Draw an irrational number on the number line (without pythagoras sentence)

Question

Let's say im a guy for ancient Greece and I only have a string and a pencil. And I want to draw a line, the width of the line is the square root of 6. And I only know how to draw a line in the width of real numbers. I've checked out the https://en.wikipedia.org/wiki/Spiral_of_Theodorus but I can't use Pythagoras theorem.

Sorry I couldn't be more specific, if you still don't understand, I'll do my best to explain it again.

Answer 1

Draw two adjacent segments of size $2$ and $3$. Using the combined segment as diameter, draw a semicircle. Now draw a perpendicular at the point where the two segments meet. That perpendicular defines a segment of length $\sqrt 6$ where it meets the semicircle.

This is the geometric equivalent of the right triangle altitude theorem that says that $h^2=mn$, where $h$ is the altitude of a right triangle with respect to the hypothenuse and $m$ and $n$ are the projections of the sides onto the hypothenuse.

Answer 2

With a string and a pencil, you can measure "one unit", that being the length of the string, or a portion of it. Then you can make a triangle "one unit by one unit." Further, this triangle can be a right triangle since you can perpendicularly bisect a line segment to make the second side. Then you have a hypotenuse side that is $\sqrt 2$ units in length. Because it is isosceles, you can cut it in half along the right angle and get two pieces that have leg lengths of $\frac 1{\sqrt 2}=\frac {\sqrt 2}2$, and so you can get legs as small as desired that are a multiple of $\sqrt 2$.

Next, you can make a second right triangle whose short leg is $\sqrt 2$, and whose hypotenuse is $2\sqrt 2$. Then the long leg has length $\sqrt 6$. Now you have your line of the given length.

The construction below attempts to demonstrate this process. Begin with one line (black), draw a line perpendicular to it (anywhere), draw a "unit" circle centered at the point of intersection, then the chord (green) between two consecutive intersections of line and circle has length $\sqrt 2$. Draw a circle having this radius (purple), then the diameter will have length $2\sqrt 2$. Draw a perpendicular to the chord (blue) that intersects the (purple) circle's center. Draw a circle (light blue) having radius equal to the diameter of the previous circle (purple), or $2\sqrt 2$. The chord where the (blue) line intersects the (light blue) circle has length $2\sqrt 6$, since the one side associated with the (green, light green, blue) triangle has length $\sqrt 6$.