I What number classification does i*e, or i*sqrt(2) belong to?

[Sorcerer]

I have been told that irrational and transcendental numbers only belong to the set of real numbers.

But what if you multiply one by i? Is it now an “imaginary irrational” or “imaginary transcendental” number?Based on a quick google while making this thread, Wikipedia says that transcendentals can be real or complex, but why would this not apply to all irrational numbers as well? Why isn’t i√2 also considered complex?
I am just trying to get a better grasp on the subset hierarchy of numbers. I appreciate any links, explanations, or research direction.

 

[Let'sthink]

For simple algebra yes i√2? can be terated as any other number with the meaning of powers of i predefined. But if you wish to call it a complex number, then it is represented by a ordered pair: (0, √2) or for algebra purpose (0+ii√2).

 

[fresh_42]

I have been told that irrational and transcendental numbers only belong to the set of real numbers.
But what if you multiply one by i? Is it now an “imaginary irrational” or “imaginary transcendental” number?

No, they are still either imaginary as $i\sqrt{2}$ and / or transcendental $i \pi$.

Based on a quick google while making this thread, Wikipedia says that transcendentals can be real or complex, but why would this not apply to all irrational numbers as well? Why isn’t i√2 also considered complex?

Because this is not how it works. Algebraically we distinguish
$$\mathbb{N}\subseteq\mathbb{Z}\subseteq\mathbb{Q}\subseteq\mathbb{R}\subseteq\mathbb{C}\subseteq\mathbb{H}\subseteq\mathbb{O}$$
and their different properties:

half group → group and ring → prime field and quotient ring → topological closure → algebraic closure → (non commutative) division ring → (non associative) division algebra.

So there is nowhere a concept of imaginary numbers. They are simply complex numbers, and as such algebraic or transcendental. Irrational literary means not rational, so it is $\notin \mathbb{Q}$ which usually means $\mathbb{R}-\mathbb{Q}$. Imaginary numbers would be a (Corr.: subring) additive subgroup $i \cdot \mathbb{R}$ which is by itself not very interesting, it doesn't even have a $1$. So the entire case of imaginary numbers is only due to the fact, that we normally identify complex numbers with the Euclidean plane and we have to name the axis. Algebraically, they are not important at all, except for the field extension $\mathbb{C} \cong \mathbb{R}(i) = \mathbb{R}[ i ]$ where only one element $i$ as the solution of $x^2+1=0$ is needed. From an algebraic point of view, we don't even need $i$, because $\mathbb{C}\cong \mathbb{R}[x]/(x^2+1)\,.$

 

[Sorcerer]

Thanks. You two have given me a lot to think about and research. This is not a topic I am very familiar with. But for now, at the risk of asking a stupid question with an obvious answer, one quick follow up:By saying $i \cdot \mathbb{R}$ doesn’t have a 1, do you mean that it doesn’t have a multiplicative identity?

Which would mean it is not a ring?

 

[fresh_42]

Thanks. You two have given me a lot to think about and research. This is not a topic I am very familiar with. But for now, at the risk of asking a stupid question with an obvious answer, one quick follow up:By saying $i \cdot \mathbb{R}$ doesn’t have a 1, do you mean that it doesn’t have a multiplicative identity?

Which would mean it is not a ring?

I made a mistake. It is no ring as multiplication leads outside of it: $ia \cdot ib = -ab \notin i \cdot \mathbb{R}$, so sorry for this error. Thus it's only an additive subgroup, a copy of $(\mathbb{R},+)$, i.e. even less. The message remains: imaginary numbers are nothing special, just a sort of artificial zero of $x^2+1=0$ to enhance our capabilities of calculation, such as the negative numbers extended counting.

 

[Sorcerer]

I made a mistake. It is no ring as multiplication leads outside of it: $ia \cdot ib = -ab \notin i \cdot \mathbb{R}$, so sorry for this error. Thus it's only an additive subgroup, a copy of $(\mathbb{R},+)$, i.e. even less. The message remains: imaginary numbers are nothing special, just a sort of artificial zero of $x^2+1=0$ to enhance our capabilities of calculation, such as the negative numbers extended counting.

That said, are irrational numbers multiplied by $i$ still considered irrational numbers? Because when I look at the wikipedia page of transcendental numbers, it says they can be either real or complex, but when I look at the wikipedia page of irrational numbers, it says they are only real numbers. So what would you call $0\cdot a+i\sqrt{2}$ ? Just a complex number? A complex irrational number?

Direct quote of wikipedia on irrational numbers:
"In mathematics, the irrational numbers are all the real numbers which are not rational numbers"

 

[fresh_42]

As said, irrational means not rational, but it is only used for real numbers. Of course $i\sqrt{2}$ is also not rational, but usually the term is reserved for real numbers only. The reason is, that $\mathbb{Q}\subseteq \mathbb{R}$ has a special role. It's a dense subset and one can directly go from rationals to reals by topological closure. On the other hand about $\mathbb{Q}\subseteq \mathbb{C}$ can only be said that the rationals are the prime field (smallest field) of the complex numbers. Other relations only come into play in specific applications. One would rather consider $\mathbb{Q}[ i ]$, i.e. all complex numbers with rational components than the rationals alone. But even those don't have a special name. Names are used to abbreviate. There is simply no need to name $i\cdot \mathbb{Q}$ or irrational numbers in $\mathbb{C}$. However, in $\mathbb{R}$ it is convenient to distinguish them, and last but not least it's historically grown. This means the ancient Greeks used ratios for their geometry, and irrational numbers are simply those, which do not correspond to ratios, e.g. diagonals. Therefore the name: ir-rational = no ratio.

 

[Sorcerer]

Thank you. That clears things up quite a bit.