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Cake day: August 9th, 2023

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  • Valthorn@feddit.nutoScience Memes@mander.xyzI just cited myself.
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    edit-2
    5 months 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.

    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.