User:Brews ohare/Pythagoras

Proof using differentials
One can arrive at the Pythagorean theorem by studying how changes in a side produce a change in the hypotenuse in the following diagram and employing a little calculus.

Side a is increased by an infinitesimal amount da, keeping side b vertical and of fixed length. To remain attached to side b, side c rotates a bit and also increases in length by dc.

A differential "triangle" is formed, bounded by da and dc. In this infinitesimal triangle the side opposite angle θ is an arc and not a straight line. However, this distinction is negligible when da is taken to be infinitesimal. (If that statement is acceptable, skip the indented aside below.)

"Here is an aside to show the differential 'triangle' is indeed a right triangle. The lower figure considers the curved side of radius c, and the chord joining the ends of this arc. The chord and the radii at either end of the chord form an isosceles triangle enclosing the angle Δθ, and with base angle φ. The base of the triangle has length 2c sin (Δθ/2), and the length of the arc is c Δθ. As Δa → 0, Δθ → 0 too. Thus, sin (Δθ/2) → (Δθ/2), so the chord length and the arc length become the same. (The chord, a straight line, is the shortest distance between the end points, so in this limit, the arc has become a straight line.) The equality of sides is sufficient to show that as  as Δa → 0, the triangle made up of da, dc and the chord of length c dθ is congruent with the differential triangle made up of da, dc and the arc of length c dθ. Also, these are right triangles because the angle at the base of the isosceles triangle φ  → π/2 as Δa → 0. That follows because the sum of the angles interior to a triangle must add to π. Applying that rule to the right triangle that is half the isosceles triangle, φ + Δθ/2 =  π/2, and so φ → π/2 as Δθ → 0. Because the angle φ and the adjoining angle are supplementary angles, the adjoining angle also becomes π/2. Thus, the differential triangle of sides da and dc and the arc of length c dθ, obtained in the limit as Δa → 0, is rigorously a right triangle, with the curved side c Δθ replaced by the chord of the same length c dθ at right angles to side dc."

The infinitesimal triangle bounded by da and dc is similar to the large triangle with sides (a + da ) and (c + dc ) because both are right triangles and the acute angles θ are the same because they are alternate angles created by the traversal line c + dc crossing two parallel lines, a + da and da. Because θ occurs in both of these right triangles, they are similar triangles; consequently, the ratios of the corresponding sides are the same:


 * $$ \frac {a + da}{c +dc} = \frac {dc}{da} \ . $$

An equivalent way to establish this equality is to note that each ratio is simply the cosine of the angle θ, so both ratios are the same.

Cross-multiplying to clear fractions:


 * $$(c+dc)\,dc = (a+da)\,da, $$

or:


 * $$ c\, dc \left(1+\frac{dc}{c}\right) = a\, da \left( 1+\frac{da}{a} \right).$$

Because the increments da and dc are infinitesimally small, both da/a and dc/c can be neglected compared to 1, resulting in:


 * $$ c\, dc = a \, da, \text{ or: }d(c^2) = d(a^2), $$

which has the integral:


 * $$ c^2 = a^2\ + \ \text{ constant}.$$

When a = 0 then c = b, so the "constant" is b 2. Hence,


 * $$c^2 = a^2 + b^2.\,$$

As can be seen, the squares are due to the particular proportions between the changes and the sides: the change in a side decreases inversely with the length of the side so c dc = a da. The same approach made by letting b increase while a is held fixed would show b db = c dc. More generally, if both a and b change by da and db respectively, the total change in c is c dc  = a da  + b db. The sum of squares therefore is an expression of the independent contributions of the changes in the sides a and b, a point not evident from the geometric proofs.

Throughout, the quantities da and dc have been referred to as infinitesimal changes in a and c respectively. A more formal approach would use instead finite incremental changes Δa and Δc. At the end of the analysis, the limit as their sizes are made infinitesimal would lead to the same results as Δa → da and Δc → dc.

Scratch
Σ Σ(aibj − biaj )2 = sin2 θ. ei× ej = Σ ckij ek.
 * a||2 ||b||2−(a · b)2 = Σ Σ(aibj − biaj )2.