Exploring the Complexity of 0: Real, Imaginary, or Both?

In summary, the conversation discusses the possible classifications of the number 0 in terms of real and imaginary numbers. The three options considered are: 1) 0 is a purely real complex number, 2) 0 is not real nor imaginary, and 3) 0 is both real and imaginary. After ruling out the option of 0 being just imaginary, the conversation leans towards 0 being both real and imaginary due to its appearance on both real and imaginary numberlines and its ability to satisfy absolute value equations as a real number. The conversation also explores other mathematical concepts related to 0, such as its role as a neutral element in addition and an absorbing element in multiplication. Ultimately, the precise definitions of real and imaginary numbers
  • #36
ehild said:
The complex numbers are defined above the real numbers, as ordered pairs of real numbers, written in the form as z=x+yi, with addition and multiplication defined.

The construction of the complex numbers from the real numbers is via ordered pars of reals.
You can define the complex numbers without any reference to the reals.

You can not define real numbers with complex numbers.
From Wikipedia
Let K be a topological field, which contains a subset P, satisfying
  1. P is closed under addition, multiplication and inverses.
  2. If ##x, y \in P## and ##x \neq y## then ##y-x \in P## or ##x-y \in P##, but not both.
  3. For all non-empty ##S \subset P## there exists ##x \in P## such that ##S+P = x+P##.
Further suppose K has a non-trivial involution such that for all non-zero ##z \in K## one has ##zz^* \in P##.
Then K with the topology generated by ##\{ y \, |\, p - (y-x)(y-x)^* \in P,\, x \in K, \, p \in P\}## is isomorphic as topological fields to the complex numbers, and P is the set of positive numbers. Once you have the positive reals, you have all of the reals.
 
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  • #37
pwsnafu said:
The construction of the complex numbers from the real numbers is via ordered pars of reals.

Also note that it is perfectly possible to construct the complex numbers in such a way that ##\mathbb{R}## actually is a subset of ##\mathbb{C}##.
 

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