Hey /sci/ if 0.999... is equal to 1, is 1.000...1 also equal to 1?
>>7749641
Technically yes but this is trivially true.
1.000...1 = 1.000... because if there are infinite zeros then you never get to the last 1 in the first place.
If that's what you make it. Yes.
>>7749641
>1.000...1
Not well defined
let x = .999999...
The 10x = 9.99999...
10x-x = 9.99999... - .99999... = 9
Thus 9x=9 or x=1.
>>7749655
Posting the algebraic proof actually does nothing to help answer OP's question.
popsci /b/tard detected.
In my opinion 0.999...=1 because 0.333...=1/3, 1/3+1/3+1/3=3/3=1
>>7749674
Except 0.333... is an extreme aproximation of 1/3, due impossibility of existing in base10 number system
Try another valid base and all this problem is gone
>>7749684
Also /thread
false - Not equal
>>7749657
let x = 1.000...1
10x = 10.0000....10
10x-x = 10.000...10-1.000...1 = 9.000...9
thus 9x = 9.000...9 or x = 1.000...1
>>7749657
Okay, then. consider the limit of 1+.1^n as n->infinity. This is then 1+limit .1^n as n->infinity. The limit goes to zero and so you get
1=limit 1+.1^n as n->infinity.
>>7749641
The problem is that saying a 1 is at the end implies it's not an infinite series of digits following the decimal and that it actually has a defined end. That's why .999 does not equal 1 but .999... does equal 1.
>>7749689
You end up exactly where you started. Why did you even posted this?
>>7749696
Now the question makes sense and the answer is still yes.
Unlike the other guy who wanted to use basic algebra, I'm going to give you the proof the way it is supposed to be. A proof that also works for the 0.999... = 1 problem.
Assume that 1 and 1+.1^n as n->infinity lie on the real number line.
We know between any two distinct numbers a and b there are infinitely many other real numbers separating them.
Can you find a number between 1 and 1+.1^n as n->infinity? The short is answer is no, you can't. Therefore both expressions are equal to 1 or to generalize this idea, they both lie in the same place in the real number line.
>>7749703
Assume it is an infinitesimal.
>>7749644
nigga ..1 implies there's not infinite zeroes.
so no
1/3 + 1/3 + 1/3 = 3/3 = 1
1/3 = 0.3333...
0.3333... + 0.3333... + 0.3333... = 0.9999....
0.9999.... = 1
>>7749712
Take it in the context of what OP said.
>If this infinitely long number is equal to 1, is this not infinitely long number equal to 1?
No, that doesn't make sense.
But if that is the case then there are infinitely many real numbers between 1.00...1 and 1 and thus it is not equal to 1.
But that is not what OP meant anyways.
>>7749641
The number 1.0000...1 is not meaningful as a real number, because it represents a non-zero infintesimal. It *does* exist over the hyperreals, but then things start getting a lot weirder and a *lot* more complicated.
>>7749707
There is no infinitesimal.
Assume K is the infinitesimal.
If 0.999... +K = 1
Then 0.333... + K/3 = 1/3
Then 0.333... + (0.333... + K/3)K = 1/3
Then 0.333... + (0.333... + (0.333... +K/3))K = 1/3
forever expanding.
>>7749684
.333 is an approximation, 0.333... is exact
>>7749744
I wrote that last one incorrectly, 0.333... +(0.333... +(0.333... + K/3)K)K
>>7749735
The number 0.999... is not meaningful as a real number, because it represents a non-zero infintesimal subtracted from 1. It *does* exist over the hyperreals, but then things start getting a lot weirder and a *lot* more complicated.
>>7749641
The error isn't in our math, but how we represent it, examples like this are why we sometimes have to resort to... fractions...
>>7749762
0.9999... is meaningful as a real number as it is the representation of the convergent geometric series [eqn]\sum^{\infty}_{n=1} 9\cdot 10^{-n}[/eqn], but that series sums to 1. Again, only over the hyperreals does 0.000...1 and friends exist, in the form of the infintesimal ε.
>>7749800
0.999.... = limit of (1 - 10^(-n)) as n goes to infinity
1.00...1 = limit of (1 + 10^(-n)) as n goes to infinity
It's just a symmetry statement.
>>7749808
Assuming that's the case, then we can see the difference between those two numbers:
[eqn]
\lim_{n \to \infty} \left( 1 + 10^{-n}\right) - \left( 1 - 10^{-n} \right
[/eqn]
That expression is equivalent to
[eqn]
\lim_{n \to \infty} \frac{2^{1-n}}{5^n}
[/eqn]
Applying the quotient rule:
[eqn]
2 \lim 10^{-n} = 2\cdot 0 = 0
[/eqn]
Thus, those two expressions are actually equal over the real numbers. Q.E.D., bitches!
>>7749641
No.
1.000...1 does not equal 1, but
1.000...0999... equals 1.000...1
>>7749704
Isn't that just indeterminate?
>>7749704
>Unlike the other guy who wanted to use basic algebra, I'm going to give you the proof the way it is supposed to be.
I'm the same one in both cases.
Using basic algebra is the simplest and clearest way that I've seen it done. I learned that technique in high school. It is very useful when converting a number with repeating digits to a fraction.
For example, convert .92153153153 --- (repeating 153) to a fraction.
let x = .92153153153...
then 100x = 92.153153153...
and 100000x = 92153.153153153...
subtract 100000x - 100x = 92153.153153153... - 92.153153153...
thus 99900x = 92061
divide by 99900,
x=92061/99900
simplifing,
x=10229/11100
Thus, .92153153153... = 10,229/11,100
...9999.0 = 9 + 90 + 900 + 9000 + ... = -1
>>7749641
Two numbers are different if there is a positive distance (difference) between them.
What is |1 - 0.9999...|?
Similarly, what is |1 - 1.000...1|?
Construct the reals as equivalence classes of Cauchy sequences.
The equivalence relation has to do with limits of these sequences.
Since obviously 1.00....1 (just take [math]\epsilon<0.00....1[/math]) does not have a limit equivalent to 1, then they are not in the same equivalence class.
>>7749641
yep.
>>7749921
Stop being a jackass and go back to /b/
1 = .999...
Thus 1 -.999... = 0.
Hence, that equation up there just shows .333... = 1/3.
