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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

enter image description here

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.)

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  • $\begingroup$ Solving this - or rather, just wanting to do this - was one of Descartes' fundamental contributions, leading to analytic geometry and paving the way for calculus. So it's hard to overstate how important this is. $\endgroup$
    – SRobertJames
    Jan 8 at 0:37

4 Answers 4

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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.

enter image description here

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    $\begingroup$ 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 ? $\endgroup$
    – temo
    May 1, 2012 at 13:12
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    $\begingroup$ 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. $\endgroup$
    – Steven-Owen
    May 1, 2012 at 19:10
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    $\begingroup$ Excuse me. What is $\lambda$? $\endgroup$
    – Red Banana
    Sep 10, 2013 at 9:37
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    $\begingroup$ @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. $\endgroup$
    – Christian Blatter
    Sep 10, 2013 at 13:28
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    $\begingroup$ Thanks! Is it then 1) Take any line 2) get the intersection point between that line and the vertical line above 1 3) get the intersection point between the vertical line above x 4) draw a line point in 2 till y 5) draw a line from point in 3, parallel to line drawn in 4? Or am I missing/confusing something? $\endgroup$
    – Matifou
    Apr 21, 2017 at 16:41
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Here is a hint and an answer ($a+b$, $a-b$, $\frac ab$ may be done too : see the link).

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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
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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)

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