Geometric model of real numbers

Question

I'm looking for a way to multiply real numbers using only geometric techniques. Suppose we already know how to halve a real number and multiply a real number with a natural in a geometric way.

For the multiplication of two real I thought to do this:

$ a=[a]+\left \{ a \right \} $

$ b=[b]+\left \{ b \right \} $

in square brackets indicate the integer part (segment equal or multiple of the whole unit) and the other indicates the fractional part (shorter segment of the unit). Then:

$ ab=[a][b]+[a]\left \{ b \right \}+\left \{ a \right \}[b]+\left \{ a \right \}\left \{ b \right \} $

The first 3 addenda can be "calculated" geometrically, the last in general no.

$ \left \{ a \right \} $ and $ \left \{ b \right \} $ are both shorter segments of the unit.

Then I begin to double both $ n_1 $ times until both have passed the unit. I'll get $ a'=2^{n_1}\left \{ a \right \} $ and $ b'= 2^{n_1}\left \{ b \right \} $

At this point I can continue building the product-segment in this way (since I know how to halve a number):

$ \left \{ a \right \}\left \{ b \right \}=\frac{a'b'}{2^{2n_1}}=\frac{[a'][b']}{2^{2n_1}}+\frac{[a']\left \{ b' \right \}}{2^{2n_1}}+\frac{\left \{ a' \right \}[b']}{2^{2n_1}}+\frac{\left \{ a' \right \}\left \{ b' \right \}}{2^{2n_1}} $

again I do not know how to "calculate" the last numerator but I can do the same thing by getting:

$ \left \{ a' \right \}\left \{ b' \right \}=\frac{a''b''}{2^{2n_2}}=\frac{[a''][b'']}{2^{2n_2}}+\frac{[a'']\left \{ b'' \right \}}{2^{2n_2}}+\frac{\left \{ a'' \right \}[b'']}{2^{2n_2}}+\frac{\left \{ a'' \right \}\left \{ b'' \right \}}{2^{2n_2}} $

with $ a''=2^{n_2}\left \{ a' \right \} $ and $ b''= 2^{n_2}\left \{ b' \right \} $ and again only the last numerator will have to be expanded and so on... (until one of the two fractional parts become null or until infinite if necessary and if I want infinite precision)

Do you have better methods?

Thanks and sorry if my English is not very good.

Answer 1

There is a better way, I think. First, you have to be given a segment of unit length, otherwise multiplication doesn't make sense. Now if you have $a$ and $b$, you can construct $h$ so that $h^2=a\cdot b$, using the geometric mean theorem. After that, you use the same $h=CD$ and your unit segment $1$ to find the $x$ satisfying $h^2=x\cdot1$: the unit segment is then $DB$, you find the circle with midpoint $M$ on the line through $B$ and $D$ ($M$ is where the perpendicular bisector of $BC$ hits the line through $D$ and $B$), the circle will have a second intersection point $A$ with the line through $BD$, so $x=a\cdot b/1$ will be $AD$.

Answer 2

So now I can also build in a similar way the inverse, and so the number $a/b$, as shown in the next image.

1/a

Right?