/sci/ answer this

[Anonymous No.4594629]

Hey /sci/, my friend is a student in the University of Zurich and he asked me this simple yet complicated question. Prove that 0.9999999999(repeating)=1

[Anonymous No.4594630]

>google.com

[Anonymous No.4594643]

1/3 = 0.333333333...
0.333333333... x 3 = 0.9999999999...
that's all I can think of

[sage No.4594666]

sage

[Anonymous No.4594671]

.99999... = lim (n=1->infinity) of (9/10) + (9/100) + ... + (9/10^n) = lim (etc) of 1 - (1/10^n) = 1

[Anonymous No.4594680]

>>4594666
I don't like that sage, it didn't even use limits or sum. You should at least make them work for the answer.

[Anonymous No.4594703]

The simple fact that there is a decimal before the first nine means that it will never be the same value as 1

[Anonymous No.4594710]

there is no number between .999..... and 1. therefore they are the same number.

[Anonymous No.4594965]

>>4594710
This is the best and it uses a fundamental property of the real number system instead of some bullshit assumption like 0.333... = 1/3. (Like someone will think that .9999... != 1, but .3333... = 1/3)
In fact I believe it would be possible (if my understanding of surreal numbers is correct, which I doubt it is because I just read the wikipedia article yesterday) to construct a surreal number that is less than one, but greater than all numbers with a zero in front of the decimal. Namely,
{ X | 1}
where <div class="math"> X = {x, x = \sum_{k = 1}^n \frac{9}{10^k} for some positive integer n}
http://en.wikipedia.org/wiki/Surreal_number
Pic related. The man's a muthafucking crazy genius!</div>

[Anonymous No.4594978]

This is the best proof that 0.9999... = 1 because it uses a fundamental property of the real number system instead of some bullshit assumption like 0.333... = 1/3. (Like someone will think that .9999... != 1, but .3333... = 1/3)
In fact I believe it would be possible (if my understanding of surreal numbers is correct, which I doubt it is because I just read the wikipedia article yesterday) to construct a surreal number that is less than one, but greater than all numbers with a zero in front of the decimal. Namely,
{ X | 1}
where
<div class="math"> X = \left{x, x = \sum_{k = 1}^n \frac{9}{10^k} \right} </div> for some positive integer n
http://en.wikipedia.org/wiki/Surreal_number
Pic related. The man's a muthafucking crazy genius!

[Anonymous No.4594998]

This is the best proof that 0.9999... = 1 because it uses a fundamental property of the real number system instead of some bullshit assumption like 0.333... = 1/3. (Like someone will think that .9999... != 1, but .3333... = 1/3)
In fact I believe it would be possible (if my understanding of surreal numbers is correct, which I doubt it is because I just read the wikipedia article yesterday) to construct a surreal number that is less than one, but greater than all numbers with a zero in front of the decimal. Namely,
{ X | 1}
where
X = {x, <span class="math"> x = \sum_{k = 1}^n \frac{9}{10^k} [/spoiler] } for some positive integer n
http://en.wikipedia.org/wiki/Surreal_number
Pic related. The man's a muthafucking crazy genius!

[Anonymous No.4595035]

1/3 + 1/3 + 1/3 = 1
1/3 = .333...
.333 + .333 + .333 = 1/3 + 1/3 + 1/3