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Solution: epsilon proof: lim 0.9999...=1
Posted Jul20-12 at 02:03 PM by
John Creighto
Updated Jul20-12 at 06:19 PM by
John Creighto
In another form I presented a proof (
post #145
)that the lim of 0.9999....=1 using epsilon delta arguments. I was asked to explain each step so I am moving it here for the latex support.
I start with what DanLanglois presents
presents in post #139
Quote:
0.999... + 0.1 = 1 + 0.099...
0.999... + 0.01 = 1 + 0.0099...
0.999... + 0.001 = 1 + 0.00099..
I do not prove this but it follow from basic arithmetic (by the way numbers carry in addition)....
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John Creighto
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