(redirected from Boxes.AGeneralizedContinuedFraction) Continued TractionsFirst, continued fractions: Definition We define {$L,R:\mathbb {N}^2\Rightarrow \mathbb {N}^2$} by: {$$L(m,n)=(m+n,n)$$} {$$R(m,n)=(m,m+n).$$} Think of ordered pairs as representing rational numbers: {$$(m,n)\equiv \frac mn$$} Every rational is equivalent to a finite composition of {$L$} and {$R$} applied to {$(0,1)$}. Definition The infinite continued fraction, {$$L^2R^3L^1R^3L^1 \ldots $$} is defined as the limit: {$$L^2(0,1),\ L^2R^3(0,1),\ L^2R^3L^1(0,1),\ L^2R^3L^1R^3(0,1),\ \ L^2R^3L^1R^3L^1(0,1),\ \ldots$$} This is just equivalent to the usual expression. Now, continued "tractions": Definition We define {$L,R,M,C:\mathbb {N}^3\Rightarrow \mathbb {N}^3$} by: {$$L(a,b,c)=(a+b+c,b,c)$$} {$$R(a,b,c)=(a,b,a+b+c)$$} {$$M(a,b,c)=(a,a+b+c,c)$$} {$$C(a,b,c)=(b+c,a+c,a+b)$$} Each of {$L,R,M,C$} can be interpreted as a map from the box triangle to itself. We call these maps, tractions and compositions of them, finite and infinite, continued tractions. All Converge: All infinite continued tractions converge. All points are tractable: All points in the box triangle are equal to some continued traction. |