While I agree that my proof is blunt, yours doesn’t prove that .999… is equal to -1. With your assumption, the infinite 9’s behave like they’re finite, adding the 0 to the end, and you forgot to move the decimal point in the beginning of the number when you multiplied by 10.

x=0.999…999

10x=9.999…990 assuming infinite decimals behave like finite ones.

Now x - 10x = 0.999…999 - 9.999…990

-9x = -9.000…009

x = 1.000…001

Thus, adding or subtracting the infinitesimal makes no difference, meaning it behaves like 0.

Edit: Having written all this I realised that you probably meant the infinitely large number consisting of only 9’s, but with infinity you can’t really prove anything like this. You can’t have one infinite number being 10 times larger than another. It’s like assuming division by 0 is well defined.

0a=0b, thus

a=b, meaning of course your …999 can equal -1.

Edit again: what my proof shows is that even if you assume that .000…001≠0, doing regular algebra makes it behave like 0 anyway. Your proof shows that you can’t to regular maths with infinite numbers, which wasn’t in question. Infinity exists, the infinitesimal does not.

x=.9999…

10x=9.9999…

Subtract x from both sides

9x=9

x=1

There it is, folks.

You have 20 apples that you put into 5 boxes. “How many apples per box?” 20/5=4. Easy.

You have 20 apples that you put into half a box. “How many apples per box?” 20/0.5=40. You can see how 20 apples in half of a box implies 40 apples per full box.

You have 20 apples that you don’t put into anything. “How many apples per box?” 20/0=? What box? What are you talking about? How can you suggest that there are a number of apples per box when there’s no box to begin with? Lunatic!