Transcendental Union Algebraic = Irrational? It is true that $\mathbb{R} = \mathbb{Q} \bigcup \overline{\mathbb{Q}}$ where $\mathbb{R}$ is the set of real numbers, $\mathbb{Q}$ is the set of rational numbers, and $\overline{\mathbb{Q}}$ is the set of irrational numbers, isn't it?
But, is it true that $\overline{\mathbb{Q}} = \mathbb{T} \bigcup\;(\mathbb{A}\backslash\mathbb{Q})$ where $\mathbb{T}$ is the set of real transcendental numbers and $\mathbb{A}$ is the set of real algebraic numbers?
 A: By definition,
$
\overline{\mathbb{Q}}
= \mathbb{R} \setminus \mathbb{Q}
$,
and so $\mathbb{R} = \mathbb{Q} \cup \overline{\mathbb{Q}}$.
By definition,
$\mathbb{T} = \mathbb{R} \setminus \mathbb{A}$,
and so $\mathbb{R} = \mathbb{T} \cup \mathbb{A}$.
Hence,
$
\overline{\mathbb{Q}}
= \mathbb{R} \setminus \mathbb{Q}
= (\mathbb{T} \cup \mathbb{A})\setminus\mathbb{Q}
= (\mathbb{T} \setminus\mathbb{Q}) \cup (\mathbb{A}\setminus\mathbb{Q}) = \mathbb{T} \cup (\mathbb{A}\setminus\mathbb{Q})$.
A: Both true. For the second one do inclusion both ways. If $x$ is irrational, it is either transcendental or algebraic, hence in the given union (being irrational is used if it isn't transcendental). Other inclusion is even easier to see.
A: Since the other two users answered “yes”, I will answer “no”. :-) The problem with your statement lies in the fact that $\mathbb A$ and $\mathbb T$ are complementary parts of $\mathbb C$, not $\mathbb R,~$ i.e., $~\mathbb A \cup\mathbb T=\mathbb C$, and $\mathbb A\cap\mathbb T=$ $=\varnothing$. This is due to the very definition of algebraic numbers as roots to polynomials with rational coefficients. Obviously, $x^2+1$, for instance, is one such polynomial, yet its two roots, $\pm i\in\mathbb A$, are not part of $\mathbb R$.
