Date
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What we discussed/How we spent our time
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Jan 18
|
Syllabus. Text.
We discussed the Pythagorean Theorem. We learned
the Bride's Chair proof found in Euclid's Elements,
and the simpler dissection proof of Thabit ibn Qurra.
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Jan 20
|
We learned more proofs of the Pythagorean Theorem:
the proof attributed to Leonardo da Vinci,
James Garfield's proof, and the proof of
Dijkstra's generalization of the Pythagorean Theorem.
Pythagorean triples!
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Jan 23
|
Quiz 1 with
956 different solutions.
Rational points on the unit circle. Chord and tangent method.
Rational parametrizations of conics.
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Jan 25
|
We proved Euclid's formula for Pythagorean Triples:
every primitive Pythagorean Triple has the form
(a,b,c)
=
(r2-s2, 2rs, r2+s2) or
(2rs, r2-s2, r2+s2),
where r and s have no common factor and have opposite parity, while
nonprimitive Pythagorean Triples have the form (ka,kb,kc) for some
primitive Pyth. Triple (a,b,c).
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Jan 27
|
Commensurable numbers:
we defined magnitudes a and b to be commensurable
if there is a magnitude c and positive integers r and s
such that a = rc, b = sc. We showed that a and b are commensurable
magnitudes iff the ratio of their lengths is a rational number.
We gave a geometric algorithm for determining commensurability.
(Construct an a×b rectangle and repeatedly delete
maximal square subregions. The algorithm terminates iff
a and b are commensurable, as we showed using the Euclidean algorithm.)
We used this algorithm to show that the Golden Ratio
is not rational.
We used reductio ad absurdum to
show that √2 is irrational.
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Jan 30
|
Quiz 2.
We discussed the difference between theorems and definitions.
We discussed the use of the deductive method in Euclid's Elements.
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Feb 1
|
We discussed the regular polyhedra,
graph representations of polyhedra,
and Euler's formula.
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Feb 3
|
Snow day.
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Feb 6
|
Quiz 3.
Straightedge and compass constructions, part 1.
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Feb 8
|
Straightedge and compass constructions, part 2.
The constructible numbers
form a Euclidean field.
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Feb 10
|
Straightedge and compass constructions, part 3.
Descartes' Theorem: the constructible numbers
are the smallest Euclidean field.
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Feb 13
|
Quiz 4.
Straightedge and compass constructions, part 4.
Constructible numbers are algebraic, hence
Liouville's number is not constructible,
e is not constructible (Hermite, 1873),
and π is not constructible (Lindemann, 1882).
The minimal
polynomial of a constructible number has degree
that is a power of 2, hence ∛2 is not constructible.
Moreover, cos(π/3) IS constructible, while
cos(π/9) is NOT constructible.
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Feb 15
|
Straightedge and compass constructions, part 5.
Solution of the classical construction problems:
squaring the circle, doubling the cube, trisecting
a general angle.
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Feb 17
|
Straightedge and compass constructions, part 6.
A regular n-gon is constructible iff
cos(2π/n) is a constructible number iff
the number of numbers relatively prime to n
that are between 1 and n is a power of 2.
(Hence a regular 17-gon is constructible, while
a regular 7-gon is not.)
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Feb 20
|
Quiz 5.
(Extended) Euclidean algorithm. Bezout's Identity.
Linear diophantine equations.
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Feb 22
|
Continued fractions from the Euclidean algorithm.
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Feb 24
|
Infinite continued fractions. Wallis's formula
for convergents. Continued fractions method for
solving linear diophantine equations.
|
Feb 27
|
Quiz 6.
Pell's equation. (Associated names: Archimedes, Diophantus,
Brahmagupta, Bhaskara II, Lord Brouncker, Lagrange.)
The cattle problem. Continued fractions method for solving Pell's Equation.
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Feb 29
|
Review for Midterm 1.
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Mar 2
|
Midterm.
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Mar 5
|
No quiz.
Dissection versus the method of exhaustion, 1.
|
Mar 7
|
Dissection versus the method of exhaustion, 2:
Proof of the Wallace-Bolyai-Gerwien Theorem.
We started explaining Archimedes'
evaluation of the area of a parabolic sector by exhaustion.
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Mar 9
|
We finished explaining Archimedes'
evaluation of the area of a parabolic sector by exhaustion.
We used the method of exhaustion to prove that
1+(1/4)+(1/42)+(1/43)+
…)=1/3.
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Mar 12
|
Quiz 7.
Number theory in Asia.
We began discussing modular arithmetic. We proved
that congruence modulo n is compatible with addition
and multiplication. We gave a necessary and sufficient
condition for a linear congruence ax≡c(mod b) to be solvable,
and we described a method of solution. (Method=reduce
to a linear diophantine equation.)
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Mar 14
|
Chinese remainder theorem. Notes.
|
6pm
Math 100
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Special talk:
Zero to Infinity: Great moments in the history of number
Professor Edward B. Burger
Baylor University & Williams College
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Mar 16
|
We explained why the solution to a a system of congruences
is unique modulo the lcm of the moduli, and pointed out
that a system with relatively prime moduli is consistent.
We spent the rest of the time working on problems
from Wednesday's sheet of notes.
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Mar 19
|
Quiz 8.
We discussed Brahmagupta's problem of determining
which rational triangles have rational area.
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Mar 21
|
We began discussing solutions of polynomial equations.
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Mar 23
|
We derived Cardano's Formula for depressed cubics, x3=px+q,
and explained how to use the formula to solve the general
cubic equation
ax3+
bx2+
cx+
d=0.
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Apr 2
|
Quiz 9.
We explained how to solve the quartic equation.
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Apr 4
|
We
practiced
solving cubic and quartic equations.
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Apr 6
|
Solutions of higher degree polynomial equations:
♦
Viete's formulas.
♦
Newton-Girard identities.
♦
Tschirnhaus transformations.
♦ Forms for the quintic:
general quintic, depressed quintic, principal quintic form,
Bring-Jerrard quintic form.
♦ Bring radical (a root of x5-x=a).
♦ Every quintic can be solved by radicals + Bring radicals.
♦ (Ruffini, Abel, Galois) Bring radicals are necessary
for the solution of an arbitrary quintic. The general equation of degree n
is not solvable by ordinary radicals if n≥5.
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Apr 9
|
Quiz 10.
We discussed the Newton-Girard Identities for
kth power sums of the roots of a polynomial of degree n
in the cases k≤n.
We surveyed some of the major developments in mathematics
during 1600's-1800's (Calculus, diff. eqns., calculus of variations,
development of analysis, non-Euclidean geometry and its effect
on the development of logic, set theory and its effect on
the development of the foundations of mathematics).
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Apr 11
|
Hilbert's Problems
Problem #1: the Continuum Hypothesis, part 1.
We discussed the definitions of |A|=|B|, |A|≤|B| and |A|<|B|.
We showed that |ℕ|=|ℤ|.
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Apr 13
|
Hilbert's Problems
Problem #1: the Continuum Hypothesis, part 2.
We defined choice functions and stated
the axiom of choice.
We stated the Cantor-Schroeder-Bernstein Theorem (=CSB Theorem).
We introduced von Neumann ordinals, and stated the well-ordering
principle (an equivalent form of the axiom of choice).
We showed that ω, ω+1, and ω+2
are equipotent.
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Apr 16
|
Quiz 11.
Hilbert's Problems
Problem #1: the Continuum Hypothesis, part 3.
Ordinal arithmetic. As an example, we showed that every
ordinal below ω2 equals
ωj+k for natural numbers j and k.
We showed that the CSB Theorem implies that
equipotence classes of ordinals are intervals.
For each ordinal α, there is an ordinal β
such that |α|<|β|.
Cardinal numbers are defined to be initial ordinals.
Every set is equipotent with a unique cardinal number.
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Apr 18
|
We discussed how to show that |ℝ|= |ℝ2|
(a HW problem). This week's HW due date was moved to Friday.)
Hilbert's Problems
Problem #1: the Continuum Hypothesis, part 4.
"Aleph" notation. The difference between
ℵα and ωα.
Cantor diagonalization.
|A|<|P(A)| for any A.
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Apr 20
|
Hilbert's Problems
Problem #1: the Continuum Hypothesis, part 5.
|ℝ|=|P(ℵ0)|.
The Continuum Hypothesis (CH) asserts that
|ℝ|=|P(ℵ0)|=ℵ1.
The Generalized Continuum Hypothesis (GCH) asserts that
|P(ℵα)|=ℵα+1
for any α. Gödel proved that the GCH cannot be refuted
from the usual axioms of set theory.
Cohen proved that CH cannot be proved from the usual axioms of set theory.
|
Apr 23
|
Quiz 12.
Hilbert's Problems
Problem #2: the consistency of arithmetic.
Meaning of "consistency", "arithmetic".
Gödel's Theorem,
Gentzen's Theorem.
|
Apr 25
|
Hilbert's Problems
Problem #3: Are two tetrahedra with equal height and base area
scissors congruent? Part 1.
We introduced the Dehn invariant and stated
Dehn's and Sydler's Theorems, which prove that
two polyhedra are scissors congruence iff they have equal volume
and equal Dehn invariant.
|
Apr 27
|
Hilbert's Problems
Problem #3: Are two tetrahedra with equal height and base area
scissors congruent? Part 2.
We discussed the arithmetic of Dehn invariants,
why Dehn's Theorem is true, and identified two
tetrahedra with equal bases and height
with different Dehn invariants.
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Apr 30
|
Hilbert's Problems
Problem #3: Are two tetrahedra with equal height and base area
scissors congruent? Part 3.
A symbol (L,α) is the zero symbol iff L=0 or
α is a rational multiple of π.
We discussed how to use vector arithmetic
to compute dihedral angles of polyhedra.
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May 2
|
Hilbert's Problems
Problem #3: Are two tetrahedra with equal height and base area
scissors congruent? Part 4.
We explained that
(i) a Dehn symbol (L,α) equals zero iff L=0 or α
is a rational multiple of π.
(ii) α is a rational multiple
of π iff the number z = cos(α)+i*sin(α) is
a root of unity.
(iii) roots of unity are algebraic integers.
We gave some techniques for showing that some numbers
are NOT algebraic integers. These can be used to show that some
angles are NOT rational multiples of π. (In particular,
we showed that if tan(α)=√2, then α
is not a rational multiple of π. Also, we showed that if
cos(α) = -1/3, then α
is not a rational multiple of π.)
I circulated the take-home final.
Remember, you may use any book while working on the
exam, but you are not allowed help from any person
other than me.
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May 4
|
Hilbert's Problems
We discussed Hilbert's 7th and 8th problems.
The discussion covered: the Gelfond-Schneider Theorem,
the consequence that eπ is transcendental,
the statements of the Riemann Hypothesis, Goldbach's Conjecture
and the Twin Prime Conjecture.
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