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by rsaarelm
690 days ago
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You could call it an incorrect answer if there was a correct answer to division by zero, but it's undefined instead with no correct answer. Sounds pedantic, but in math pedantic stuff matters, and apparently you can expand things to define division by zero as zero and not break math, https://www.hillelwayne.com/post/divide-by-zero/ |
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You really can't though.
> If x/0 is a value, then the theorem should extend to c=0, too.” This is wrong. The problem is not that 1/0 was undefined. The problem was that our proof uses the multiplicative inverse, and there is no multiplicative inverse of 0. Under our modified definition of division, we still don’t have 0⁻, which means our proof still does not work for dividing by zero. We still need the condition. So it is not a theorem that a * (b / 0) = b * (a / 0).
This is like saying there's nothing wrong with defining 2 + 2 = 5, and addition will still be associative because (a + b) + c still = a + (b + c) unless b = 2. Like, sure, you can redefine division to not have the normal properties that it does, and then argue that your redefinition is sound because the theorems only apply to things that have the normal properties of added numbers. But that's not what + means!
If these people really believed the arguments they're making, they would actually define x/0 = 5, or 19, or something on those lines.