Math Rendering Test Case
Inline Math
This is Inline Math using \( ... \)
\( x + y = z \in \left\{ -1, 1 \right\} \) and now using $ ... $
\( x + y = z \in \left\{ -1, 1 \right\} \) and it should handle the case of giving my friend 2$.
We can also do the same without gaps.
This is Inline Math using \(...\)
\(x + y = z \in \left\{ -1, 1 \right\}\) and now using $...$
\(x + y = z \in \left\{ -1, 1 \right\}\) and it should handle the case of giving my friend 2$.
It can also take care of cases of one sided gaps.
This is Inline Math using \( ...\)
\( x + y = z \in \left\{ -1, 1 \right\}\) and now using $ ...$
\( x + y = z \in \left\{ -1, 1 \right\}\) and it should handle the case of giving my friend 2$.
This is Inline Math using \(... \)
\(x + y = z \in \left\{ -1, 1 \right\} \) and now using $... $
\(x + y = z \in \left\{ -1, 1 \right\} \) and it should handle the case of giving my friend 2$.
Here is a dollar test. My friend has $2.00 and I have $10.00. \( 2 + 10 = 13 \).
Display Math
Well, this should be less challenging:
\[ x + y = z \in \left\{ -1, 1 \right\} \]
\[x + y = z \in \left\{ -1, 1 \right\}\]
\[ x + y = z \in \left\{ -1, 1 \right\}\]
\[x + y = z \in \left\{ -1, 1 \right\} \]
\[ x + y = z \in \left\{ -1, 1 \right\} \]
\[x + y = z \in \left\{ -1, 1 \right\}\]
\[ x + y = z \in \left\{ -1, 1 \right\}\]
\[x + y = z \in \left\{ -1, 1 \right\} \]
Environments
\[\begin{align*}
x + y & = z \in \left\{ -1, 1 \right\} \\
g + h & = l
\end{align*}\]
\[ \begin{align*}
x + y & = z \in \left\{ -1, 1 \right\} \\
g + h & = l
\end{align*} \]
\[ \begin{align*}
x + y & = z \in \left\{ -1, 1 \right\} \\
g + h & = l
\end{align*}\]
\[\begin{align*}
x + y & = z \in \left\{ -1, 1 \right\} \\
g + h & = l
\end{align*} \]
\[\begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*}\]
\[\begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*}\]
\[\begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*}\]
\[\begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*}\]
\[\begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*}\]
\[ \begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*} \]