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Subject: "Calculus Proof of PT"     Previous Topic | Next Topic
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sbrodie
Member since Dec-28-10
Dec-28-10, 12:38 PM (EST)
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"Calculus Proof of PT"
 
   The recent posting of a "Calculus-based" proof of PT draws me into this discussion. I'm afraid I am still skeptical. The problem is deeper than whether or not it is possible to phrase the proof in such a way that the term "distance" never appears.

Rather, the problem is the fundamental one of how to connect "analytic geometry" with the "synthetic geometry" of Euclid, Archimedes, and their direct successors.

Historically, "analytic geometry" evolved as an enhancement of synthetic geometry, not a replacement for it. The idea was to "draw" the coordinate axes "on" the Euclidean ("synthetic geometry") plane, and use the knowledge thus inherited from synthetic geometry to pose geometrical questions as equations to be solved and to thus infer geometrical conclusions from the analytic properties of the equations and their solutions. In many cases, the "analytic" proofs add little insight to the synthetic treatment (say, the proof that the medians of a triangle are concurrent). In other cases, the analytic approach has afforded important clarification to the synthetic geometry (say, the proof of the "two-circles" theorem -- that two circles whose centers are separated by less than the sum of their radii intersect in two points, which clarifies the role of an "axiom of completeness" in the geometry). In still other cases, the analytic approach connects the geometry to powerful tools which allow the solution of otherwis inaccessable problems, such as the "impossible" constructions of doubling the cube, trisecting the angle, or constructing a regular polygon of 7 sides.

Of course, in this approach, the parallel postulate and the Pythagorean theorem are "built-in" to the geometry from the start, and little is gained by seeing PT "pop out" again from time to time.

The alternative to this "classical" interpretation of analytic geometry is to try to use the Cartesian plane as "model" for synthetic geometry -- that is, to start with only the algebraic formalism of ordered pairs of real numbers, and interpret the definitions and predicates which arise in this context in terms of the concepts of synthetic geometry. If one can then prove the axioms of synthetic geometry "from scratch," the Cartesian plane becomes a model for synthetic geometry. This is hard work, seldom carried out in practice.

Some of the concepts are easy -- indeed, perhaps easier than in the synthetic context. A "point" is simply an ordered pair of real numbers. A "line" is the solution set of a linear equation

Ax + By + C = 0

Parallel lines are lines which do not intersect, etc.

The challenge is now to connect the objects of this interpretation with the objects and measurements of traditional synthetic geometry. This is easy enough at the traditional textbook level -- we define "distance" on the Cartesian plane by means of the traditional "distance formula" based on PT:

distance(A to B) = Sqrt((ax - bx)^2 + (ay - by)^2)

and angle measure by means of trigonometry -- to measure an angle, draw a circle at the vertex, measure the length of the chord subtended by the angle and the segment the chord cuts off from the initial ray, and use the inverse tan function. One then proves that these measurement procedures satisfy the appropriate axioms, and the program is complete -- though, to be sure, having "built in" the PT-based definition of distance, there is, again, not much accomplished by confirming PT as a result.

In contrast, the "program" of the recent Calculus-based proof of PT seems to hover somewhere else, with definitions and assumptions unmoored in the context. To be sure, one can set up a coordinate system and define lines as solution sets to linear equations without reference to distance, but this is not enough structure to give the discussion a geometric meaning. For example, suppose you lay out your coordinate plane with the y-axis pointing "northeast" instead of "north"? Points are still identifiable as ordered pairs; lines are still identifiable as solutions to linear equations. But what is the meaning of "distance" in this context? What is the meaning of "perpendicular"? What defines a "right triangle"? How can you tell that this layout of the plane is not correct?

A simple example of the difficulty is the assertion that the slopes of perpendicular lines are negative reciprocals. The fact is so ingrained in our thinking that it is difficult (for me, anyway) to recall a proof. A quick check of my old textbooks finds only proofs rooted in the "historical" approach mentioned above.

It is worth noting in passing that the negative reciprocal formulation of perpendicularity and PT are very closely related, quite apart from the differential equation under discussion. See my Cut-the-Knot page,

https://www.cut-the-knot.org/pythagoras/PTcom3.shtml

Comments welcome.

Scott


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alexbadmin
Charter Member
2694 posts
Dec-28-10, 01:03 PM (EST)
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1. "RE: Calculus Proof of PT"
In response to message #0
 
   >Of course, in this approach, the parallel postulate and the
>Pythagorean theorem are "built-in" to the geometry from the
>start, and little is gained by seeing PT "pop out" again
>from time to time.

The parallel postulate is indeed built-in. I fail to see why this is true for the PT.

>A simple example of the difficulty is the assertion that the
>slopes of perpendicular lines are negative reciprocals. The
>fact is so ingrained in our thinking that it is difficult
>(for me, anyway) to recall a proof. A quick check of my old
>textbooks finds only proofs rooted in the "historical"
>approach mentioned above.

But this is right on the page. You have two perpendicular lines with slopes y/x and -x/y giving -1 as the product.


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sbrodie
Member since Dec-28-10
Dec-28-10, 11:39 PM (EST)
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2. "RE: Calculus Proof of PT"
In response to message #1
 
   1. If a notion of "length" or "distance" is not built-in to the geometry, then please explain to me what is the meaning of the label "c" on the hypotenuse of the right-triangle with legs a and b in the figure.

2. Yes, it is simple to derive the negative reciprocal formula for the slopes of perpendicular lines if you know in advance enough Euclidean geometry to understand Alex's figure at the bottom of the page... but how do you know in this context when the lines are "perpendicular" in the first place" And, how is it clear that this notion of perpendicular and the Euclidean notion coincide?

Scott.

Scott


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alexbadmin
Charter Member
2694 posts
Dec-28-10, 11:58 PM (EST)
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3. "RE: Calculus Proof of PT"
In response to message #2
 
   1. c is the length of the line segment from the origin to point (a, b)

2. This proof only depends on the synthetic definition of perpendicularity. I actually do not understand your quandary. You draw a couple of lines and say, these two lines are perpendicular. If so, such and such triangles are similar. The similarity implies that the slopes (which are nothing but two ratios) are reciprocals of each other (give or take the minus sign.)

Returning to 1., Euclid did not have a 2d distance as we understand it today. But he was able to measure the lign segments however oriented in the plane. Euclid I.2 is exctly this: place a segment of given length on a given line starting from a given point.

Actually, I am not sure what is the essence of that assertion of mine. Of course he did not have a 2d metric as we understand it today because what we have in mind comes out of analytic geometry which he was unaware of.


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