Descartes’ Geometry of Square Roots

[kuruman]

It's not quite clear why this bears the name of Plato who was a philosopher rather than a a mathematician ...

The word philosophy literally means love of wisdom. In Plato's time, the distinction between a mathematician and a philosopher was not as sharp as it is today. The perceived connection of Plato to mathematics, which in pre-algebra days meant geometry, is illustrated by the traditional belief in an inscription at the entrance of Plato's Academy that read "Let no one ignorant of geometry enter."

 

[bob012345]

Note that if $h$ has unit length, $a$ and $b$ are reciprocals.

 

[kuruman]

Note that if $h$ has unit length, $a$ and $b$ are reciprocals.

I wouldn't put it quite that way. Starting with $h^2=ab,$ which we assume to be known, we can rewrite it in two ways: $$\begin{align} & \frac{h}{a}=\sqrt{\frac{b}{a}} \\
& \frac{h}{b}=\sqrt{\frac{a}{b}}. \end{align}$$ Simply put, it doesn't matter which of $a$ and $b$ you call the unit and which the given segment. Segment $h$ will be the square root of whatever segment you chose as "given" measured with a ruler subdivided in the units of whatever segment you chose as "unit". The two equations above say that "$h$ expressed in units of $a$ is the reciprocal of $h$ expressed in units of $b.$ You can see why that is if you multiply the two equations.

The only way for $h$ to have unit length, i.e. equal to one of the other two segments, is if both $a$ and $b$ are unit segments. This is demanded by the construction (see red circle in post #25.) So I would say "Note that $h$ has unit length if and only if $a$ and $b$ are unit segments in which case all three segments are equal to the radius of the unit circle."

 

[neilparker62]

The word philosophy literally means love of wisdom. In Plato's time, the distinction between a mathematician and a philosopher was not as sharp as it is today. The perceived connection of Plato to mathematics, which in pre-algebra days meant geometry, is illustrated by the traditional belief in an inscription at the entrance of Plato's Academy that read "Let no one ignorant of geometry enter."

https://archive.org/details/nicolaicopernici00cope_1/page/n7/mode/2up

 

[bob012345]

I wouldn't put it quite that way. Starting with $h^2=ab,$ which we assume to be known, we can rewrite it in two ways: $$\begin{align} & \frac{h}{a}=\sqrt{\frac{b}{a}} \\
& \frac{h}{b}=\sqrt{\frac{a}{b}}. \end{align}$$ Simply put, it doesn't matter which of $a$ and $b$ you call the unit and which the given segment. Segment $h$ will be the square root of whatever segment you chose as "given" measured with a ruler subdivided in the units of whatever segment you chose as "unit". The two equations above say that "$h$ expressed in units of $a$ is the reciprocal of $h$ expressed in units of $b.$ You can see why that is if you multiply the two equations.

The only way for $h$ to have unit length, i.e. equal to one of the other two segments, is if both $a$ and $b$ are unit segments. This is demanded by the construction (see red circle in post #25.) So I would say "Note that $h$ has unit length if and only if $a$ and $b$ are unit segments in which case all three segments are equal to the radius of the unit circle."

In the case I just mentioned, neither $a$ or $b$ are unit length except in the case where the semi-circle is bisected. Here, $a$ and $b$ are in a forced relationship so the position of $h$ is fixed. In this example $a=1/2$, $b=2$.

 

[kuruman]

In the case I just mentioned, neither $a$ or $b$ are unit length except in the case where the semi-circle is bisected. Here, $a$ and $b$ are in a forced relationship so the position of $h$ is fixed. In this example $a=1/2$, $b=2$.

It looks like that you lost track of what this construction is all about. It finds the square root of one of the segments, $a$ or $b$ that, together, make up the diameter.

Note that $\sqrt{1/2}=0.707$ and $\sqrt{2}=1.41.$ Clearly, $h=1$ is the square root of neither $a$ nor $b$.

One of the two segments $a$ or $b$, must be the unit of length otherwise the construction will not work.

Here is the correct way to look at your construction.

Case I - $a=$ "0.5" is the unit segment.
In this system of units, $h$ is twice as long as $a$, i.e. $2$ units and $b$ is four times as long As $a$, i.e. $4$ units.
Clearly $h$ is the square root of $b$.

Case II - $b=$ "2" is the unit segment.
In this system of units, $h$ is half as long as $b$, i.e. $1/2$ units and $a$ is one quarter times as long as $b$, i.e. $1/4$ units.
Clearly $h$ is the square root of $a$.

See how it works?

 

[bob012345]

It looks like that you lost track of what this construction is all about. It finds the square root of one of the segments, $a$ or $b$ that, together, make up the diameter.

Note that $\sqrt{1/2}=0.707$ and $\sqrt{2}=1.41.$ Clearly, $h=1$ is the square root of neither $a$ nor $b$.

One of the two segments $a$ or $b$, must be the unit of length otherwise the construction will not work.

Here is the correct way to look at your construction.

Case I - $a=$ "0.5" is the unit segment.
In this system of units, $h$ is twice as long as $a$, i.e. $2$ units and $b$ is four times as long As $a$, i.e. $4$ units.
Clearly $h$ is the square root of $b$.

Case II - $b=$ "2" is the unit segment.
In this system of units, $h$ is half as long as $b$, i.e. $1/2$ units and $a$ is one quarter times as long as $b$, i.e. $1/4$ units.
Clearly $h$ is the square root of $a$.

See how it works?

We seem to have different perspectives on this thread which is ok. From my perspective, we were discussing Descartes’ problem where the goal is to construct $h=\sqrt{b}$ but also the general case where $a$ and $b$ can be different lengths not considered unity and $h=\sqrt{ab}$ where $h$ is the geometric mean of the lengths as in the Wallis version. The last example I posted was an example of the latter. I agree with your comments above when interpreted as discussing Descartes version. Sorry for any confusion.

 

[kuruman]

We seem to have different perspectives on this thread which is ok. From my perspective, we were discussing Descartes’ problem where the goal is to construct $h=\sqrt{b}$ but also the general case where $a$ and $b$ can be different lengths not considered unity and $h=\sqrt{ab}$ where $h$ is the geometric mean of the lengths as in the Wallis version. The last example I posted was an example of the latter. I agree with your comments above when interpreted as discussing Descartes version. Sorry for any confusion.

I think we can both agree on the following:
1. Using any number of methods, it can be shown that $h=\sqrt{ab}.$
2. If either one of the segments, $a$ or $b$, is defined as the unit segment, then $h$ represents the square root of the other segment in the chosen system of units.

 

[lemonfrostt]

Euclid constructs the arrangement in Book 6, Prop 13 of Elements.

Book 6
Prop 13​

To find the (straight-line) in mean proportion to two given straight-lines.

Let AB and BC be the two given straight-lines. So it is required to find the (straight-line) in mean proportion to AB and BC.

Let (AB and BC) be laid down straight-on (with respect to one another), and let the semi-circle ADC have been drawn on AC [Prop. 1.10]. And let BD have been drawn from (point) B, at right-angles to AC [Prop. 1.11]. And let AD and DC have been joined.

And since ADC is an angle in a semi-circle, it is a right-angle [Prop. 3.31]. And since, in the right-angled triangle ADC, the (straight-line) DB has been drawn from the right-angle perpendicular to the base, DB is thus the mean proportional to the pieces of the base, AB and BC [Prop. 6.8 corr.].

Thus, DB has been found (which is) in mean proportion to the two given straight-lines, AB and BC. (Which is) the very thing it was required to do.

Spoiler: Elements, Book 3, Prop 31

Book 3
Prop 31​

In a circle, the angle in a semi-circle is a right-angle, and that in a greater segment (is) less than a right-angle, and that in a lesser segment (is) greater than a right- angle. And, further, the angle of a segment greater (than a semi-circle) is greater than a right-angle, and the angle of a segment less (than a semi-circle) is less than a right-angle.

Let ABCD be a circle, and let BC be its diameter, and E its center. And let BA, AC, AD, and DC have been joined. I say that the angle BAC in the semi-circle BAC is a right-angle, and the angle ABC in the segment ABC, (which is) greater than a semi-circle, is less than a right-angle, and the angle ADC in the segment ADC, (which is) less than a semi-circle, is greater than a right-angle.

Let AE have been joined, and let BA have been drawn through to F.

And since BE is equal to EA, angle ABE is also equal to BAE [Prop. 1.5]. Again, since CE is equal to EA, ACE is also equal to CAE [Prop. 1.5]. Thus, the whole (angle) BAC is equal to the two (angles) ABC and ACB. And F AC, (which is) external to triangle ABC, is also equal to the two angles ABC and ACB [Prop. 1.32]. Thus, angle BAC (is) also equal to F AC. Thus, (they are) each right-angles. [Def. 1.10]. Thus, the angle BAC in the semi-circle BAC is a right-angle.

And since the two angles ABC and BAC of triangle ABC are less than two right-angles [Prop. 1.17], and BAC is a right-angle, angle ABC is thus less than a right-angle. And it is in segment ABC, (which is) greater than a semi-circle.

And since ABCD is a quadrilateral within a circle, and for quadrilaterals within circles the (sum of the) opposite angles is equal to two right-angles [Prop. 3.22] [angles ABC and ADC are thus equal to two right-angles], and (angle) ABC is less than a right-angle. The remaining angle ADC is thus greater than a right-angle. And it is in segment ADC, (which is) less than a semi-circle.

I also say that the angle of the greater segment, (namely) that contained by the circumference ABC and the straight-line AC, is greater than a right-angle. And the angle of the lesser segment, (namely) that contained
by the circumference AD[C] and the straight-line AC, is less than a right-angle. And this is immediately apparent. For since the (angle contained by) the two straight-lines BA and AC is a right-angle, the (angle) contained by the circumference ABC and the straight-line AC is thus greater than a right-angle. Again, since the (angle contained by) the straight-lines AC and AF is a right-angle, the (angle) contained by the circumference AD[C] and the straight-line CA is thus less than a right-angle.

Thus, in a circle, the angle in a semi-circle is a right-angle, and that in a greater segment (is) less than a right-angle, and that in a lesser [segment] (is) greater than a right-angle. And, further, the [angle] of a segment greater (than a semi-circle) [is] greater than a right-angle, and the [angle] of a segment less (than a semi-circle) is less than a right-angle. (Which is) the very thing it was required to show.

Spoiler: Elements, Book 6, Prop 8

Book 6
Prop 8​

If, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another.

Let ABC be a right-angled triangle having the angle BAC a right-angle, and let AD have been drawn from A, perpendicular to BC [Prop. 1.12]. I say that triangles ABD and ADC are each similar to the whole (triangle) ABC and, further, to one another.

For since (angle) BAC is equal to ADB—for each (are) right-angles—and the (angle) at B (is) common to the two triangles ABC and ABD, the remaining (angle) ACB is thus equal to the remaining (angle) BAD [Prop. 1.32]. Thus, triangle ABC is equiangular to triangle ABD. Thus, as BC, subtending the right-angle in triangle ABC, is to BA, subtending the right-angle in triangle ABD, so the same AB, subtending the angle at C in triangle ABC, (is) to BD, subtending the equal (angle) BAD in triangle ABD, and, further, (so is) AC to AD, (both) subtending the angle at B common to the two triangles [Prop. 6.4]. Thus, triangle ABC is equiangular to triangle ABD, and has the sides about the equal angles proportional. Thus, triangle ABC [is] similar to triangle ABD [Def. 6.1]. So, similarly, we can show that triangle ABC is also similar to triangle ADC. Thus, [triangles] ABD and ADC are each similar to the whole (triangle) ABC.

So I say that triangles ABD and ADC are also similar to one another.

For since the right-angle BDA is equal to the right-angle ADC, and, indeed, (angle) BAD was also shown (to be) equal to the (angle) at C, thus the remaining (angle) at B is also equal to the remaining (angle) DAC [Prop. 1.32]. Thus, triangle ABD is equiangular to triangle ADC. Thus, as BD, subtending (angle) BAD in triangle ABD, is to DA, subtending the (angle) at C in triangle ADC, (which is) equal to (angle) BAD, so (is) the same AD, subtending the angle at B in triangle ABD, to DC, subtending (angle) DAC in triangle ADC, (which is) equal to the (angle) at B, and, further, (so is) BA to AC, (each) subtending right-angles [Prop. 6.4]. Thus, triangle ABD is similar to triangle ADC [Def. 6.1].

Thus, if, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another. [(Which is) the very thing it was required to show.]

Corollary​

So (it is) clear, from this, that if, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the (straight-line so) drawn is in mean proportion to the pieces of the base. (Which is) the very thing it was required to show.