## A very light-weight question

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• A light moment for the team, please I know I am wrong but I am just having fun.   OK, observe that:   a) 0.111... = 1/9 = 1/10 + 1/100 + 1/1000 + ...
Message 1 of 3 , Jan 15, 2009
A light moment for the team, please I know I am wrong but I am just having fun.

OK, observe that:

a) 0.111... = 1/9 = 1/10 + 1/100 + 1/1000 + ...
0.222... = 2/9 = 2/10 + 2/100 + 2/1000 + ...
0.333... = 3/9 = 3/10 + 3/100 + 3/1000 + ...
0.444... = 4/9 = 4/10 + 4/100 + 4/1000 + ...
0.555... = 5/9 = 5/10 + 5/100 + 5/1000 + ...
0.666... = 6/9 = 6/10 + 6/100 + 6/1000 + ...
0.777... = 7/9 = 7/10 + 7/100 + 7/1000 + ...
0.888... = 7/9 = 7/10 + 7/100 + 7/1000 + ...
Extrapolating from the above, would it be true for me to say:
0.999... = 9/9 = 9/10 + 9/100 + 9/1000 + ... = 1
If I am wrong, then what is the rational fractions of:
0.999999999.....

Thanks.

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• ...   Yes. The decimal representation of terminating rationals are not unique. This has to be adjusted for when doing things like using Cantor s diagonal
Message 2 of 3 , Jan 15, 2009
> Extrapolating from the above, would it be true for me to say:
>   0.999... = 9/9 = 9/10 + 9/100 + 9/1000 + ... = 1 If I am wrong, then what is the rational fractions of:
>   0.999999999.....

Yes. The decimal representation of "terminating" rationals are not unique. This has to be adjusted for when doing things like using Cantor's diagonal argument to prove the reals are not countable.

So indeed 1.99999999999... is an even prime.

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