{}p) equals (p =\textgreater{} q) \tn 
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Vacuous proof & if we can show that p is false, then we have a proof, called a vacuous proof, of the conditional statement p → q \tn 
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Proof by contradiction & Assume \textasciitilde{}p is true, find contradiction, therefore \textasciitilde{}p is true. prove that p is true if we can show that ¬p → (r ∧ ¬r) is true for some proposition r \tn 
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Counterexample & to show that a statement of the form ∀xP (x) is false, we need only find a counterexample. \tn 
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Proof by exhuastion & ex: Prove that (n + 1)3 ≥ 3n if n is a positive integer with n ≤ 4. Prove by doing n = 1,2,3,4 \tn 
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\begin{tabularx}{3.833cm}{x{1.51052 cm} x{1.92248 cm} }
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Proofs (cont)}}  \tn
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Proof by cases & ex: Prove that if n is an integer, then n2 ≥ n. Case (i): When n = 0. Case (ii): When n ≥ 1. Case (iii): In this case n ≤ −1 \tn 
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Constructive Existence Proof & ∃xP(x). To find if P(x) exists, show an example P(c) = True \tn 
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Nonconstructive Existence Proof & Assume no values makes P(x) true. Then contradict. \tn 
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UNIQUENESS proof & When asked for unique, prove exists, then unique. ex: x exists, x=/=y , so y doesnt have that property, therefore x is unique. \tn 
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without loss of generality & an assumption in a proof that makes it possible to prove a theorem by reducing the number of cases to consider in the proof \tn 
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\begin{tabularx}{3.833cm}{x{1.61351 cm} x{1.81949 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Sets}}  \tn
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Element of set & a ∈ A, a ∈/ A \tn 
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roster method & V = \{a, e, i, o, u\}, O = \{1, 3, 5, 7, 9\}. \tn 
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set builder notation & ex: the set O of all odd positive integers less than 10 can be written as: O = \{x ∈ Z+ | x is odd and x \textless{} 10\}. ex) A = \{𝑥|𝑥 ≥ −1 ∧ 𝑥 \textless{} 1\}. ex) \{x | P(x)| \tn 
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Interval Notation & {[}-2,8) \tn 
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Natural numbers N & N = \{0, 1, 2, 3, . . .\} \tn 
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Integers Z & Z = \{. . . , −2, −1, 0, 1, 2, . . .\} \tn 
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Positive Integers Z+ & Z+ = \{1, 2, 3, . . .\} \tn 
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Rational Numbers Q & Q = \{p/q | p ∈ Z, q ∈ Z, and q =/= 0\} \tn 
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Real Numbers R & All previous sets (N, Z, Q) \tn 
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R+ & positive real numbers \tn 
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Complex numbers C & \{a+bi, ...\} \tn 
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Equal Sets & Two sets are equal if and only if they have the same elements. Therefore, if A and B are sets, then A and B are equal if and only if ∀x(x ∈ A ↔ x ∈ B). We write A = B if A and B are equal sets. Dont matter if its \{1,3,3,3,2,2,3,\}, still \{1,3,2\}. Also dont matter order. \tn 
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\begin{tabularx}{3.833cm}{x{1.61351 cm} x{1.81949 cm} }
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Sets (cont)}}  \tn
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Null/ Empty Set & ∅, nothing. \{\}. \tn 
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\{∅\} & 1 element \tn 
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Singleton set & One element. \tn 
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Universal Set U & Universe in context of statement. Example vowels in alphabet: U = \{z,y,x,w, ...\}, A = \{a,e,i,o,u\} A is a subset of U. \tn 
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Subset & ∀x(x ∈ A → x ∈ B). Ex) the set A is a subset of B if and only if every element of A is also an element of B. We use the notation A ⊆ B to indicate that A is a subset of the set B. Ex) A = \{1,2,3\}, B = \{1,2,3,4\}, A ⊆ B. \tn 
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Showing that A is a Subset of B & To show that A ⊆ B, show that if x belongs to A then x also belongs to B. \tn 
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Showing that A is Not a Subset of B & To show that A ⊆/ B, find a single x ∈ A such that x ∈/ B. \tn 
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Showing Two Sets are Equal & To see if A = B, Show A ⊆ B and B ⊆ A \tn 
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Sets (cont)}}  \tn
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proper subset & ∀x(x ∈ A → x ∈ B) ∧ ∃x(x ∈ B ∧ x  ∈ A)A ⊆ B but A =/= B. B contains an element not in A. Ex) A = \{1,2,3\}, B= \{1,2,3,4\}. 4 makes it proper subset. \tn 
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Cardinality & |A| Distinct elements of set. A = \{1,2,3,3,4,4\} |A| = 4 \tn 
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Power Set & the power set of S is the set of all subsets of the set S. The power set of S is denoted by P(S). Ex) A = \{1,2,3\}. P(A) = \{∅, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{0, 2\}, \{1, 2\}, \{0, 1, 2\}\}. Ex) P(\{∅\}) = \{∅, \{∅\}\} \tn 
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Cardinality of Power Set & 2\textasciicircum{}n, n is elements. \tn 
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Tuple & (a1,a2,a3, ..., an) Ordered. Ex) (5,2) =/= (2,5) \tn 
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Cartesian Product & \{(a, b) | a ∈ A ∧ b ∈ B\}. The Cartesian product of A and B, denoted by A × B, is the set of all ordered pairs (a, b), where a ∈ A and b ∈ B. Ex) A = \{0,1\} B = \{2,3,4\}, A x B = \{(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)\} \tn 
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Sets (cont)}}  \tn
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Truth Set & P(x): abs(x) = 3. Truth Set of P(x) = \{3,-3\} \tn 
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Set Operations}}  \tn
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Union & A ∪ B = \{x | x ∈ A ∨ x ∈ B\}. Ex) A = \{1,4,7\} B = \{4,5,6\}. A ∪ B = \{1,4,5,6,7\} \tn 
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Intersection & A ∩ B = \{x | x ∈ A ∧ x ∈ B\}. Ex) A = \{1,4,7\} B = \{4,5,6\}. A ∩ B = \{4\}. \tn 
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disjoint & If A ∩ B = nothing, A and B are disjoint. \tn 
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principle of \seqsplit{inclusion–exclusion} |A ∪ B| = |A| + |B| − |A ∩ B| & ex) A = \{1,2,3,4,5\}, B =\{4,5,6,7,8\}. \}A u B| = |A| + |B| -|A n B\} = 5 + 5 - 2 = 8 \tn 
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A − B, difference of A and B & A - B = A ∩ B. \{x | x ∈ A ∧ x /∈ B\} Elements in A that are not in B. Ex) \{1, 3, 5\} − \{1, 2, 3\} = \{5\}. This is different from the difference of \{1, 2, 3\} and \{1, 3, 5\}, which is the set \{2\}. \tn 
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Complement of A, A\textasciicircum{}c & \{x ∈ U | x /∈ A\} Everything in the universe context thats not in A. Ex) U = \{1,2,3,4\}. A = \{2\} B = \{3\}. A\textasciicircum{}c = \{1,3,4\} \tn 
% Row Count 34 (+ 7)
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\vfill
\columnbreak
\begin{tabularx}{3.833cm}{x{1.7165 cm} x{1.7165 cm} }
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Set Operations (cont)}}  \tn
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U = ℝ A = \{𝑥|𝑥 ≥ −1 ∧ 𝑥 \textless{} 1\}, B =\{𝑥|𝑥 \textless{} 0 ∨ 𝑥 ≥ 2\} & 𝐴 ∪ 𝐵 = \{𝑥|𝑥 \textless{} 1 ∨ 𝑥 ≥ 2\}. 𝐴 ∩ 𝐵 = \{𝑥|𝑥 \textless{} 0 ∧ 𝑥 ≥ −1\}. 𝐴\textasciicircum{}c = \{𝑥|𝑥 \textless{} −1 ∨ 𝑥 ≥ 1\}. \tn 
% Row Count 8 (+ 8)
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Identity, , , , , , , absorbtion, & A ∩ U = A. A ∪ ∅ = A. \tn 
% Row Count 10 (+ 2)
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domination & A ∪ U = U. A ∩ ∅ = ∅ \tn 
% Row Count 12 (+ 2)
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idempotent & A ∪ A = A. A ∩ A = A \tn 
% Row Count 14 (+ 2)
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complementation & (A\textasciicircum{}c)\textasciicircum{}c = A \tn 
% Row Count 15 (+ 1)
% Row 11
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commutative & A ∪ B = B ∪ A. A ∩ B = B ∩ A \tn 
% Row Count 17 (+ 2)
% Row 12
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associative & A ∪ (B ∪ C) = (A ∪ B) ∪ C. A ∩ (B ∩ C) = (A ∩ B) ∩ C \tn 
% Row Count 21 (+ 4)
% Row 13
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distributive & A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) \tn 
% Row Count 26 (+ 5)
% Row 14
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de morgans & (A n b)\textasciicircum{}c = A\textasciicircum{}c u B\textasciicircum{}c. (A U B)\textasciicircum{}c = A\textasciicircum{}c n B\textasciicircum{}c \tn 
% Row Count 29 (+ 3)
% Row 15
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absorption & A ∪ (A ∩ B) = A. A ∩ (A ∪ B) = A. \tn 
% Row Count 32 (+ 3)
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\vfill
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\begin{tabularx}{3.833cm}{x{1.7165 cm} x{1.7165 cm} }
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\mymulticolumn{2}{x{3.833cm}}{\bf\textcolor{white}{Set Operations (cont)}}  \tn
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complement & A ∪ A\textasciicircum{}c = U. A ∩ A\textasciicircum{}c = ∅. \tn 
% Row Count 2 (+ 2)
% Row 17
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countable & a set that either is finite or can be placed in one-to-one correspondence with the set of positive integers. To be countable, there must exist a 1-1 and onto (bijection) between the set and ℕ! (i.e. ℤ+) \tn 
% Row Count 13 (+ 11)
% Row 18
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Ex) Show that the set of positive even integers E is countable set. & Let f(x) = 2x. Then f is a bijection from N to E since f is both one-to-one and onto. To show that it is one-to-one, suppose that f( n) = f( m). Then 2 n = 2 m, and so n = m. To see that it is onto, suppose that t is an even positive integer. Then t = 2k for some positive integer k and f(k) = t \tn 
% Row Count 28 (+ 15)
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\end{tabularx}
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% That's all folks
\end{multicols*}

\end{document}