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My question is very simple. How would one take the floor of a complex number and what is the floor of i?

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  • $\begingroup$ Floor function is defined via ordering in the real numbers. But there is no such (total) ordering in complex numbers. So doesn't make sense. $\endgroup$ – NeedForHelp Jan 13 '17 at 2:42
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    $\begingroup$ Absolute value was originally defined for real numbers. Yet Absolute value can be extended to the complex plane $\endgroup$ – Brothersquid Jan 13 '17 at 2:44
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    $\begingroup$ Total order was defined for real numbers. But unfortunately you won't be able to extend it to complex numbers. But you can define it the way you want, e.g. by taking floor function of real and imaginary parts. I don't know if it's useful or not. $\endgroup$ – NeedForHelp Jan 13 '17 at 2:45
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    $\begingroup$ @TheGreatDuck, you may want to be more precise with that remark; the complex numbers cannot be ordered in a way that logically extends algebraic properties. It's easy enough to arbitrarily order them. This is different from e.g. the way that real numbers cannot be counted, but algebraic numbers can be. $\endgroup$ – Wildcard Jan 14 '17 at 1:44
  • $\begingroup$ @TheGreatDuck actually I had to check, since what you say seemed counter-intuitive. The link I provided is what clarified it for me. Before that I didn't understand; the literal statement you made (that Euler proved the complex numbers cannot be ordered) isn't literally true. This is a math site; lots of literal people around. ;) And the word "obviously" is the arch-enemy of mathematical proofs. :D $\endgroup$ – Wildcard Jan 14 '17 at 5:09
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The floor function is defined as the largest integer less than or equal to the real number given.

Since "less than or equal to" isn't defined for complex numbers you can't have a floor function for them.

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Wolfram Alpha defines $\lfloor a+bi\rfloor$ as $\lfloor a\rfloor+\lfloor b]i$. With this convention you may say that $\lfloor i\rfloor = i$

Taka a look also here

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    $\begingroup$ Wolfram Alpha has in fact a rigorous definition of the floor of a complex number. Whether it is useful or not, it is a different issue. $\endgroup$ – Momo Jan 19 '17 at 20:30
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Momo's answer is probably more official, but there could be another definition as follows:

$$\lfloor a + bi \rfloor = n(a+bi)$$ where $n$ is a positive real number $\le 1$ such that $$|\lfloor a+bi \rfloor | = \lfloor |a+bi| \rfloor$$

In other words, keep the same angle and reduce the absolute value to the next integer.

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    $\begingroup$ Your constriction gives floor of -1.5 to be -1 $\endgroup$ – Q the Platypus Jan 13 '17 at 23:18
  • $\begingroup$ @QthePlatypus, good point. The basic idea of having a floor function which would change the absolute value but not the argument, could still be useful, though. I think it would make sense to take the floor of the absolute value if either the real component is positive, or if the number is a positive multiple of $i$, and take the ceiling of the absolute value otherwise. (Thus the floor of a complex number would always either be somewhere to the left of the original number on the complex plane, or directly below it.) $\endgroup$ – Wildcard Jan 14 '17 at 0:52

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