# 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.)

## 4 Answers

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. • Nice! It took me a moment to work that out, but it's actually pretty simple. – MathematicalOrchid May 1 '12 at 12:15
• Beautiful! But now I'm a little bit confused: What is the underlying cause that enables us to do addition without the choice of a unit, but not multiplication ? – temo May 1 '12 at 13:12
• I saw this come up before when doing constructions with ruler and compass but didn't understand why a unit had to be picked until I thought about it in terms of groups. It's necessary because a priori we don't know if $x$ is bigger or smaller than the multiplicative identity, or unit. Let us consider only positive numbers for now. If $x$ is bigger than the unit, then multiplication by $x$ must make the number bigger. Likewise, if it is smaller, multiplication makes the number smaller. For addition, the identity of the group is 0 and it is easier to say in some sense a number's relation to 0. – Steven-Owen May 1 '12 at 19:10
• Excuse me. What is $\lambda$? – Billy Rubina Sep 10 '13 at 9:37
• @Gustavo Bandeira: $\lambda$ is an arbitrary scaling factor. The identity $(\lambda x)+(\lambda y)=\lambda(x+y)$ means that addition of lengths behaves properly under scaling. – Christian Blatter Sep 10 '13 at 13:28

Here is a hint and an answer ($a+b$, $a-b$, $\frac ab$ may be done too : see the link).

Maybe the Greek could also come up with it, not necessarely refering to area, but in the disguise of continued ratio (not the same as continued proportion):

A : B : C = A : C


So for the multiplication, we would start with:

1 : x

1 : y


By geometric reasoning, for example in Blatters construction, because we have similar triangles, we could see that 1:y is in equal proportion as x:xy:

1 : x

x : xy


So if we remove the middle term we get:

1 : xy


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)