Comment on I just cited myself.
yetAnotherUser@discuss.tchncs.de 6 days agoUnfortunately not an ideal proof.
It makes certain assumptions:
- That a number 0.999… exists and is well-defined
- That multiplication and subtraction for this number work as expected
Similarly, I could prove that the number which consists of infinite 9’s to the left of the decimal is equal to -1:
...999 = x ...990 = 10x Calculate x - 10x: x - 10x = ...999 - ...990 -9x = 9 x = -1
And while this is true for 10-adic numbers, it is certainly not true for the real numbers.
Valthorn@feddit.nu 6 days ago
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.
a0=b0, thus
a=b, meaning of course your …999 can equal -1.
yetAnotherUser@discuss.tchncs.de 6 days ago
Yes, but similar flaws exist for your proof.
The algebraic proof that 0.999… = 1 must first prove why you can assign 0.999… to x.
My “proof” abuses algebraic notation like this - you cannot assign infinity to a variable. After that, regular algebraic rules become meaningless.
The proper proof would use the definition that the value of a limit approaching another value is exactly that value. For any epsilon > 0, 0.999… will be within the epsilon environment of 1 (= the interval 1 ± epsilon), therefore 0.999… is 1.