All Converge: All continued tractions converge.Proof: Let {$F_0 F_1 F_2 F_3\ \ldots$} be an infinite sequence of tractions. We require a metric for the box triangle. Definition: ''The distance between {$T=\langle a,b,c\rangle$} and {$T'\langle a',b',c'\rangle$} ({$a+b+c=a'+b'+c'=1$}) is the maximum of {$|a-a'|$}, {$|b-b'|$}, {$|c-c'|$}. We write this as {$d(T,T')$}. Definition: ''For any function {$F:\mathbb {N}^3\Rightarrow \mathbb {N}^3$} we define the "squeeze of {$F$}", written {$$\mathcal{S}(F),$$} as the greatest distance {$d(F(T),F(T'))$} over all pairs of points {$T,T'$} in the box triangle. First Squeeze Claim: For any traction {$F$} and any finite composition of tractions, {$G$}, {$$\mathcal S(FG)\le\mathcal S(G)$$} We need to show that {$\mathcal S(F_0 F_1 F_2 F_3\ \ldots\ F_n)\longrightarrow 0$}. This enough to prove convergence. Second Squeeze Claim: For any tractions {$F\neq G$}, {$$\mathcal S(FG)\le\frac23\mathcal S(G).$$} The two claims together show that if more than one traction is represented infinitely often in the infinite sequence, then {$\mathcal S(F_0 F_1 F_2 F_3\ \ldots\ F_n)\longrightarrow 0$}. Also, from the proof of the [First Squeeze Claim]], if {$C$} is represented infinitely often in the infinite sequence, then {$\mathcal S(F_0 F_1 F_2 F_3\ \ldots\ F_n)\longrightarrow 0$}, since {$\mathcal S(CG)\le \frac12\mathcal S(G)$}. One more claim suffices. Third Squeeze Claim: If for some {$m$}, we have {$F_j=F_k$} for all {$j,k>m$}, then {$\mathcal S(F_0 F_1 F_2 F_3\ \ldots\ F_n)\longrightarrow 0$}. |