Don't make me REPETEND myself...

[Pete]

@Stuart

Great post! Woohoo, my routine was able to identify .000000000000001184237892933500309785207112630208333333 as an eventual infinite repetend of 3s.

.000000000000001184237892933500309785207112630208333333

So for fun, let's turn that decimal into a fraction...

10^48 * .000000000000001184237892933500309785207112630208333333 = 1184237892933500309785207112630208.333333
10^ 50 * .000000000000001184237892933500309785207112630208333333 = 118423789293350030978520711263020833.3333

118423789293350030978520711263020833.333333
- 1184237892933500309785207112630208.333333
------------------------------------------------------------
117239551400416530668735504150390625

99000000000000000000000000000000000000000000000000 / 117239551400416530668735504150390625 =

1 / 844424930131968

I bring this up because since I'm not onboard with resorting to bit turning math, or whatever it's called, I'm getting curious about converting to fractions to round repetends. As such...

1 / 3 * 3

digital = 1 / 3 = .3... but * 3 = .999... unless the repetend is handled properly in multiplying back to 1.

So with fractions,

1 3
- * - = 1 No problem.
3 1

so if I write an algorithm that does the following...

.3... * 3

10x * .3... = 3.3...
1000x * .3 = 333.3
333.3
- 3.3 = 330
1000x
- 10x = 990x
990x = 330
330 / 990 reduced = 1 /3
1 / 3 * 3 = 3 /3 = 1
So converting to fraction multiplication now 1 / 3 * 3 = 1 instead of .9...

Well, easy in theory, but when we get into really big number repetends, it might be difficult and slow to implement.

Pete