Question

Alright, so this thing works, but like how can it break?

Answer 1

Let's assign this doubly infinite sum to be a number, and then solve for it:
\[n=\cdots+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\cdots\]
The way we solve for n is notice that multiplying it by x does nothing to the number, it shifts each term around to the next one, so we have this identity:
\[x*n=n\]
But hey, 0 also has this property
\[x*0=0\]
so
\[n=0\]

Answer 2

i was hoping i would be something i can answer. :D. Can you do this @zepdrix?

Answer 3

Now what's this mean? Well in base 10 we could write zero as:
\[0=...9999.9999...\]
Or alternatively we can subtract \(.9999...\) from both sides to get:
\[...9999 = -.9999... = -1\]
So we have an infinite repeating integer to the left form of representing the same number. And hey, it still works like -1 because if we add 1 to it we get 0, just look:
\[...9999 + 1 = ...0000\]
since it carries over haha...

Answer 4

wouldnt infinity also break it

i mean the whole reason why zero doesnt work is the issue of dividing by zero

Answer 5

Yeah kinda seems like it would, wouldn't it? But really we're not dealing with infinite valued numbers, we're just looking at an alternative representation of finite numbers that happens to have an infinite repeating to the left form. It doesn't seem to be that anything breaks, since you can freely change back and forth and it's consistent.

Answer 6

whats the .....9999.9999....

Answer 7

From earlier:
\[0=\cdots+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\cdots\]

So plug in x=1 and multiply this whole thing by 9.

Answer 8

ahh okay i gotcha

Answer 9

weird

Answer 10

im sorry im in 7th grade this isway to hard so bye... good luck tho:)

Answer 11

.

Answer 12

@Astrophysics AGREED

Answer 13

o_O

Answer 14

Haha

Answer 15

Most rules of arithmetic are not defined for infinite sums like this, such as commutativity, and maybe distribution? so when we use them and get weird things, it's because we are assuming we know the result, but we don't. :) i think

Answer 16

Seems like you've rediscovered two's complement.
https://en.wikipedia.org/wiki/Two%27s_complement

Answer 17

or something equivalent i guess

Answer 18

Here's how I look at it.
Given:
\(n=\cdots+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\cdots\)
then 1+n=\(\color{blue}{\cdots+x^2+x+1}\)+\(1+\frac{1}{x}+\frac{1}{x^2}+\cdots\)
which has a sum of \(+\infty\) in all three cases:
1. x>1, 1+n=\(+\infty\)+1/(1-x)=\(+\infty\)
2. x=1, 1+n=\(+\infty\)
3. x<1, 1+n=1/(1-x)+\(+\infty\)=\(+\infty\)
So in all cases, we cannot do arithmetic with n.

Answer 19

Well, I believe there's an isomorphism between the infinite to the left representation of a rational number to the infinite to the right hand version, for instance let's take this:

\[...3434=n\]\[...3434*100=n*100\]Subtracting the first from the second gives:\[34=100n-n\]\[\frac{34}{99} = n\]

But we could do this for any periodic or rational number, I guess to really show this is an isomorphism we need to show that the result of addition of these strings of things done with the regular algorithm we know will give the result we expect in terms of rational numbers.