Is real number system combination of rational and irrational numbers?

Is real number system combination of rational and irrational numbers?
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Freddy Doyle
Yes, that covers all of them, by definition.
We take the real numbers $\mathbb{R}$ and the rational numbers $\mathbb{Q}$. We note that $\mathbb{Q}$ is a proper subset of $\mathbb{R}$, so we take all the real numbers that aren't rational and call them irrational numbers, i.e. we define$\mathbb{I}=\mathbb{R}\setminus \mathbb{Q}$.
There are other splits possible. In particular, we can take all the real numbers that are roots of polynomials with integer coefficients, and call them the algebraic numbers (sometimes notated as $\mathbb{A}$, although that also sometimes includes the complex algebraic numbers). So we can also consider the set of what's left, and that's what we call the transcendental numbers $\mathbb{R}\setminus \mathbb{A}$. Since any rational number $\frac{p}{q}$ is the solution to $qx-p=0$, we can soon confirm that $\mathbb{Q}\subset \mathbb{A}$, i.e. the rational numbers form a proper subset of the algebraic numbers, and conversely the transcendental numbers form a proper subset of the irrational numbers.