Editing Square root (section)

==Geometric construction of the square root==
[[File:SqrtGeom.gif|thumb|[[Straightedge and compass construction|Constructing]] the length <math>x = \sqrt{a}</math>, given the <math>a</math> and the unit length]]
[[File:Spiral_of_Theodorus.svg|thumb|The [[Spiral of Theodorus]] up to the triangle with a hypotenuse of {{math|{{radic|17}}}}]]
[[File:Root_rectangles_Hambidge_1920.png|thumb|[[Jay Hambidge]]'s construction of successive square roots using [[root rectangle]]s]]
The square root of a positive number is usually defined as the side length of a [[square]] with the [[area]] equal to the given number. But the square shape is not necessary for it: if one of two [[similarity (geometry)|similar]] [[Euclidean plane|planar Euclidean]] objects has the area ''a'' times greater than another, then the ratio of their linear sizes is <math>\sqrt{a}</math>.

A square root can be constructed with a compass and straightedge. In his [[Euclid's Elements|Elements]], [[Euclid]] ([[floruit|fl.]] 300&nbsp;BC) gave the construction of the [[geometric mean]] of two quantities in two different places: [http://aleph0.clarku.edu/~djoyce/java/elements/bookII/propII14.html Proposition II.14] and [http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI13.html Proposition VI.13]. Since the geometric mean of ''a'' and ''b'' is <math>\sqrt{ab}</math>, one can construct <math>\sqrt{a}</math> simply by taking {{math|1=''b'' = 1}}.

The construction is also given by [[Descartes]] in his ''[[La Géométrie]]'', see figure 2 on [http://historical.library.cornell.edu/cgi-bin/cul.math/docviewer?did=00570001&seq=12&frames=0&view=50 page 2]. However, Descartes made no claim to originality and his audience would have been quite familiar with Euclid.

Euclid's second proof in Book VI depends on the theory of [[Similar triangles#Similar triangles|similar triangles]]. Let AHB be a line segment of length {{math|''a'' + ''b''}} with {{math|1=AH = ''a''}} and {{math|1=HB = ''b''}}. Construct the circle with AB as diameter and let C be one of the two intersections of the perpendicular chord at H with the circle and denote the length CH as ''h''. Then, using [[Thales' theorem]] and, as in the [[Pythagorean theorem#Proof using similar triangles|proof of Pythagoras' theorem by similar triangles]], triangle AHC is similar to triangle CHB (as indeed both are to triangle ACB, though we don't need that, but it is the essence of the proof of Pythagoras' theorem) so that AH:CH is as HC:HB, i.e. {{math|1= ''a''/''h'' = ''h''/''b''}}, from which we conclude by cross-multiplication that {{math|1= ''h''<sup>2</sup> = ''ab''}}, and finally that <math>h = \sqrt{ab}</math>. When marking the midpoint O of the line segment AB and drawing the radius OC of length {{math|(''a'' + ''b'')/2}}, then clearly OC > CH, i.e. <math display=inline>\frac{a + b}{2} \ge \sqrt{ab}</math> (with equality if and only if {{math|1=''a'' = ''b''}}), which is the [[inequality of arithmetic and geometric means|arithmetic–geometric mean inequality for two variables]] and, as noted [[Square root#Computation|above]], is the basis of the [[Greek Mathematics|Ancient Greek]] understanding of "Heron's method".

Another method of geometric construction uses [[right triangle]]s and [[Mathematical induction|induction]]: <math>\sqrt{1}</math> can be constructed, and once <math>\sqrt{x}</math> has been constructed, the right triangle with legs 1 and <math>\sqrt{x}</math> has a [[hypotenuse]] of <math>\sqrt{x + 1}</math>. Constructing successive square roots in this manner yields the [[Spiral of Theodorus]] depicted above.