Irrationals yet as ordered pairs of rationals?

Let's consider the naturals $\mathbb{N}$ as being given. Then we have the following sequence of extensions:
  1. The whole numbers $\mathbb{Z}$ are defined as ordered pairs of naturals (: Wikipedia)
  2. The rational numbers $\mathbb{Q}$ are defined as ordered pairs of whole numbers (: Wikipedia)
  3. The real numbers $\mathbb{R}$ are not defined as ordered pairs of rational numbers (: Wikipedia)
  4. The complex numbers $\mathbb{C}$ are defined as ordered pairs of real numbers (: Wikipedia)
All extensions of the number system are such that many important properties of the previous number type are inherited. Three out of four extensions are done by the formation of ordered pairs, with proper definitions of the four basic operations: addition, substraction, multiplication and division. The step from the rational numbers to the real numbers, however, differs essentially from the other extensions. According to standard mathematics, it cannot be done by considering pairs of rational numbers. But let's be stubborn and nevertheless try the following
Definition. A real number $r$ is an ordered pair of rationals $(a,b)$ such that $a < b$. The absolute difference $|a-b|$ is called the error of the real number $r$. The main problem is to minimize the error (i.e. preferrably there is a limit that makes it zero). The idea is that irrational numbers can be understood as the number squeezed between ever more converging pairs of rational numbers. It is noticed, however, that this idea is rather an idealization of a not-so-nice reality, where errors do really exist.
Example 1. Let the real number $\,e\,$ for any natural $n$ be defined by $$e = \left(\left[1+\frac{1}{n}\right]^n , \left[1+\frac{1}{n}\right]^{n+1}\right)$$ Where it is noticed that there is a lot of theory in $\mathbb{Q}$ preceding this. The error can be made as minimal as desired with help of a theory about limits in $\mathbb{Q}$, which is assumed to be fully developed: $$\lim_{n\to\infty} \left[ \left(1+\frac{1}{n}\right)^{n+1}-\left(1+\frac{1}{n}\right)^n \right] =\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n\frac{1}{n} = \lim_{n\to\infty}\frac{e}{n}=0$$ Example 2. Let the real number $\,\pi\,$ for any natural $n$ be defined by Leibniz formula: $$ \pi = \left( 4\sum_{k=0}^{2n-1} \frac{(-1)^k}{2k+1} , 4\sum_{k=0}^{2n} \frac{(-1)^k}{2k+1}\right) $$ Again, the error can be made as minimal as desired: $$ \lim_{n\to\infty} \left[ 4\sum_{k=0}^{2n} \frac{(-1)^k}{2k+1} - 4\sum_{k=0}^{2n-1} \frac{(-1)^k}{2k+1} \right] = \lim_{n\to\infty} 4 \frac{(-1)^{2n}}{4n+1} = 0 $$ Does the above make sense?