# Compass-and-straightedge construction of the square root of a given line?

Given

1. A straight line of arbitrary length
2. The ability to construct a straight line in any direction from any starting point with the "unit length", or the length whose square root of its magnitude yields its own magnitude.

Is there a way to geometrically construct (using only a compass and straightedge) the a line with the length of the square root of the arbitrary-lengthed line? What is the mathematical basis?

Also, why can't this be done without the unit line length?

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I cannot understand your last remark. If the lenght of a segment is 4 units, its square root is 2 units, that is its half; but if it is 16 units, its square root is 4 units, that is a quarter. Thus to make sense of the notion of a square root you must specify which is the unit you use, which is tantamount to have a unit length segment. –  mau Jul 26 '10 at 7:32
@mau I guess it was a bad way of asking why the square root along one dimension is fundamentally different, geometrically, then, say, bisection, which can be done without a unit-length segment. –  Justin L. Jul 26 '10 at 7:38
Maybe because a square root does not have a geometric meaning in one dimension (and maybe not an inherent geometric meaning at all), whereas bisection has to do with ratios, which have geometric interpretation. Square roots are related to the geometric mean, which does have geometric meaning. –  Isaac Jul 26 '10 at 7:40
So probably the best answer is that square root is not e dimensional invariant; you are mixing lengths (the square root itself) and areas (the original number, that you must see as an area). Greek geometry was very strict in it; only with the rise of algebra such distinctions were lost. –  mau Jul 26 '10 at 7:43
This is also a proposition of Euclid. Isaac's and mau's constructions are similar. –  Larry Wang Jul 26 '10 at 8:00

If you have a segment $AB$, place the unit length segment on the line where $AB$ lies, starting with $A$ and in the direction opposite to $B$; let $C$ be the other point of the segment. Now draw a semicircle with diameter $AC$ and the perpendicular to $B$; this line crosses the semicircle in a point $D$. $AD$ is then the square root of $AB$.

$\triangle BCD$ is a right triangle, like $\triangle ACD$ and $\triangle ABD$; all of these are similar, so you find out that $AC/AD = AD/AB$. But $AC=1$, so $AD = \sqrt{AB}$.

See the drawing below:

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yup, I transliterate from Italian (where we talk about "triangolo rettangolo"). Thanks for pointing it out. –  mau Jul 26 '10 at 8:52