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by missingET
811 days ago
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No, there’s an abstract algebra extension of real numbers to have an extra symbol h such that h^2=0. This is not a real number so you cannot apply the argument h^2=0 implies h=0, much like complex numbers don’t obey all properties of real numbers. (For example for real numbers, x!=0 implies x^2>0 but i^2=-1) https://en.m.wikipedia.org/wiki/Grassmann_number |
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