4
$\begingroup$

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

  • 0
    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. – 2017-01-13
  • 1
    Absolute value was originally defined for real numbers. Yet Absolute value can be extended to the complex plane – 2017-01-13
  • 2
    Total order was defined for real numbers. But unfortunately you [won't be able to extend it to complex numbers](http://math.stackexchange.com/questions/487997/total-ordering-on-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. – 2017-01-13
  • 0
    I already asked this question here: http://math.stackexchange.com/questions/1764832/what-is-lfloor-i-rfloor Please clarify and give precise reasoning why your question is different than mine. – 2017-01-13
  • 1
    On another note, euler proved that the complex numbers cannot be ordered. This means one cannot use the definition n <= x < n in any context. Therefore, the floor function cannot exist in any sense within complex numbers in the same way we think of it in real numbers. i.e. just make something up. – 2017-01-13
  • 2
    @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.](http://math.stackexchange.com/a/1786628/276406) This is different from e.g. the way that real numbers cannot be counted, but algebraic numbers can be. – 2017-01-14
  • 0
    @Wildcard Not really. I figure it is reasonable enough to gather that I mean euclid proved that the concept of greater and lesser cannot be extended. That comes with the logical need for algebraic rules to apply, obviously. After all... you understood and in this context I'm sure the op can understand. Besides, this is a blatant duplicate of my question. – 2017-01-14
  • 0
    @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 – 2017-01-14
  • 1
    @Wildcard to be fair, I was speaking colloquially not making a formal proof. So I figured the op would conclude I meant > and < by "ordering" and that it would invalidate any regular definition of floor. One would have to twist the meaning as the meaning becomes nonsense. – 2017-01-14
  • 1
    To those who voted to close this question, I find that to be horribly egregious. This is a duplicate question. It needs to be closed as duplicate not **as unclear**. It is only two sentences. There's nothing to become unclear. Please vote as a duplicate so that I might receive credit as the original asker. This is deceptive and makes it appear as if questions like mine are not allowed which is not at all right. – 2017-01-14

3 Answers 3

6

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.

4

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

  • 0
    true, but wolfram alpha cannot actually assert that to be true by definition. Wolfram alpha is just trying to best guess what it would be. Also, that is only true when a or bi is an integer. Clearly wolfram alpha is guessing horribly. – 2017-01-13
  • 1
    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. – 2017-01-19
  • 0
    What I mean is that floor is defined via inequalities and they don't exist for complex numbers. Furthermore it is a mapping of real numbers to integers. So then must ask what "integer" means for complex numbers. It's a horribly vague idea. Wolfram alpha cannot claim to have THE definition. – 2017-01-19
0

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.

  • 2
    Your constriction gives floor of -1.5 to be -1 – 2017-01-13
  • 0
    @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.) – 2017-01-14
  • 0
    @Wildcard you could make it piecewise defined so that the negative coordinates are treated differently. Make them bulge out. – 2017-01-19