B Is 0 a Real Number, an Imaginary Number, or Both?

[Ssnow]

sure it is real ;-)

 

[fresh_42]

sure it is real ;-)

Yep. The complete discussion is nonsense. It would have been much more interesting to discuss whether 0 can be considered natural, although this might be rather philosophic or at best historic.

 

[PeroK]

Yep. The complete discussion is nonsense. It would have been much more interesting to discuss whether 0 can be considered natural, although this might be rather philosophic or at best historic.

... says the self-appointed arbiter of what it is sensible to discuss, insulting all those who made the effort to post on this thread.

Physics Forums is about helping others, not condescending to pass judgement on them. There are plenty of places on line where you can do that.

 

[ehild]

@Isaac0427 said: ↑
Ok, I'm confused. Is this statement true: 0=0i=0+0i?

Yes.

A real number is a complex number whose imaginary part is 0.
An imaginary number is a complex number whose real part is 0.

What does that make 0?

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.
"x" is considered the real part of z and "y" is considered the imaginary part.
The addition is defined similarly as in case of two dimensional vectors, z₁+z₂ =(x₁+x₂)+i(y₁+y₂)
There is a neutral (null) element of addition, a complex number denoted by 0_c so as z+0_c=z.
As 0_c is element of the complex numbers, it has to be written in the form of a+ib. So z+0_c=z means that (x+iy)+(a+ib)=(x+a)+ i(y+b) = x+iy which means that 0_c=0+0i.
You can not define real numbers with complex numbers. But you can define an imaginary number as a complex number whose real part is 0. So you can consider 0i an imaginary number.

You can not state that the zero element of the real numbers is identical with the zero element of the complex numbers, as they belong to different sets, as @Number Nine and @fresh_42 and others pointed out. 0 ≠ 0+0i. The left 0 is real, the right side is a complex number.
There are a lot of other sets where you can define addition and zero element. Think about the null-vector or null matrix, or an empty set which is also denoted by ∅. You can not say that they are the same, and 0 is equivalent with the empty set or a null-vector.

Some people, including you, lost his nerves in this thread. You can show that somebody was wrong, even to say that what he wrote had no sense, but you must not hurt other members personally.

 

[jbriggs444]

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.

As has been pointed out, this is not correct. There are other foundational approaches that do things differently. The real number zero and the complex number zero may or may not end up being the same entitity, depending on the chosen approach.

The statement

0 ≠ 0+0i

is not justified.

One could argue that the above statement is actually incorrect, but that would open a different can of worms -- the resolution of overloaded notation in mathematical statements.

 

[pwsnafu]

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.

 

[micromass]

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 $\mathbb{C}$.