Is my classification of transcendental correct?

[JustinzCruiz]

From what I've recently read, Transcendental numbers are those irrational numbers which cannot be formed by sqrt or any root of any order. So If the irrational numbers which are formed by taking roots of different orders on numbers which are not perfect roots of that order, it is called as a surd. So transcendental would have to be those irrational numbers which are not surds. So Transcendental could be classified as a type of irrational number which are not surds. More simply speaking, Transcendentals are irrational numbers that are not Surds. Am I right with this?

 

[fresh_42]

From what I've recently read, Transcendental numbers are those irrational numbers which cannot be formed by sqrt or any root of any order. So If the irrational numbers which are formed by taking roots of different orders on numbers which are not perfect roots of that order, it is called as a surd. So transcendental would have to be those irrational numbers which are not surds. So Transcendental could be classified as a type of irrational number which are not surds. More simply speaking, Transcendentals are irrational numbers that are not Surds. Am I right with this?

Not quite. The definition goes the other way around. Transcendental numbers $t$ are those which do not follow an equation $0=t^n+c_1t^{n-1}+c_2t^{n-2}+ \ldots + c_{n_1}t + c_n$ with all $c_i \,{=}_{e.g.}\, \mathbb{Q}$ and some arbitrary $n\in \mathbb{N}\,.$

This way one gets rid of the requirement of written roots, which is ambiguous. E.g. there are numbers which satisfy a polynomial equation of degree five (or higher), which means they are algebraic = not transcendental, but cannot be written by root expressions either.

Irrational only means no quotient of integers. Thus $\sqrt{2}$ is irrational, but not transcendental. Therefore there are actually two different categories of numbers: rational / irrational and algebraic / transcendental where rational numbers are algebraic, and transcendental numbers are irrational.

 

[mfb]

All real numbers that are not algebraic are transcendental.
There are both transcendental and algebraic irrational numbers. And there are rational numbers, all of them are algebraic.

Here is a Venn diagram:

 

[JustinzCruiz]

So I was right except for saying just surds, I should have said using surds which includes things like (1+√5)/2 which perform operations on surds to form irrationals which are not transcendental(Algebraic). Restating : Irrationals that cannot be formed using surds are Transcendentals.
But I do know that algebraic numbers are not just irrational numbers but also include whole of Real numbers except Transcendentals.
Exception of my statement : Raising algebraic number to the power of an irrational algebraic number (Eg:2^√2) which indeed uses a surd to form a Transcendental.

If I am wrong, please inform me of those numbers that are algebraic irrationals which are not formed using surds.

According to me Irrationals are of 2 types:
1. Surds ( Including those formed by operations on surds)
2. Transcendentals.

If I am wrong, there needs to exist 3 types of Irrationals:
1. Surds( Including those formed by operations on surds ) [Algebraic]
2. Non Surds( Not formed by operations on surds or surds itself) but [Algebraic]. {Please inform me if this category exists and give me some examples of it}
3. Transcendentals. [Not Algebraic]

 

[mfb]

The second. The roots of $x^5 - x +1$ are an example, there is no way to express them as finite sum/product/... of nth roots. of integers.
The existence of these numbers is shown by the Abel–Ruffini theorem.

 

[JustinzCruiz]

Thank you so much. You made my day.

 

[fresh_42]

What you named surds is usually called "can be expressed by radicals". Radical is a general term for "root".

 

[Leo Authersh]

I have asked a similar question. You might find it useful.
https://math.stackexchange.com/q/2385349/469594