# Part: Real Numbers

### Motivation for the set of real numbers $$\mathbb R$$

In the set of rational numbers $$\mathbb Q=\left\{q:=\frac xy, x\in \mathbb Z, y\in \mathbb Z\setminus\{0\}\right\}$$ it is possible to solve any linear equation of the form

$ax + b = c,\quad\quad a,b,c\in\mathbb Q,~a\neq 0.$

The major motivation for further extending the realm of numbers is the solvability of the equation $x^2=a,~~~~~~~(a\in \mathbb Q, a > 0)~~~~~~~~~~~~~~~~~~( * )$ which cannot be always solved by a rational number. For instance, while $$x^2=\frac {25}{16}$$ has the two rational solutions $$x_1=\frac 54,~x_2=-\frac 54$$, there is no rational solution of the equation $$x^2=2$$, which was discovered in 5th century BC in Ancient Greece. The main reason for the generally missing solution of the equation $$( * )$$ is the topological structure of rational numbers $$(\mathbb Q, + ,\cdot)$$. On the one hand, rational numbers are a dense set, that is for any rational number $$q\in\mathbb Q$$ and for any rational $$\epsilon > 0$$ (no matter how small), there exists at least another rational number $$p$$ such that the distance between $$p$$ and $$q$$ is smaller then $$\epsilon$$, formally $$|q - p| < \epsilon$$. On the other hand, there are surprisingly many "gaps between" rational numbers. One such gap is the solution of $$( * )$$ and there are infinitely many others, called irrational numbers. What is even more surprising from the topological point of view, the cardinality of the rational numbers is, (in some strict sense called countability), smaller then the cardinality of the irrational numbers. The rational numbers, together with the irrational numbers, constitute the set of real numbers.

If $$a > 0$$ is any real number, then the equation $$( * )$$ has always two real numbers as solutions. This is why: Consider the sequence $$(x_n)_{n\in\mathbb N}$$ of real numbers

$\begin{array}{ll} x_0 > 0 \in\mathbb Q\quad\text{(arbitrarily chosen)}\\ x_{n+1}:=\frac 12\left(x_n + \frac a{x_n}\right) \end{array}$

It can be easily shown that this sequence is monotonically decreasing and bounded and as such, it is convergent in the set of real numbers. The limit is called the square root of the number $$a$$ and denoted by $$\sqrt a$$. Because $$(-\sqrt a)^2=(\sqrt a)^2$$, also $$-\sqrt a$$ is a solution of $$( * )$$. Thus, the equation $$( * )$$ is always solvable and has at two real solutions: $$x_1=\sqrt a,~x_2=-\sqrt a$$.

Chapters: 1
Explanations: 2
Parts: 3 4
Sections: 5 6

Thank you to the contributors under CC BY-SA 4.0!

Github: