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One of my colleagues (Silver) gave a general purpose talk last week. You can't untie a knot A $A$ by tying another knot B $B$ into the rope.

Proof. First show that the connect-sum of long knots is abelian by sliding A $A$ through B. $B$. Then form the decreasing-in-size connect-sum (AB)(AB)(AB) ... . $(AB)(AB)(AB) \dots$. Regroup: A(BA)(BA) ... . $A(BA)(BA) \dots$. If $AB = 0 --- 0$, i.e. the unknot, then the left-hand grouping is the unknot. The right-hand-grouping is A $A$ since BA=AB=0. $BA=AB=0$. Since A $A$ is a knot, then A=0 -- $A=0$, a contradiction. QED

The nice thing about the argument is that it uses an infinitely long piece of rope, an infinite sequence of knots, and the knots shrink to being infinitesimally small.

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One of my colleagues (Silver) gave a general purpose talk last week. You can't untie a knot A by tying another knot B into the rope.

Proof. First show that the connect-sum of long knots is abelian by sliding A through B. Then form the decreasing-in-size connect-sum (AB)(AB)(AB) ... . Regroup: A(BA)(BA) ... . If AB =0 --- the unknot, then the left-hand grouping is the unknot. The right-hand-grouping is A since BA=AB=0. Since A is a knot, then A=0 -- a contradiction. QED

The nice thing about the argument is that it uses an infinitely long piece of rope, an infinite sequence of knots, and the knots shrink to being infinitesimally small.