if you can prove him wrong
No "if" about it.
On page 15 he attempts to divide squares into three classes, but since he ignores units, his classification system has no value.
Example: if a square has a side of 4 meters, then it's area is 16 meters² and its perimeter is 16 meters. Because the author doesn't understand units, he concludes the perimeter equals the area and puts that square in a certain class. However, if measured in centimeters, the same square has side of 400 centimeters, area 160,000 centimeters² and perimeter 1600 centimeters, which the author would classify the square in a difference class for a choice of units that cannot make a physical difference. The mistake is thinking one can compare apples (units of area) with oranges (units of linear measure). The ratio DIVIDED by perimeter is seen to be 1 meter = 100 centimeters in both cases because this is a sensible operation on these physical magnitudes.
Thus he doesn't have a classification of squares with respect to other squares, but rather squares with respect to a choice of "units of length", abbreviated u.l.
On page 21 he attempts to classify circles by the same valueless system
"When
π was used for calculating values of the circle with the diameter 4 u.l. the circumference was larger than the area".
P = 2 π r = π D
A = π r² = π D²/4
If D = 4, then A = 4 π = P, so he is wrong.
What is this magic where D=4 means A = P for both circles and squares? This comes from a study of regular polygons.
Look at an regular polygon of N sides and side of a certain side length L. All the angles and sides are identical. So if you draw the perpendicular bisectors of all the sides, they must meet at a single point in the interior. Let's call this point the center. Let's call the length from the center to the middle of any side, r, and the length from the center to any vertex, R. Because r is perpendicular to a side, we can use the Pythagorean theorem and compute the area of this triangle.
r² = R² – (L/2)² = (4 R²-L²)/4
A₁ = 1/2 r L = 1/2 L √( R² – (L/2)² ) = 1/4 L √(4 R²-L²)
Because the polygon has N such sides, we have the total area, A = N A₁, and the total perimeter, P = NL
So the ratio A/P = A₁/L = r/2 = 1/4 √(4 R²-L²)
This ratio holds for equilateral triangles, squares, regular pentagons, regular hexagons, etc. And since D = 2r for squares , circles, and any regular polygon with an even number of sides, the A/P = D/4 which unsurprisingly equals 1 u.l. when D = 4 u.l.
So how do we go from N-sided polygons to a circle? With the concept of geometric limits.
For a regular polygon, L/2 = R sin (π/N) and r = R cos(π/N) so L = 2 r tan(π/N) and R = r sec(π/N).
So P = NL = 2 N r tan(π/N) = 2 N R sin (π/N) and A = N A₁ = N r² tan(π/N) = N R sin (π/N) √( R² – (R sin (π/N))² ) = N R^2 sin( 2 π /N ) / 2
Thus the behavior of $$\lim_{N\to\infty} N \tan \frac{\pi}{N} = \lim_{x\to 0} \frac{\tan (\pi x)}{x} = \pi \left. \frac{d \tan x}{d x} \right|_{x=0} = \pi \sec^2 0 = \pi$$ and $$\lim_{N\to\infty} \frac{N}{2} \sin \frac{2 \pi}{N} = \lim_{x\to 0} \frac{\sin (\pi x)}{x} = \pi \left. \frac{d \sin x}{d x} \right|_{x=0} = \pi \cos 0 = \pi$$ prove the formulas for a circle: $$A = \pi r^2 = \pi R^2, \; P = 2 \pi r = 2 \pi R$$.
