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GeneralRe: 1 = 0 Pin
Richard Deeming2-Dec-15 2:34
mveRichard Deeming2-Dec-15 2:34 
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HobbyProggy2-Dec-15 2:39
professionalHobbyProggy2-Dec-15 2:39 
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Richard Deeming2-Dec-15 2:48
mveRichard Deeming2-Dec-15 2:48 
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HobbyProggy2-Dec-15 2:51
professionalHobbyProggy2-Dec-15 2:51 
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Nelek2-Dec-15 3:40
protectorNelek2-Dec-15 3:40 
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kiLLe_5123-Dec-15 0:17
MemberkiLLe_5123-Dec-15 0:17 
GeneralRe: 1 = 0 Pin
Nelek2-Dec-15 3:37
protectorNelek2-Dec-15 3:37 
GeneralRe: 1 = 0 Pin
yiangos20-Mar-16 12:57
professionalyiangos20-Mar-16 12:57 
Well...

0.999... is

0+(9/10) + (9/100) + (9/1000) +...

which is 0+9*((1/10)+(1/100)+(1/1000)+...)

which is 9* (sum(1/10^n))

where the sum runs over n for n=1 to infinity.

Now for the same sum, if n ran from 0 to infinity, there is a convenient formula, that says it is (I'll mark this with a capital S to distinguish it from the one we're trying to calculate):

Sum(1/10^n)=1+1/10+1/100+...=1/(1-(1/10))=1/(9/10)=10/9

therefore the sum we want is

1/10+1/100+...=(10/9)-1=1/9

therefore

0.999...=9*(1/9)=1

So it's math.

On the other hand, computer calculated numbers are approximations, and in the binary system at that, and this is why you get all those rounding errors.

But I get the feeling you already know all this, and that you are just toying with us obsessive compulsive types.
Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων!
(Alas! We're devoured by lamb-guised wolves!)

GeneralRe: 1 = 0 Pin
Nagy Vilmos1-Dec-15 23:44
professionalNagy Vilmos1-Dec-15 23:44 
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HobbyProggy1-Dec-15 23:47
professionalHobbyProggy1-Dec-15 23:47 
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Florgenator3-Dec-15 0:31
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Member 121837503-Dec-15 2:39
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willichan3-Dec-15 5:46
professionalwillichan3-Dec-15 5:46 
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yiangos20-Mar-16 13:02
professionalyiangos20-Mar-16 13:02 
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willichan31-Mar-16 12:27
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yiangos6-Apr-16 10:17
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jschell3-Dec-15 12:47
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obermd3-Dec-15 17:11
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maze33-Dec-15 22:49
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James Curran3-Dec-15 5:51
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CPallini1-Dec-15 23:07
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Rage1-Dec-15 23:28
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Marc Clifton2-Dec-15 1:25
mvaMarc Clifton2-Dec-15 1:25 
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Corporal Agarn2-Dec-15 3:44
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