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

\[\begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*}\]
\[ \begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*} \]
\[ \begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*}\]
\[\begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*} \]
\[\begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*}\]
\[\begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*}\]
\[\begin{align*} x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*}\]
\[\begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l \end{align*}\]
\[\begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*}\]
\[ \begin{align*}x + y & = z \in \left\{ -1, 1 \right\} \\ g + h & = l\end{align*} \]