Representing the multiplication of two numbers on the real line There is a simple way to graphically represent positive numbers $x$ and $y$ multiplied using only a ruler and a compass: Just draw the rectangle with height $y$ in top of it side $x$ (or vice versa), like this



But is there a way to draw the number $xy$ directly on the real line (i.e. not as an area on top of the real line) by using only some standard drawing means like using a compass, a ruler, a straightedge etc. (i.e. not multiplying $x$ and $y$ out and then putting the number $xy$ at right spot), like indicated above ?

(I think this question actually asks if the multiplication is representable as a composition of (the mathematical translation of) operations of drawing circles using a straightedge etc.)
You can do so, but if you want to represent the result of the multiplication again as a length you have to choose a unit. The ancient Greeks didn't come up with this idea; whence their products were always areas.

Responding to temo's comment: When the point $O\in g$ representing the number $0$ has been chosen the geometric construction for the sum of two numbers is scale invariant, as a consequence of $(\lambda x)+(\lambda y)=\lambda(x+y)$; but a similar identity for multiplication does not hold: $(\lambda x)\cdot(\lambda y)\ne\lambda(x\cdot y)$. You can test the effect in the figure by keeping $x$ and $y$ but choosing a new unit $1'$. The point $xy$ will now be at a different location.
This is an old question that I myself asked; I saw this thread and then thought about the answer, and here is what I came up with.

Suppose I have three segments that have common endpoint A: a segment with length 1 and endpoint B, a segment of length x with endpoint X, and a segment of length y with endpoint Y. All of these points are collinear.

Construct three circles centered at A that have a radius 1, x, and y. Construct C on the unit circle so that AC is perpendicular to AB. The points where line containing A and C intersect the other circles are called Cx and Cy in the obvious way.

Now create a point M on the line containing A and Y so that angle ACxX is congruent to ACyM. Since ACxX and ACyM are similar triangles, you can show that the length of AM is x*y. (This was inspired by the proportion guy posting elsewhere in this thread)