Typesetting Multiline Inequalities If we write
$$\begin{align}
a&<b\\
&=c\end{align}$$
does this mean
$$a<b=c$$
or
$$a<b; a=c$$
?
How do we write the other in multiline format?
(Edited to Add)
What about
$$\begin{align}
a&<b\\
&=c\\
&=d\\
&=f\end{align}$$
?
 A: It means $a<b=c$. Basically, this notation is due to lack of space, so terms are separated into multi lines.
A: Generally,
\begin{align*}
a &< b \\
&= c
\end{align*}
is to be interpreted as $a < b = c$. This can be convenient for a series of derivations or inequalities, when putting them on the same line could be ugly or hard to follow. To pull an example from a homework of mine:
\begin{align*}
f(z)
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \frac{1}{(z+ia)^2}  \\
&= \frac{1}{(z-ia)^2}  \frac{1}{(2ia)^2} \ \sum_{k=0}^\infty \binom{2+k-1}{k} \left( - \frac{z-ia}{2ia} \right)^k  \\
&= \frac{1}{(z-ia)^2} \frac{1}{(2ia)^2} \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^k} (z-ia)^k   \\
&=  \sum_{k=0}^\infty \frac{(-1)^k (k+1)}{(2ia)^{k+2}} (z-ia)^{k-2} \\
&=  \sum_{n=-2}^\infty \frac{(-1)^n (n+3)}{(2ia)^{n+4}} (z-ia)^{n} 
\end{align*}
(Usually I accompany these with minor notes on the right side, or number the equal signs and then explain the equalities afterwards, it simply depends. The former looks ugly on MSE most of the time so I left it out.)

If you want to write $a < b; a = c$, then you could just use
\begin{align*}
a &< b \\
a &= c
\end{align*} 

which yields
\begin{align*}
a &< b \\
a &= c
\end{align*}
Perhaps throw a curly brace on the left-hand side to indicate they should considered together or a group?
