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Example: We can't quite achieve all points in the discrete box triangle as continued tractions. We miss the rims. But here, for example, is the point {$\left(\frac 37,\frac 15,\frac {13}{35}\right)$}, or, {$\left(15,7, 13\right)$}.  This is in the center, so the first function we will use is {$C$}. If I apply the inverse of {$C$}, {$$C^{-1}(a,b,c)=(b+c-a,a+c-b,a+b-c),$$} we get {$(5,21,9)$}  This is in the upper fourth, so we use {$M$}. Applying the inverse of {$M$}, {$$M^{-1}(a,b,c)=(a,b-a-c,c),$$} we get {$(5,7,9)$}  That's {$C$} again. We keep going,  with the functions: {$L$}, $M$}, {$C$}, {$L$}, {$L$}. This checks: {$$V[CMCLMCLL(1,1,1)]=V[(120,56,104)]=V[(15,7,13)]= \left(\frac 37,\frac 15,\frac {13}{35}\right).$$} |