Equivalence of rational knots

Question

I am trying to understand theorem 1.2 from http://homepages.math.uic.edu/~kauffman/VegasAMS.pdf:

Theorem 1.2. Suppose that rational tangles with fractions $\frac{p}{q}$ and $\frac{p'}{q'}$ are given ($p$ and $q$ relatively prime. Similarly for $p'$ and $q'$.) If $K(\frac{p}{q})$ and $K(\frac{p'}{q'})$ denote the corresponding rational knots obtained by taking numerator closures of these tangles, then $K(\frac{p}{q})$ and $K(\frac{p'}{q'})$ are topologically equivalent if and only if

1. $p = p'$ and
2. either $q \equiv q'\ (\text{mod } p)$ or $qq' \equiv 1 \ (\text{mod } p)$.

According to the theorem, the knots $K(\frac{3}{4})$ and $K(\frac{3}{5})$ should not be equivalent. However, after using SnapPy (https://www.math.uic.edu/t3m/SnapPy/spherogram.html), I obtain that both knots are the trefoil:

In: RationalTangle(3,4).numerator_closure().exterior().identify()
Out: [3_1(0,0), K3a1(0,0)]

In: RationalTangle(3,5).numerator_closure().exterior().identify()
Out: [3_1(0,0), K3a1(0,0)]

I also get the same result after drawing the knots by hand using the conventions from the above paper.

If I use the denominator closure in SnapPy instead, then the knots from the fractions $\frac{3}{4}$ and $\frac{3}{5}$ are different. However, according to the theorem, the closures of the fractions $\frac{5}{3}$ and $\frac{5}{2}$ should be equivalent. Using SnapPy, the denominator closures of $\frac{5}{3}$ and $\frac{5}{2}$ are different, but the numerator closures of $\frac{5}{3}$ and $\frac{5}{2}$ are equivalent (the figure-eight knot).

Am I doing anything wrong?

Answer 1

I did the calculation of the two knots by hand myself following the conventions of the paper, and I get that they are both trefoil knots. However, $K(3/4)$ is a right-handed trefoil knot, and $K(3/5)$ is a left-handed trefoil knot. Max Dehn showed these were not equivalent in 1914. I assume Snappy does not keep track of the orientation of the exterior.