x=.9999...
10x=9.9999...
Subtract x from both sides
9x=9
x=1
There it is, folks.
Valthorn
joined 1 year ago
Two candidates that support genocide, but one is a christofascist. No matter who you vote for, genocide support wins. But you think it's better to give the christofascist better odds than to inconvenience yourself with a vote you don't 100% agree with, and possibly abstain from your chance to ever vote again. Not voting won't fix the issue, since there's no threshold on voter turnout for the election to count. The struggle against genocide must be fought in other ways. So unfortunately, this fall you're getting genocide, so please make sure you don't get fascism too.
Except, it's not an attempt to end democracy, as fascism is.
I feel like the caffeine is the least of your problems if you drink 70 cups of coffee in a day. ~15 liters of fluid is no joke.
It's a march by Czech composer Julius Fučík called "Entry of the Gladiators", written in 1897.
Edit: It gets memeier! From Wikipedia:
He originally titled it "Grande Marche Chromatique", reflecting the use of chromatic scales throughout the piece, but changed the title based on his personal interest in the Roman Empire.
Acknowledging that they are obese is stating a fact. Claiming that they wouldn't or couldn't take good care of their dog because of their obesity is fat shaming. They're not 600+lbs-can't-leave-the-bed obese.
You have given Dobby clothes. Dobby is free!
Removes sunglasses My god! It's so crazy it might actually work!
So some kind of "dark, real and gritty" version Wile E. Coyote and the Road Runner?
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.