Multiplication of variables with a compass when a segment of length 1 is not given.

Question

"Suppose that length a and b are given. Construct sqrt(ab) (Not that a segment of length 1 is not given)." I know how to multiply ab together and then take the square root of ab if a length of segment 1 is given but not when it is not.

Answer 1

It seems you’re familiar with constructing a length that is the product of two lengths, and with constructing a length that is the square root of another length. Those constructions are fine, but they need to refer to a given unit length to be meaningful.

The product of two lengths is more naturally represented as an area. (“Naturally” in the sense that it doesn’t depend on a unit length being given. Also in that it’s familiar to most people from their primary education.) Conversely, a square root is more naturally represented as obtaining a length from an area—specifically, the side length of a square with the given area.

So an alternative construction would be:

1. Construct $ab$ as a rectangle.
2. Square the rectangle.
3. The side length of the square is $\sqrt{ab}$.

Here’s a demonstration. The lengths $a$ and $b$ can be adjusted freely, although my construction seems to have relied on $a$ being shorter than $b$, so the square disappears if that stops being true!

Answer 2

Without a unit length, the square root of a length is undefined.^[1] So let’s throw our hands up in the air and say we might as well make up our own unit length. Then we can construct the lengths $\sqrt a$ and $\sqrt b$ according to this unit.

Now imagine that your teacher/professor has secretly kept the “real” unit length for this problem, and it isn’t the same as ours. Theirs is 1; ours is $x$. When we show the professor the lengths we constructed, they say we “actually” constructed $\sqrt{xa}$ and $\sqrt{xb}$.^[2]

But hang on… if we multiply our constructed lengths together to get (what we thought was) $\sqrt {ab}$ and show that to the professor, they’ll measure it and say it’s “really” $\sqrt{xa}\times\sqrt{xb}=x\sqrt{ab}$… right?

Nope! At this point we remember that, like the square root, multiplication of lengths isn’t defined without a unit length either. If we multiply two lengths using unit $x$, and the professor looks at the answer using their “real” unit 1, they’d tell us our answer is too short by a factor of $x$.^[3] (Or, well, too long if $x<1$.) So in this case, what we thought was going to look (to them) like $x\sqrt{ab}$ actually turns out to measure exactly $\sqrt{ab}$ using their “real” unit!

In short, the product $\sqrt{ab}$ comes out the same, no matter what arbitrary unit length we chose.

A visual demonstration. Adjust lengths $a$, $b$, and the unit as you like. The product $ab$ and the square roots $\sqrt{a}$ and $\sqrt{b}$ vary depending on the unit, but $\sqrt{ab}$ doesn’t.

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[1] Consider: I have a 25 cm line, and using cm as my unit, I say its square root is 5 cm. But if you measure the same line to be 0.25 m, you’ll say its square root is 0.5 m… ten times longer than the original line! (And let’s not mention our American friend who insists the line is 10″…)

[2] If our unit lengths are in the ratio $1:x$, our square roots will be in the ratio $1:\sqrt x$.

[3] For example, I have my 2.5 cm line, and I multiply it by another line that’s 1 cm long. Since my units are cm, I get 1 × 2.5 = 2.5 cm. But you, using mm as your unit this time, insist that it should be 10 × 25 = 250 mm (25 cm). Our units are in the ratio $10:1$, but our products are in the ratio $1:10$.

Answer 3

Note that the question does not ask you to construct a length $ab.$ It merely asks for $\sqrt{ab}.$

The expression $\sqrt{ab}$ is also known as the geometric mean of $a$ and $b.$

Usually when we say "mean" of two numbers we are talking about the arithmetic mean, which is a number exactly halfway between the other two numbers in an arithmetic sequence: $m$ is the arithmetic mean of $a$ and $b$ if and only if there is an amount $d$ such that $m = a + d$ and $b = m + d$; that is, $$ (a, m, b) = (a, a + d, a + 2d). $$

"Geometric mean of $a$ and $b$" is a bit different: it describes a number that is halfway between $a$ and $b$ in a geometric sequence. So $g$ is the geometric mean of $a$ and $b$ if and only if there is a ratio $r$ such that $g = ra$ and $b = rg$; that is, $$ (a, g, b) = (a, ra, r^2 a). $$

Note that you don't need to have access to a unit length in order for the geometric mean of two lengths to have meaning, because the ratio of lengths of two segments is determined by those two segments alone, without reference to the length of a unit segment.

Now suppose you knew how to construct the geometric mean of $a$ and $b$ given segments of length $a$ and $b,$ without referring to any other previously given segment. Then, given a unit length segment and a segment of length $b,$ you could use this exact same procedure to construct a segment of length $\sqrt b,$ because $\sqrt b$ is the geometric mean of $1$ and $b.$

This would work because $1, \sqrt b, b$ is a geometric sequence.

Now what you can think about is this: knowing a construction for the square root of a length, how does it work? Does it, in fact, take whatever two input lengths you give it and find a length between the other two in geometric sequence, so that if one of the lengths happens to be a unit length then the result will be the square root of the other length? What if neither of the lengths you give to the construction is the unit length; does it still construct a geometric sequence of lengths, and does it find the geometric mean of the two input lengths?