constructions using only compass and ruler

Question

Given a line segment of say length $L$ is it possible

1. Is it possible to create a line segment of length $\sqrt{L}$ ?
2. Is it possible to create a line segment of length $\frac{a}{b}L$, where $a,b$ are positive integers ?

The constructions only permit to use a ruler and a compass. My question arises due to a statement in the book Hardy and Wright, which is as follows :

Euclidean constructions by ruler and compass are equivalent analytically to solutions of a series of linear or quadratic equations.

Answer 1

For the square root , you have got sufficient answers in the comments itself.

For the ${a\over b}L$ part:

$1^{st}$ divide $L$ into $b$ equal parts . This is done by:

Take the line $L$ and draw $2$ parallel straight lines at any angle (use $60^\circ$ for convenience) going in opposite directions at the two ends.

<p>Take any length on the compass,  and make $b-1$ cuts on each of the lines, starting from either end of $L$. On line $A$ say the points are $A_1,A_2,\cdots, A_{b-1}$ and on line $B$ the points are $B_1,B_2,\cdots, B_{b-1}$.</p>

<p>Join $A_1\&amp;B_{b-1}\ ,\ A_2\&amp;B_{b-2}\ ,\cdots, A_{b-1}\&amp;B_1$. These $b-1$ lines divide $L$ into $b$ equal divisions. Now choose $a$ of these divisions to get ${a\over b}L$</p>

Answer 2

"Euclidean constructions by ruler and compass are equivalent analytically to solutions of a series of linear or quadratic equations."

In which case 1 is a solution to $x^2=L$ and 2 is a solution to $bx=aL$ both have to be possible. ... How? that's another question altogether.

The tradition way to do 1)

is to make extend L by a unit and construct a perpenduclar at the point where we began the extension. As the midpoint of the line L + 1 long, create a semi-circle with radius (L + 1)/2. Mark where the circle intersect the perpendicular we drew. The length to the base of the perpendicular is $\sqrt{L}$.

2) Extend your line b-fold so that you have a line of length $aL$. (Just copy it endpoint to endpoint $a$ times.) Then at an endpoint create a line $b$ units long and mark each unit. Connect the to endpoints and call the resulting line M. At each unit mark construct a line parallel to M. These lines will intersect the line that is $aL$ units long. The points of intersections cut the line $bL$ units long into $a$ equal lenght portions. Each portion is $\frac ab L$ long.

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My original stupendous error on constructing $\sqrt{L}$ got me to thinking how could one figure this out without being told and I realized:

As $\sqrt{L}$ is a solution to a quadratic equation we must involve a circle.

Let a circle be described by $x^2 + y^2=r^2$ where $x,y,r$ are all constructible values.

We want, say, $y = \sqrt{L}$ so $x = \sqrt{r^2 - L}$. We need to find an $r$ were this is a square of a previously constructable value. It'd be circular (pun unintended) if $r$ involve $\sqrt{L}$ so lets assume $r$ is something linear. i.e. $r = a + bL$.

So we need to have $r^2 - L = (a+bL)^2 - L = a^2 + (2ab - 1)L + b^2L^2$ be a square of constructable values. As we don't really know how split $L$ into $L = MN$ factors (that's pretty much the point of what we are attempting to solve) the only way $a^2 + VL + b^2L^2$ to be a perfect square is for $V = \pm 2ab$.

So $2ab - 1 = \pm 2ab$ or in other words $ab = 1/4$. The simplest and must symmetric solution would be $a = b = 1/2$.

Then we have $r = (L+1)*1/2$, $x = (L-1)*1/2$ and $y = \sqrt{L}$.

Now the construction seems apparent. If the radius of the circle is $(L + 1)*1/2$ the diameter is $L+1$. So take the line of length $L$ and extend it by one unit. Take the midpoint of that line and construct a circle with center at the midpoint and radius determined by the endpoint of the extended line.

$x = (L-1)*1/2$ is the distance from the center of the circle to the end of the original line of length $L$ just at the point where we extended it by a unit. Construct the perpendicular to the circle and that height is $y =\sqrt{L}$.

Answer 3

The answer in general is, starting from an unit length, and by ruler and compass alone no. For this to be possible you would need $a$, $b$, and $L$ to be of a special type of algebraic number termed a constructible number. Algebraic numbers are complex numbers that are nonzero roots of univariate polynomials in $x$ in the polynomial ring $\mathbb{Q}[x]$ (or $\mathbb{Z}[x]$ after clearing denominators). The constructible numbers then consist of rational numbers (roots of $bx-a$, where $b\neq0$, and $a$, $b\in\mathbb{Z}$, hence rationals are algebraic numbers of degree $1$), quadratic irrationals (irrational roots of a quadratic polynomial $ax^2 + bx + c$ with $a$, $b$, $c\in\mathbb{Z}$, hence algebraic numbers of degree $2$), plus any combination of these by addition, subtraction, multiplication, division, or extraction of square roots: so if $a$, $b$, and $c$ are constructible numbers with $b\neq0$, then $$ a\pm b,\quad\frac{a}{b},\quad\sqrt{a},\quad\frac{a}{b}\sqrt{c}$$ are constructible. The complex numbers can be shown to be constructible iff the real and imaginary parts are constructible, and then so are their conjugates.

This is the reason the Greeks couldn't `square the circle' as it involved constructing $\sqrt{\pi}$, and $\pi$ is a transcendental number; such numbers are unconstructible as are not the roots of polynomials in $\mathbb{Z}[x]$. The set of all constructible numbers $C$ forms a subfield of the field of algebraic numbers, with $C$ being the quadratic closure of $\mathbb{Q}$.