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Scriveyn · 02-08-2012, 11:52 AM

Hi Zen,

If 0.̅9̅ were not equal to 1, there would have to be a difference d.

For any value d you can find an exponent n, so that 1/10^(n) >= d > 1/10^(n+1)

e.g. d = 0.0002
0.001 >= d > 0.0001

But if that value d were the difference, then it is obvious that

(1 - d) < 1 - 10^(n+1) < 0.̅9̅

e.g. 0.9998 < 0.9999 < 0.̅9̅

So, obviously the difference is less than d=0.0002 or any d you choose.

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02-08-2012, 11:52 AM	#11
Scriveyn amnesic-confabulatory opsimath Join Date: Oct 2006 Location: Between my ears Posts: 739	Hi Zen, If 0.̅9̅ were not equal to 1, there would have to be a difference d. For any value d you can find an exponent n, so that 1/10^(n) >= d > 1/10^(n+1) e.g. d = 0.0002 0.001 >= d > 0.0001 But if that value d were the difference, then it is obvious that (1 - d) < 1 - 10^(n+1) < 0.̅9̅ e.g. 0.9998 < 0.9999 < 0.̅9̅ So, obviously the difference is less than d=0.0002 or any d you choose.

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