On Constructible Numbers and Angles Original Notes Adopted from March 5, 2002 (Week 21)

More on Constructible Numbers and Angles Original Notes adopted from March 5, 2002 (Week 21)

Lemma: If x0 is a root of a polynomial with coefficients in F(r), then x0 is a root of a polynomial with coefficients in F.

n n-1 Proof: Given (an + bn √r) x0 + (an-1 + bn-1 √r) x0 + ... +(a1 + b1 √r)x0 + a0 + b0 √r =0 with ai , bi ∈ F. n By Dividing through by an + bn √r, we can assume it is poly monic. (ie. coefficients of x0 is 1).

n n-1 x0 (an-1 + bn-1 √r) x0 + ... +(a1 + b1 √r)x0 + a0 + b0 √r =0 Then, n n-1 n-1 x0 + an-1 x0 +....+ a1x0 + a0 = √r (bn-1x0 + ... + b1x0 + b0 ) Square both Sides n n-1 2 n-1 2 (x0 + an-1 x0 +....+ a1x0 + a0 ) - r (bn-1x0 + ... + b1x0 + b0 ) =0

Recall: x0 algebraic if p(x0)= 0 some polynomial p with rational coefficients.

Theorem: Every constructible number is algebraic.

Proof: Suppose x0 is constructible.

There exists Q = F0 ⊂ F1 ⊂ F2 ⊂...... ⊂ Fk with Fi = Fi-1 (√ri ) some ri ∈ Fi-1 , ri = 0, √ri ∉ Fi-1 such that x0 ∈ Fk

Fk = Fk-1 ( ¥ rk ) x0 = a + b ¥ rk, a, b ∈ Fk x0 root of x – ( a + b ¥U k ) = 0.

Lemma ⇒ x0 root of polynomial with coefficients in Fk-1 ⇒ x0 root of polynomial with coefficients in Fk-2 ∴ x0 root of polynomial with coefficients in Q.

Corollary: The set of constructible numbers is countable. Proof: It's contained in set of algebraic numbers which we proved countable.

Corollary: Angle θ can't be constructed with straightedge and compass if cosθ is transcendental

Eg. 370 We know that cos (37/3) is transcendental ⇒ (37/3) 0 is not constructible. Can you trisect 37 degrees if it is just given ( We presume you are given the angle , it is irrelevant if you can construct it or not)

Theorem: If cosθ is transcendental, then θ can't be trisected with straightedge and compass.

Proof: Can construct number cosθ from θ given:

Let F0 = {p(c)/q(c) , p, q polynomials with rational coefficients, q(c) ≠ 0} where c = cosθ.

F0 number field. Before we started with Q. Here, we start with F0, which is the smallest field containing Q & c. Here, with straightedge and compass , we can make towers starting with F0 F0 ⊂ F0( ¥ r0 ) ⊂ (F0( ¥ r0 )) ( ¥ r0 ) cos3δ = 4cos3δ - 3cosδ. if 3δ = θ, δ = 1/3θ.

Show: cosδ is not in any such tower. If it were, cos3δ = c, so c = 4x3 – 3x would have a solution in a tower.

Preceding as before(if root of such a cubic in F(¥U LWVFRQMXJDWHLVD lso a root, go down until set root in F0. 3 ∴it suffices to prove: 4x – 3x – c =0 has no root in F0. Suppose x = p(c)/q(c) is a root with p, q polynomials with coefficients in Q.

4 (p(c)) 3 - 3p(c) - c =0 (q(c)) 3 q(c)

4 (p(c)) 3 - 3p(c) (q(c)) 2 - c(q(c)) 3 =0 Polynomial in c, rational coefficients = 0, contradicts c being transcendental unless the polynomial(after collecting coefficients) has all its coefficients 0; eg. 2(π3- π) + 2π - 2π3 = 0. But this can't happen: 1) if degree p > degree q, 4 (p(c)) 3 has only terms of highest degree 2) if degree p <= degree q. c(q(c)) 3 has only term of highest degree. ∴ Leading coefficients of the polynomial (after collecting terms) is not 0.

Corollary: Only countable numbers of angles can be trisected.

Showing we can Trisect ANY angle given we have a "marked" straightedge.

It is clear from the diagram that x is a third of θ. x + x + w = π w + y = π y = 2x 2x + 2x + z = π z = π - 4x x+ π - 4x + θ = π θ = 3x x = θ/3 x + z + θ = π

Polyhedron: solid, faces are polygons. A regular polyhedron(platonic solid) is a polyhedron all of whose faces are regular polygons with same number of sides as each other and all of whose vertices lie on the same number of faces.

1) Cube 2)Tetrahedron ± 4 faces which are triangles 3) Dodecahedron ± 12 sided pentagon 4) Icosahedron ± 20 sides b) Octagon

All these have only 3,4,5 sides per face.

Theorem: There are only 5 regular polyhedron Proof: Let n be the number of sides of a face and k, the number of faces on which a vertex lies.

Note that k >= 3 For regular polynomial of n sides, angle is (n-2)π, so each angle is (n-2)π/n = π - 2π/n

k(π - 2π/2) < 2π n>=3(Polygons have at least 3 sides). ⇒ 1/n <= 1/3 so ±1/n>= 1/3

π - 2π/n >= π - 2π/3 = π/3 k(π - 2π/n) >= kπ/3.

∴ kπ/3 < 2π ∴ k<6 ∴k =3,4, or 5 k =3... 3(π - 2π/2) < 2π 3(n-2/n)π < 2π 2n ± 6 <2 n<6.

If k =3; n =3 tetrahedron k=3 & n =4 cube k =3, n=5 dodehedron.