Take a <Hopf link> with its two <unknot> components labeled by <integer surgery coefficients> $2,3$.

\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-141-surgery.png]
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The <slam-dunk move> eliminates a meridional component with coefficient $b$ and replaces the other coefficient $a$ by $a-1/b$. Eliminating the component with coefficient three gives
$$
\boxed{[2,3]^-:=2-\frac13=\frac53.}
$$
Thus this <Dehn surgery on a framed link> becomes $5/3$ <Dehn filling> on the <unknot>, with the required orientation. As a check, its <surgery linking matrix> is
$$
\begin{pmatrix}2&1\\1&3\end{pmatrix},
$$
whose <determinant> is five.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 141 / 1 / d / Solution

Let $H_+$ be the positively oriented <Hopf link> obtained by closing a two-strand braid with two positive crossings. Switching one crossing gives the two-component <unlink>; the <oriented smoothing> gives the <unknot>. The printed <skein relation> gives
$$
qV(U_2)-q^{-1}V(H_+)=q^{1/2}-q^{-1/2},
\qquad V(H_+)=q^{5/2}+q^{1/2}.
$$
Represent the right-handed <trefoil knot> by the closure of the two-strand braid with three positive crossings. Switching one crossing gives the <unknot>, while its <oriented smoothing> gives $H_+$. Consequently
$$
q-q^{-1}V(T_+)=(q^{1/2}-q^{-1/2})(q^{5/2}+q^{1/2}),
$$
so the two <Jones polynomials> are
$$
\boxed{V(T_+)=q+q^3-q^4,\qquad V(T_-)=q^{-1}+q^{-3}-q^{-4}.}
$$
The second expression follows from the <mirror of a link> identity. They are unequal <Laurent polynomials>, so invariance under <link isotopy> proves that \b[the right-handed and left-handed trefoils are not isotopic].

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 141 / 2 / d / Solution

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The Hopf link consists of two <unknots> with <linking number> $\pm1$. Each component is a <meridian of a knot> of the other.

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 Incoming links: Hopf link