JT
Posts:
1,388
Registered:
4/7/12


Re: Can a fraction have none noneending and nonerepeating decimal representation?
Posted:
Aug 2, 2013 3:06 PM


Den fredagen den 2:e augusti 2013 kl. 20:51:01 UTC+2 skrev quasi: > jonas.thornvall wrote: > > > > >But one 1/phi is not a rational ratio, but the 1/6 hexagon > > >(center to vertex/sum of sides) is and so is all polygons > > >derived from multiplying the vertices. > > > > Let n be an integer with n >= 3, and let p be the perimeter > > of a regular ngon inscribed in a circle of radius r. Then > > p/r is rational if and only if n = 6. > > > > In other words, the regular hexagon is the _only_ regular > > ngon for which p/r is rational. > > > > Thus, your claim that the ratio p/r remains rational for regular > > polygons derived from "multiplying the vertices" (which > > presumably just means "increasing the number of vertices") is a > > false claim. > > > > With the exception of the case n = 6, all of those ratios are > > irrational. > > > > But as has been pointed out by others, even if those ratios were > > all rational, that would _not_ automatically imply that the limit > > is rational. > > > > For example, let x be an irrational number between 0 and 1. > > > > Define x_1,x_2,x_3,... by > > > > x_1 = (x rounded to the nearest 1/10) > > x_2 = (x rounded to the nearest 1/10^2) > > x_3 = (x rounded to the nearest 1/10^3) > > ... > > x_k = (x rounded to the nearest 1/10^k) > > ... > > > > The numbers x_1,x_2,x_3,... are clearly rational. > > > > It's also clear that the limit of the sequence x_1,x_2,x_3,... > > exists and equals x. > > > > Thus it's possible for a sequence of rational numbers to have > > a limit which is irrational. > > > > quasi
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