Dual bundle of a stable vector bundle. Given a vector bundle $E \to X$ over a complex curve $X$, let its rank and degree be $r$ and $d$ respectively. Then the slope of $E$ is defined to be 
$
\mu(E):= \frac{d}{r}.
$
$E$ is defined to be stable if $\mu(U) < \mu(E)$ for every non-trivial proper subbundle $U$ of $E$.
I'm trying to decide whether the following is true:
$E$ is stable if and only if its dual bundle $E^*$ is stable.
If would like the following to hold: $\mu(U) < E \iff \mu(E/U) > \mu(E)$ because then given a subbundle $U$ I can dualize the ses
$0 \to U \to E \to E/U \to 0$
to give ses
$0 \to (E/U)^* \to E^* \to U^* \to 0$, 
from which it follows that 
$E^*$ stable $\implies \mu((E/U)^*) < \mu(E^*) \implies \mu((E/U))>\mu(E) \implies \mu(U) < \mu(E).$
However, this does not seem like it would hold in general, but I can't think of a counter example (i.e stable $E$ with $E^*$ not stable). 
Thanks.
 A: This follows from the fact that both degree and rank are additive with respect to exact sequences. More explicitly, for any short exact sequence
$$
0 \to E \to F \to G \to 0
$$
one has $\deg F = \deg E + \deg G$ and $\operatorname{rk} F = \operatorname{rk}E + \operatorname{rk} G$. Thus everything can be deduced to the following fact about rational numbers: $\frac{d_1}{r_1}<\frac{d_1+d_2}{r_1+r_2}$ if and only if $\frac{d_1+d_2}{r_1+r_2}<\frac{d_2}{r_2}$.
Indeed let $d_1$ (resp. $d_2$ be the degree of $U$ (resp.  $E/U$) and $r_1$ (resp. $r_2$) its rank.
By definition $\mu(U) < \mu (E)$ means $\frac{d_1}{r_1} < \frac{d_1+d_2}{r_1+r_2}$ which can be re-written as $$
d_1r_1+d_1r_2<r_1d_1+r_1d_2
$$
canceling $d_1r_1$ we get $d_1r_2<r_1d_2$.
By the same considerations the condition $\mu(E)<\mu(E/U)$ is equivalent to $d_1r_2+d_2r_2<r_1d_2+r_2d_2$ which leads to the same $d_1r_2<r_1d_2$.
A: This follows from the fact that stability can also be checked with respect to quotients of $E$, that is $E$ is stable if and only if for every quotient bundle $E\rightarrow Q$ we have $\mu(Q)>\mu(E)$.
