Divergent stuff

[Anonymous No.7105580]

...99999999
+1
----------------
...00000000

So ...9999 = -1.

Also, ...9999 = 9*(...1111) = 9* (1/(1 - 10)) = 9*(-1/9) = -1.

Would someone please tell me, in terms of legit or even rigorous math, why this stuff works?

[Anonymous No.7105583]

>>7105580
whoops the +1 should have been on the very right, but the spaces got cut off.

[Anonymous No.7105587]

>>7105580
...it's only a model.

[Anonymous No.7105590]

its defined that way.

think of it in terms of philosophy.

its like saying, can you run faster than a car?
probably not. but what if you were the same speed as the car? Then running = car.

its a result derived from a definition

[Anonymous No.7105593]

>>7105580
What's this ...99999999? I'm not familiarized with that notation.

[Anonymous No.7105600]

>>7105593
It just means that the 9 digits go off infinitely far to the left.

[Anonymous No.7105601]

>>7105590
How is this a result from a definition?

[Anonymous No.7105603]

>>7105600
Alright... it looks like shit, but then how you arrive at ...9999 = -1 from adding 1 and ...99?

[Anonymous No.7105605]

google p-adic evaluation or something

[Anonymous No.7105606]

>>7105603
1 + 9 = 10.
Put zero, carry the one.
1 + 9 + 0 = 10.
Put zero, carry the one.
And so on...

[Anonymous No.7105607]

>>7105606
All I see is you are adding zeroes and nothing more

[Anonymous No.7105662]

>>7105603
the number becomes so big it wraps around and becomes negative

[Anonymous No.7105665]

If your ...999 is same thing as 0.999, then it's not the same. In 0.999..., or 0.(9), there is infinte number of nines, thus there is no last digit on the far right, there is infinity of nines, not just very many nines.

[Anonymous No.7105666]

>>7105662
No. Unless you are in Zp, it doesn't.

[Anonymous No.7105700]

>>7105580
>Would someone please tell me, in terms of legit or even rigorous math, why this stuff works?

It doesn't.

The sum of the series of 9*10^n does NOT equal -1, and if you add 1 to it, you do not get 0.

[Anonymous No.7105801]

>>7105580
it works very well, you can write all your pretend "...9999999" in a more rigorous way using all infinite sums. Like, you did this with 1/(1-10) being ....11111, the series 1/(1-x) you took and extended to regions outside its region of convergence, but you'll find it still has all the algebraic properties of 1/(1-x) (pretty much) so it'll satisfy some interesting rules I guess, even if most of them seem nonsensical

[Anonymous No.7105806]

>>7105801
like, maybe you can define log (...XXXX) as log (1/(1-X)) = -log(1-X)=X+1/2X^2+1/3X^3... where X is any set of digits. Again, just playing around.

[Anonymous No.7105829]

>>7105580
it works for the same reason this does

[Anonymous No.7105837]

>>7105829
try that again with an 8 I want to see what happens.

[Anonymous No.7105843]

>>7105580
Because p-adic numbers.

The sum <span class="math">\sum_{n=1}^{\infty} 9\cdot10^n[\math] converges because
<span class="math">\bigg\| \sum_{n=1}^k 9 \cdot 10^n + 1 \bigg\|_{10} = \big\| 10^{k+1} \big\|_{10}
[\math] converges to 0 in the 10-adic norm.[/spoiler][/spoiler]

[Anonymous No.7105848]

>>7105580
Because p-adic numbers.

The sum <span class="math">\sum_{n=1}^{\infty} 9\cdot10^n[/spoiler] converges because
<span class="math">\bigg\| \sum_{n=1}^k 9 \cdot 10^n + 1 \bigg\|_{10} = \big\| 10^{k+1} \big\|_{10}
[/spoiler] converges to 0 in the 10-adic norm.

[Anonymous No.7105905]

Not sure if thats a good idea.

[Anonymous No.7106024]

>>7105665
>If your ...999 is same thing as 0.999
It isn't, obviously.

>>7105700
Obviously the sum diverges,
but I was able to get the -1 answer using 2 different methods. Why?

>>7105848
>>7105605
>p-adic
What is this, and how much math do I need to know to understand it?

[Anonymous No.7106088]

>>7106024
Some basic real analysis and metric spaces for the analytic approach, which I think is the easiest way in.

[Anonymous No.7106128]

>>7106088
Thanks, I'll check it out.