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.