Can the degree of dissociation of a weak acid be more than 1?
The $\sqrt K/\sqrt C$ ratio is only an approximation.  Let's do something more exact.  (Why do I say &quot;more exact&quot; instead of &quot;completely exact&quot;?  Read on, nature has a nasty curveball in store for us!)
Let $C$ be the nominal concentration of the acid, assumed monoprotic ($\ce{HA}$) for simplicity, and $K$ be its dissociation constant.  So, recognizing all ions are of course solvated, we have
$[\ce{H^+}][\ce{A^-}]=K[\ce{HA}]$
$[\ce{HA}]=C-[\ce{H^+}]$
$[\ce{A^-}]=[\ce{H^+}]$
So
$[\ce{H^+}]^2=K(C-[\ce{H^+}])$
$[\ce{H^+}]^2+K[\ce{H^+}]-CK=0$
From the quadratic formula
$[\ce{H^+}]=\dfrac{-K+\sqrt{K^2+4CK}}{2}=\dfrac{2CK}{K+\sqrt{K^2+4CK}}$
And our dissociation is
$\dfrac{[\ce{H^+}]}{C}=\dfrac{2K}{K+\sqrt{K^2+4CK}}=\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}$
Now, if $C$ is large compared with $K$ then the terms in the denominator with $K$ become small compared with $\sqrt C$ and we simplify to $\sqrt K/\sqrt C$.  But if we dilute the acid, then the terms with $K$ in the denominator become dominant and we find a  limiting value of
$\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{C+K/4}}&lt;\dfrac{\sqrt K}{\sqrt{K/4}+\sqrt{K/4}}=1$.
Thereby the dissociation remains limited to no greater than 100%.
There is one little thing we forgot.  If we keep diluting the acid so its concentration approaches the ion concentrations from water autoionization, then we also have to count the hydrogen ions from the autoionization into the equilibrium.  This extra source of hydrogen ions cuts down on the acid dissociation, so the limiting amount of dissociation is less than 100%.  For acetic acid it's still about 99.4%, but if we pick a really weak acid like $\ce{HCN}$ with a $pK_a$ above $7$ at ambient temperature, we never get more than a small fraction to dissociate unless we force the issue by reacting with a base.