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 $...$
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$.
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} ]
\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*}