#### Ben Gooding

**Registered Member**

(a) 0.9999... is not 1

(b) 0.3333... is not 1/3

however, the limit of (a) and (b) are 1 respectively 1/3

Proof.

If you are familiar with proofs of induction, the follwing will be easy to grasp.

Hence, 0.9999... is not one.

Similar proof with 9/10 + 9/100 + 9/1000 + ... (same as 0.9999...)

and with 3/10 + 3/100 + 3/1000 + ... (same as 0.3333...)

**Step 1.**0.9 is not one, it differs from one by 0.1; "0."+["9"*n] where n=1 thus differs from one by 1 - 0.9 = one unit of value at 1 digit position after the decimal point**Step 2.**If "0."+["9"*n] differs from one by 1 unit of value of position n, then also "0."+["9"*(n+1)] differs from one (by one unit of value at position n+1), since, 1 unit of value at position n correspond to 10 units of value at position n+1 and, 10 units of value at position n+1 is required to get to one (the difference between 1 and "0."+["9"*n]), and only 9 units are added at the n+1 position, thus 10-9=1 units difference from one at n+1 digit position after the decimal point (which is the difference between 1 and "0."+["9"*(n+1)] )**Step 3.**Therefore, since Step 1 (n=1) is true and Step 2 is true, it is true for all n=1,2,3,... as n goes to infinityHence, 0.9999... is not one.

Similar proof with 9/10 + 9/100 + 9/1000 + ... (same as 0.9999...)

and with 3/10 + 3/100 + 3/1000 + ... (same as 0.3333...)

What does 0.9999... mean?

0.9999... means an infinitely many 9's are followed after zero dot.

**It does not introduce a meaning of implied limit.**If one wants to imply a limit, to that number, then one has to explicitly state that, eg,

lim (0.9999...)

Now that limit is one.

This does not mean that 0.9999... is one, just because the limit is one. The above proof is not limited to a finite n, n can be infinity, and the proof is still valid.

Thus 0.9999... is not one, since there is no implied limit.