We already know that a quadratic extension has degree [math]2[/math].  Then how about iterated quadratic extension?[br][br]It turns out that the degree of any iterated quadratic extension must be a power of [math]2[/math] i.e. the number of parameters needed to describe all elements in an iterated quadratic extension is [math]2^k[/math] for some positive integer [math]k[/math].[br][br]([b]Note[/b]: Some linear algebra is needed to prove the above fact.)

Given a real number [math]\beta[/math], if it is a root of a polynomial equation [math]f\left(x\right)=0[/math] whose coefficients are in [math]\mathbb{Q}[/math], we say that [color=#0000ff][b][math]\beta[/math] is algebraic over [math]\mathbb{Q}[/math][/b][/color].[br][br]Now we are going to simply the polynomial equation by factorising so as to lower its degree. Now, we may assume that [math]f\left(x\right)[/math] has the lowest degree. It is usually called the [b][color=#0000ff]irreducible polynomial of [math]\beta[/math] over [math]\mathbb{Q}[/math][/color][/b].[br][br]The degree of such irreducible polynomial is called the [b][color=#0000ff]degree of [math]\beta[/math] over [math]\mathbb{Q}[/math][/color][/b].

Here is a very useful result about degrees:[br][br]If [math]E[/math] and [math]F[/math] be two field extensions over [math]\mathbb{Q}[/math] such that [math]E\subset F[/math], then the degree of [math]F[/math] over [math]\mathbb{Q}[/math] is divisible by the degree of [math]E[/math] over [math]\mathbb{Q}[/math].[br][br]Let [math]\beta[/math] be a real number that is algebraic over [math]\mathbb{Q}[/math]. If it is a constructible number, it must lie in an iterated quadratic extension [math]F[/math] over [math]\mathbb{Q}[/math]. Let [math]E[/math] be the field extension just large enough to contain [math]\beta[/math] and [math]\mathbb{Q}[/math]. It can be shown that the degree of [math]E[/math] over [math]\mathbb{Q}[/math] equals the degree of [math]\beta[/math] over [math]\mathbb{Q}[/math]. Therefore, [math]E\subset F[/math] and by above, the degree of [math]F[/math] over [math]\mathbb{Q}[/math], which is a power of [math]2[/math], is divisible by the degree of [math]\beta[/math] over [math]\mathbb{Q}[/math]. In other words, the degree of [math]\beta[/math] over [math]\mathbb{Q}[/math] is also a power of [math]2[/math].[br][br]Now, we can rephrase the above important results as [b][color=#0000ff]the main theorem[/color][/b]:[br][br][b][color=#ff7700]Given a real number [math]\beta[/math] that is algebraic over [math]\mathbb{Q}[/math]. If the degree of [math]\beta[/math] over [math]\mathbb{Q}[/math] is not a power of [math]2[/math], then [math]\beta[/math] is not a constructible number.[br][br][/color][/b][br]([b]Note[/b]: the detailed proof of the above theorem is beyond the scope of this course.)