\[\newcommand{\LetThereBe}[2]{\newcommand{#1}{#2}} \newcommand{\letThereBe}[3]{\newcommand{#1}[#2]{#3}}\]\[\letThereBe{\addTag}{2}{\cssId{#1-#2}{\tag{#2}}} \letThereBe{\refTag}{2}{\href{###1-#2}{(\text{#2})}}\]\[\letThereBe{\eqT}{1}{\addTag{eq}{#1}}\letThereBe{\secT}{1}{\addTag{sec}{#1}}\]\[\letThereBe{\eqRef}{1}{\refTag{eq}{#1}} \letThereBe{\secRef}{1}{\refTag{sec}{#1}}\]\[\LetThereBe{\R}{\mathbb{R}} \LetThereBe{\N}{\mathbb{N}}\]\[\letThereBe{\set}{1}{\left\{ #1 \right\}}\]\[\letThereBe{\rcases}{1}{\left.\begin{align}#1\end{align}\right\}}\]\[\letThereBe{\rcasesAt}{2}{\left.\begin{alignat}{#1}#2\end{alignat}\right\}}\]\[\letThereBe{\lcases}{1}{\begin{cases}#1\end{cases}}\]\[\letThereBe{\lcasesAt}{2}{\left\{\begin{alignat}{#1}#2\end{alignat}\right.}\]\[\LetThereBe{\foo}{\mathrm{foo}}\]

Math test simple

This does not render, \(\foo\)!

And

\[\set{x^2 \mid x \in \R \cap \N}. \eqT{square}\]

After some text we reference it as \(\eqRef{square}\)

\[\begin{align} x &= 1 \\ y &= 2x + 3 \end{align}\]

now

\[\rcases{x &= 1 \\ y &= 2x + 3} and \lcases{x &= 1 \\ y &= 2x + 3}\]

are we here???

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