Solve N(U_{f} + \frac{0}{1}) = N(\frac{1}{0}) = unknot, N(U_{f} + \frac{t}{w}) = N(\frac{0}{1}) = unlink of two components. Method 1: 1.) Since N(\frac{0}{1}) is a (2, p) torus link, N(\frac{1}{0}) = unknot, and E is rational, U_{f} is rational. Suppose U_{f} = \frac{a}{b}. Then N(U_{f} + \frac{0}{1}) = N(\frac{a}{b} + \frac{0}{1}) = N(\frac{a}{b}) = N(\frac{1}{0}). Hence a = 1 since we can take a to be nonnegative (\frac{a}{b} = \frac{−a}{−b}). Thus U_{f} = \frac{1}{b} . N(U_{f} + \frac{t}{w}) = N(\frac{1}{b} + \frac{t}{w}) = N(\frac{w+tb}{t}) = N(\frac{0}{1}) Hence w+tb = 0. Thus w = −tb. Since gcd(w, t) = 1 and we can take t to be nonnegative, t = 1. Hence w = −b. Thus the solutions to the above system of tangle equations is U_{f} = \frac{1}{b} and \frac{t}{w} = −\frac{1}{b} where b is an arbitrary integer. Method 3: Use KnotPlot Method 2: Solve N(U_{f} + \frac{0}{1}) = N(\frac{1}{0}) = N(\frac{a}{b}) = unknot, N(U_{f} + \frac{t}{w}) = N(\frac{0}{1}) = N(\frac{z}{v}) = 2 component unlink. a = 1, b = 0, x = 0, y = −1, z = 0, v = 1, and v′ is any integer such that v′v^{±1} = 1mod z By corollary 2 if w ≅ ±̸1 mod t or if U_{f} is rational, then \frac{t}{w} = \frac{xz−av′}{bv′−yz−kt} = \frac{v′}{−kt} = and U = \frac{a}{b+ka} = \frac{1}{k} Or \frac{t}{w} = \frac{bz−av′}{xv′−yz−kt} = \frac{v′}{−kt} = and U = \frac{a}{x+ka} = \frac{1}{k} By Thm 3, if w ≅ ±1 mod t, N(\frac{z}{v}) = N(\frac{tp(pb−qa)±a}{tq(pb−qa)±b}) Hence tp(pb − qa) ± a = z or −z. Thus tp(pb − qa) = z ∓ a or −z ∓ a. Hence tp|(z ∓ a). IF t ≠ ±1, then U = (\frac{da−jb}{pb−qa} + \frac{j}{p}) ∘ (h, 0) or (\frac{j}{p} + \frac{da−jb}{pb−qa})∘(h, 0). Hence if |z∓a| is prime, U is rational. If t = ±1, then U is rational by Hirasawa and Shimokawa.

Solve N(U′_{f} + \frac{f_{1}}{g_{1}} ) = N(\frac{1}{0}) = N(\frac{a}{b}) = unknot, N(U′_{f} + \frac{f_{2}}{g_{2}} ) = N(\frac{0}{1}) = N(\frac{z}{v}) = 2 component unlink.

given that U_{f} = −\frac{1}{k} and \frac{t}{w} = \frac{1}{k} are the solutions to N(U_{f} + \frac{0}{1}) = N(\frac{1}{0}) = N(\frac{a}{b}) = unknot, N(U_{f} + \frac{t}{w}) = N(\frac{0}{1}) = N(\frac{z}{v}) = 2 component unlink.

Suppose \frac{f_{1}}{g_{1}} = (c_{1}, ..., c_{n}), f_{1} = E[c_{1}, ..., c_{n}], g_{1} = E[c_{1}, ..., c_{n−1}], e_{1} = E[c_{2}, ..., c_{n}], i_{1} = E[c_{2}, ..., c_{n−1}].

Then \frac{f_{2}}{g_{2}} = \frac{te_{1}+wf_{1}}{ti_{1}+wg_{1}} = (b_{1}, ..., b_{k} + c_{1}, ..., c_{n}) where \frac{t}{w} = (b_{1}, ..., b_{k}).

Since \frac{t}{w} = \frac{1}{k} = (k, 0), \frac{f_{2}}{g_{2}} = \frac{e_{1}+kf_{1}}{i_{1}+kg_{1}} = (k, 0 + c_{1}, ..., c_{n}) = (k, c_{1}, ..., c_{n}).

U′_{f} = U_{f} ∘ (−c_{1}, ..., −c_{n}) = (−k, 0) ∘ (−c_{1}, ..., −c_{n}) = (−k, 0 + −c_{1}, ..., −c_{n}) = (−k, −c_{1}, ..., −c_{n})

If U_{f} = \frac{a}{b′}, then U′_{f} = U_{f}∘(−c_{1}, ..., −c_{n}) = \frac{−f_{1}b′+e_{1}a}{g_{1}b′−i_{1}a}.

Since U_{f} = \frac{1}{−k}, U′_{f} = \frac{f_{1}k+e_{1}}{g_{1}k−i_{1}}

Note: (1) = (0, 1, 1), but (3) = (2) ∘ (1) ≠ (2) ∘ (0, 1, 1) = (2, 1, 1)