0
$\begingroup$

Are there any numbers that are not elements of the complex field?

Follow-up questions: Are p-edic fields subfields of the complex field?

Can quaternions be viewed as a complex vector space in three dimensions (as opposed to a superfield of the complex field)?

$\endgroup$
12
  • $\begingroup$ Well, there are things like p-adic numbers $\endgroup$
    – lulu
    Apr 18, 2016 at 12:49
  • 1
    $\begingroup$ What do you mean? Do you regard quaternions as numbers? $p$-adic numbers? Do you regard $\infty$ as a complex number? $\endgroup$
    – almagest
    Apr 18, 2016 at 12:49
  • 2
    $\begingroup$ There is no such thing as two-dimensional numbers. You can view $\mathbb{C}$ as a real vector space in which case it has dimension 2, but it doesn't make sense to speak of the dimension of a number itself. $\endgroup$
    – Mathematician 42
    Apr 18, 2016 at 12:53
  • 1
    $\begingroup$ There is no such thing as "all numbers". $\endgroup$
    – Ivan Neretin
    Apr 18, 2016 at 12:55
  • 1
    $\begingroup$ @barakmanos : No, it says that we can represent complex numbers in the 2-dimensional plane $\mathbb{R}^2$, it doesn't say that a complex number is 2-dimensional. $\endgroup$
    – Mathematician 42
    Apr 18, 2016 at 12:57

1 Answer 1

Reset to default
2
$\begingroup$

In order to make your question more precise, you would need to tell us what you mean by "number". If you are not familiar with mathematics then this may seem strange. However, there are many "numbers" that are not complex numbers; such a p-adic and quaternions as others pointed out.

There are senses where complex numbers are "complete". For instance, every polynomial can be completely factored in the complex numbers. So the polynomial $x^2+1=(x+i)(x-i)$ is factored using complex numbers but is not factorable under the reals. If we wanted to factor $x^2+1$ under the reals, we cannot. Intuitively, the graph of $y=x^2+1$ doesn't intersect the $x$ axis and has not real roots. To factor it, we must add a new number to the reals. This number is $i$ which satisfies $i^2=-1$ (in other words $i^2+1=0$). For complex numbers, there is no number that must be added to factor any complex polynomial.

Also the complex numbers are complete in the sense that sequences that should converge (so called Cauchy sequences) do converge. There are no holes in the complex numbers like there are in the rationals. For instance $\pi$ can be approximated arbitrarily well by rational numbers, but is not rational. If a number can be approximated arbitrarily well by complex numbers then it is complex.

Finally any rational or real number is a complex number with imaginary part equal to zero.

$\endgroup$
3
  • 1
    $\begingroup$ I'll research p-adic and quaternion numbers. This research will likely answer the question. I had taken a substantial amount of master level classwork about 20 years ago, but it was mostly analytical in R. I'm trying to remedy some ignorance. $\endgroup$
    – Mark Kosmowski
    Apr 19, 2016 at 16:46
  • $\begingroup$ I read a little about p-adic fields. Is each p-adic field a sub-field of the complex field? $\endgroup$
    – Mark Kosmowski
    Apr 19, 2016 at 23:23
  • $\begingroup$ No, I do not believe so. The p-adic numbers are a completion of the rationals that results in something different than the real numbers. One way to think of them is for real numbers you have number of the form $$123.44523443234....$$ That is to the right of the decimal you have infinitely many digits. For p-adic numbers it's the left side that has infinitely many digits. Thus, they are a different animal than the real or complex numbers. $\endgroup$
    – Joel
    Apr 20, 2016 at 19:00

Not the answer you're looking for? Browse other questions tagged .