Why is $\mathbb T\cup\mathbb A = \mathbb Q \cup \mathbb I =\mathbb R$? Where $\mathbb T $ is the set of transcendental numbers, and $\mathbb I $ is the set of irrational numbers and $\mathbb A $ is the set of algebraic numbers. The sets $\mathbb Q$ and $\mathbb R$ have their usual meanings. 
I'm having trouble understanding how the union of the transcendental $\mathbb T $ and algebraic numbers $\mathbb A $ is the same as the real numbers $\mathbb R$ since $\mathbb I ,\mathbb Q\subset\Bbb R$ as well.
Put in another way, I think it should be $( \mathbb T \cup \mathbb A) \subset \mathbb R $ and $(\mathbb Q \cup \mathbb I) \subset \mathbb R$ 
So I'm basically questioning the equality in $\mathbb T \cup \mathbb A  = \mathbb Q \cup \mathbb I =\mathbb R$. 
The reason I'm asking this question is because those unions I mentioned above don't seem to 'cover' all the Reals $\mathbb R$ .
Thanks in advance.
Kindest Regards.
 A: We define an irrational number to be a (real) number that isn't rational, which is equivalent to defining
$$\Bbb I := \Bbb R - \Bbb Q,$$
and hence
$$\Bbb Q \cup \Bbb I = \Bbb Q \cup (\Bbb R - \Bbb Q) = \Bbb R.$$It is correct, then, to say that $\Bbb Q \cup \Bbb I \subseteq \Bbb R$, but this statement contains less information than our equality.
The relationship between $\Bbb T$ and $\Bbb A$ is analogous.
A: Given a real number, if it is rational then it is in $\Bbb Q$, otherwise it is irrational and in $\Bbb I$. So $\Bbb{Q\cup I=R}$.
Similarly, if a real number is not algebraic then it is transcendental. 
The set $\Bbb I$ is essentially defined as $\Bbb {R\setminus Q}$ and $\Bbb T$ is defined as $\Bbb{R\setminus A}$.

This is analogous to the following situation: Let $P$ be the set of prime numbers, and $Q$ the set of composite natural numbers. Let $E$ and $O$ denote the sets of even and odd name natural numbers respectively. 
Then $\Bbb N=P\cup Q=E\cup O$. Just two partitions of the same sets. 
A: It's essentially the very definitions of numbers involved.
$\Bbb Q$ is the set of all real numbers that can be expressed as the ratio if two integerss. We define a real number to be irrational if it's not rational, so that $\Bbb I = \Bbb R \setminus \Bbb Q$ by definition, hence $\Bbb R = \Bbb Q \cup \Bbb I$.
Similarly, a real number is called algebraic (and is an element of $\Bbb A$) if it's the root of a polynomial with integer coefficients. Those real numbers that aren't the root of any such polynomial are called transcendental, so that $\Bbb T = \Bbb R \setminus \Bbb A$.
Note: many people consider some complex numbers to be algebraic as well (e.g., the imaginary unit $i$ is a solution to $x^2 + 1 = 0$), so it might be more correct to say that $\Bbb R = (\Bbb A \cap \Bbb R) \cup \Bbb T$, considering $\Bbb A$ to be a subset of $\Bbb C$. Generally these people aren't terribly concerned with $\Bbb C \setminus \Bbb A$, so I'm not sure if these too are considered transcendental numbers, or just nameless. (I have character theory of finite groups in mind here.)
