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Topic: Can a fraction have none noneending and nonerepeating decimal representation?
Replies: 108   Last Post: Aug 16, 2013 5:22 PM

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 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 n-gon 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
>
> n-gon 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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