As we know, every Euclidean construction starts with [math]\left(0,0\right)[/math] and [math]\left(1,0\right)[/math]. Therefore, by definition, [math]0[/math] and [math]1[/math] are constructible numbers i.e. they are in [math]X[/math].[br][br]Since [math]X[/math] is a field, it must contain all rational numbers i.e. [math]\mathbb{Q}[/math] is a subset of [math]X[/math].[br][br]Now, we obtain [math]\sqrt{2}[/math] by constructing the diagonal of a unit square. Therefore, [math]\sqrt{2}[/math] is constructible i.e. it is in [math]X[/math].[br][br]Obviously, [math]\sqrt{2}[/math] is not a rational number. We can "mix" it with rational numbers using rational operations to generate a field just large enough to contain [math]\sqrt{2}[/math] and [math]\mathbb{Q}[/math]. It is called the [color=#0000ff][b]field extension of [math]\mathbb{Q}[/math] by adjoining [math]\sqrt{2}[/math][/b][/color], denoted by [math]\mathbb{Q}\left(\sqrt{2}\right)[/math].

It turns out that it is very easy to describe the real numbers in [math]\mathbb{Q}\left(\sqrt{2}\right)[/math]: It must be of the form [math]a+b\sqrt{2}[/math], where [math]a[/math] and [math]b[/math] are in [math]\mathbb{Q}[/math].[br][br]For example, we consider [math]\frac{1+\sqrt{2}}{3-4\sqrt{2}}[/math] in [math]\mathbb{Q}\left(\sqrt{2}\right)[/math]. We can easily bring it to the form [math]a+b\sqrt{2}[/math] by multiplying both the numerator and denominator by [math]3+4\sqrt{2}[/math].[br][br]Note that there are two free parameters [math]a[/math] and [math]b[/math] that describe that whole field extension [math]\mathbb{Q}\left(\sqrt{2}\right)[/math]. We say that the [b][color=#0000ff]degree[/color][/b] of the field extension [math]\mathbb{Q}\left(\sqrt{2}\right)[/math] is 2.  Sometimes it is called a [b][color=#0000ff]quadratic extension[/color][/b] over [math]\mathbb{Q}[/math].[br][br]More generally, given any field [math]F[/math], if [math]\alpha[/math] is in [math]F[/math] such that [math]\sqrt{\alpha}[/math] is not in [math]F[/math], then [math]F\left(\sqrt{\alpha}\right)[/math] is a [b][color=#0000ff]quadratic extension over [math]F[/math][/color][/b] and all numbers in such quadratic extension are in the form [math]a+b\sqrt{\alpha}[/math], where [math]a[/math] and [math]b[/math] are in [math]F[/math].[br][br]([b]Note[/b]: If you know a bit linear algebra, it is not hard to see that a quadratic extension can be regarded as a two-dimensional vector space over [math]F[/math].)

We already know that we can do rational operations and taking square root by Euclidean constructions. A natural question to answer is: Can Euclidean constructions do more? [br][br]It turns out that with the help of coordinate geometry, it can be shown that these are pretty much all Euclidean constructions can do and nothing more...[br][br]Take any constructible number [math]x[/math], the point [math]\left(x,0\right)[/math] can be produced by a Euclidean construction. Suppose such construction has a number of steps. We can associate each step with a field that contains the coordinates of points constructed in such step as follows:[br][br]Initially, we are given the points [math]\left(0,0\right)[/math] and [math]\left(1,0\right)[/math]. So the field that just contains [math]0[/math] and [math]1[/math] is [math]\mathbb{Q}[/math].[br][br]Suppose an intersection point obtained in a certain construction step that has coordinate [math]\sqrt{\alpha_1}[/math] where [math]\alpha_1[/math] in [math]\mathbb{Q}[/math] and it is not in [math]\mathbb{Q}[/math] (for example, [math]\sqrt{2}[/math]). Then we consider the quadratic field extension [math]F_1=\mathbb{Q}\left(\sqrt{\alpha_1}\right)[/math] that can contain the coordinates of newly-constructed point.[br][br]Subsequently, if another intersection point obtained that has coordinate [math]\sqrt{\alpha_2}[/math] where [math]\alpha_2[/math] in [math]F_1[/math] and it is not in [math]F_1[/math], we consider the quadratic field extension [math]F_2=F_1\left(\sqrt{\alpha_2}\right)[/math] that can contain the coordinates of newly-constructed point.[br][br]This procedure continues until the last construction step, whose associated field extension is large enough to contain [math]x[/math]. Moreover, we obtain a sequence of fields [math]\mathbb{Q},F_1,F_2,\ldots,F_n[/math] such that each field is the quadratic extension of the previous one i.e. the fields are nested like the diagram shown below.

We can express the sequence of fields mentioned above as follows:[br][br][math]F_1=\mathbb{Q}\left(\sqrt{\alpha_1}\right)[/math][br][math]F_2=F_1\left(\sqrt{\alpha_2}\right)=\mathbb{Q}\left(\sqrt{\alpha_1},\sqrt{\alpha_2}\right)[/math][br][math]F_3=F_2\left(\sqrt{\alpha_3}\right)=\mathbb{Q}\left(\sqrt{\alpha_1},\sqrt{\alpha_2},\sqrt{\alpha_3}\right)[/math][br][math]\vdots[/math][br][math]F_n=F_{n-1}\left(\sqrt{\alpha_n}\right)=\mathbb{Q}\left(\sqrt{\alpha_1},\sqrt{\alpha_2},\ldots,\sqrt{\alpha_n}\right)[/math][br][br]There are the so-called [color=#0000ff][b]iterated quadratic extensions of [math]\mathbb{Q}[/math][/b][/color].[br][br]Here is the main result:[br][br][b][color=#ff7700]Any constructible number is an element of some iterated quadratic extension of [math]\mathbb{Q}[/math]. Conversely, each element of any iterated quadratic extension is a constructible number. Therefore, the set of all constructible number [math]X[/math] is the union of all iterated quadratic extensions of [math]\mathbb{Q}[/math].[/color][/b]