Griego

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El Alfabeto Griego

¡lll! ¡lll! lll
& Nombre & & Nombre & & Nombre
A\(\alpha\) & alfa & I\(\iota\) & iota & P\(\rho\) & rho
B & amp\(\beta\); beta & K &\(\kappa\) & kappa\(\Sigma\) y\(\sigma\) y sigma
\(\Gamma\)\(\gamma\) y gamma y\(\Lambda\)\(\lambda\) y lambda y T\(\tau\) y tau
\(\Delta\)\(\delta\) y delta y M\(\mu\) y mu y \(\Upsilon\)&\(\upsilon\) & upsilon
E &\(\epsilon\) & & épsilon & N\(\nu\) & nu &\(\Phi\) &\(\phi\) & & phi
Z\(\zeta\) & & zeta\(\Xi\) &\(\xi\) & & xi & X\(\chi\) & chi
H & eta\(\eta\) & O & omicron\(o\) & & &\(\Psi\) & psi\(\psi\)
\(\Theta\) & & theta\(\theta\) & & & &\(\Pi\)\(\pi\) & pi & &\(\Omega\) &\(\omega\) & omega

Notación matemática

@c! ¡l! l@
Símbolo y significado y ejemplo
\(\Rightarrow\) & si... entonces; implica\(\abs{x} > 1 ~\Rightarrow~ x^2 > 1\)
\(\Leftrightarrow\) & & si y solo si; implicación bidireccional &\(\abs{x} > 1 ~\Leftrightarrow~ x^2 > 1\)
iff & si y solo si; bidireccional implicación &\(\abs{x} > 1\) iff\(x^2 > 1\)
\(\nRightarrow\) & no implica &\(\abs{x} > 1 ~\nRightarrow~ x > 1\)
\(\exists\) & existe &\(\exists\) un número\(c > 0\)
\(\nexists\) y no existe &\(\nexists ~x\) tal que\(x^2 < 0\)
\(\exists !\)& existe un único &\(\exists !~x\) tal que\(2x-1=3\)
\(\forall\) & para cada &\(\forall x \ge 0\),\(\sqrt{x}\) es un número real
\(\equiv\) & es idénticamente igual a &\(f \equiv 0 ~\Rightarrow~ f(x)=0\) para todos\(x\)
\(\propto\)& es proporcional a &\(y ~\propto x^2 ~\Rightarrow~ y=kx^2\) para algunos\(k\)
\(\subseteq\) & es un subconjunto de &\(\lbrace 0,1 \rbrace \subseteq \lbrace 0,1,2 \rbrace\)
\(\in\) & es un elemento de &\(1 \in \lbrace 1,2,3 \rbrace\)
\(\notin\) & no es un elemento de &\(1 \notin \lbrace 2,3 \rbrace\)
\(\cup\)& unión de conjuntos &\(\lbrace 0,1 \rbrace \cup \lbrace 2,3 \rbrace = \lbrace 0,1,2,3 \rbrace\)
\(\cap\) & intersección de conjuntos & &\(\lbrace 0,1 \rbrace \cap \lbrace 1,2 \rbrace = \lbrace 1 \rbrace\)
\(\varnothing\) & conjunto vacío &\(\lbrace 0,1 \rbrace \cap \lbrace 2,3 \rbrace = \varnothing\)
\(\therefore\) & por lo tanto &\(\therefore\)\(n\) debe existir

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