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Re: 0.9 repeating = 1?
Just for curiosity, what would happen if you approached the problem backwards, starting at 1, then subtracting:
1 - .1 = .9
1 - .01 = .99
1 -.001 = .999
...
1 - 10^-n = .9~ as n approaches infinity
You could argue the 10^(-infinity) approaches 0 and thus 1 = .9~
But then again, the equivalent expression, 1/10^n divides the number 1 into smaller and smaller parts. Just because the parts are smaller doesn't mean they are nonexistent. To me, this expression states that you can always divide 1 into smaller parts, therefore you can never really "reach" infinity, and therefore never "reach" .9~
It's just like 1/x . What happens as x --> infinity? "The limit is 0." Yes, but does the function itself every EQUAL 0? It gets closer to it, and closer, and closer, .01 .001 .00001 .00000000000000000000001 But there's always that 1 at the end, no matter how many zeros you throw in there. And you can't really put in "an infinite" amount of zeros in between, because then the 1 would be coming after "infinity" which nulls its definition.
I may be wrong, but at least presents an argument different from the standard 1/3 stuff.
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