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Apr 12, 2023 at 11:18 comment added Alessandro Della Corte "Differentiate to obtain...", like Josh says, is really masterful math writing 😁 As he observes, this "implies" that no two real (complex, btw) numbers add up to a third one. Luckily multiplication still resists.
Nov 3, 2021 at 13:56 comment added Zach Teitler @joro $0^0=1$ and anyway even if that were an issue it would hardly be the only issue with this proof. :-)
Nov 3, 2021 at 7:47 comment added joro Don't you need additional work to show $0^0 \ne 0$?
Nov 2, 2021 at 10:09 history edited Zach Teitler CC BY-SA 4.0
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S Jan 10, 2019 at 9:10 history answered Zach Teitler CC BY-SA 4.0
S Jan 10, 2019 at 9:10 history made wiki Post Made Community Wiki by Zach Teitler