Schedule

Create an annotated bibliography for your term paper (deadline).

Download a copy of Euclid's Elements. That link has both the Ancient Greek and an English translation side-by-side. For Tuesday, read the sections entitled

• Definitions,
• Postulates,
• Common Notions, and
• Proposition 1.

• First statements of abstract theorems.
• Didn't leave any of his own writings.
• Marks a shift (in his locale) towards abstract reasoning and proof.

• Cult leader who liked geometry.
• His school was a major influence in the development of education through the middle ages (e.g., their four subjects of study are the first four liberal arts).
• Connections between numbers and geometric concepts.

We proved some simple number-theoretic results about triangle and square numbers using historical methods.

Turn in the annotated bibliography for your term paper.

Write a preliminary report on the mathematics for your term paper (deadline).

Let \( A \) and \( B \) be distinct points. There is a point \( C \) for which \( \triangle{ABC} \) is an equilateral triangle.

Let \( A \), \( B \), and \( C \) be points. There is a point \( D \) for which \( \mSegment{AD} = \mSegment{BC} \).

Let \( A \), \( B \), \( C \), and \( D \) be points. If \( \mSegment{AB} \leq \mSegment{CD} \), then there is a point \( E \in \segment{CD} \) with \( \mSegment{CE} = \mSegment{AB} \).

Let \( \triangle{ABC} \) and \( \triangle{DEF} \) be given. If \( \mSegment{AB} = \mSegment{DE} \), \( \mSegment{AC} = \mSegment{DF} \), and \( \mAngle{BAC} = \mAngle{EDF} \), then \( \triangle{ABC} \cong \triangle{DEF} \).

Let \( A \), \( B \), and \( C \) be non-collinear points. If \( \mSegment{AB} = \mSegment{AC} \), then \( \mAngle{ABC} = \mAngle{ACB} \).

Let \( A \), \( B \), and \( C \) be non-collinear points. If \( \mAngle{ABC} = \mAngle{ACB} \), then \( \mSegment{AB} = \mSegment{AC} \).

Let \( A \), \( B \), \( C \), and \( D \) be points. Assume \( \mSegment{AC} = \mSegment{AD} \) and \( \mSegment{BC} = \mSegment{BD} \). If \( C \) and \( D \) are on the same side of \( AB \), then \( C = D \).

Let \( \triangle{ABC} \) and \( \triangle{DEF} \) be given. If \( \mSegment{AB} = \mSegment{DE} \), \( \mSegment{BC} = \mSegment{EF} \), and \( \mSegment{CA} = \mSegment{FD} \), then \( \mAngle{ABC} = \mAngle{DEF} \).

Let \( A \), \( B \), and \( C \) be non-collinear points. There is a point \( D \) in the interior of \( \angle{BAC} \) with \( \mAngle{BAD} = \mAngle{CAD} \).

Let \( A \) and \( B \) be points. There is a point \( C \in \segment{AB} \) with \( \mSegment{AC} = \mSegment{BC} \).

Let \( A \), \( B \), and \( C \) be distinct points. If \( C \in \segment{AB} \), then there is a point \( D \) with \( CD \perp AB \).

Let \( A \), \( B \), and \( C \) be non-collinear points. There is a point \( D \) with \( AB \perp CD \).

Let \( A \), \( B \), \( C \), and \( D \) be distinct points. If \( B \in \segment{CD} \), then the sum of the angles \( \angle{CBA} \) and \( \angle{ABD} \) is the same in measure as the sum of two right angles.

Let \( A \), \( B \), \( C \), and \( D \) be distinct points. If \( C \) and \( D \) are on opposite sides of \( AB \) and the sum of \( \angle{ABC} \) and \( \angle{ABD} \) is the same in measure as the sum of two right angles, then \( B \in \segment{CD} \).

Let \( A \), \( B \), \( C \), and \( D \) be distinct points. If \( E \in \mSegment{AB} \cap \mSegment{CD} \), then \( \mAngle{AEC} = \mAngle{BED} \).

Let \( \triangle{ABC} \) be given and \( D \neq C \). If \( C \in BD \), then \( \mAngle{ACD} > \mAngle{CBA} \) and \( \mAngle{ACD} > \mAngle{BAC} \).

Let \( \triangle{ABC} \) be given. The sum of \( \angle{ABC} \) and \( \angle{ACB} \) is less in measure than the sum of two right angles.

Let \( \triangle{ABC} \) be given. If \( \mSegment{AB} < \mSegment{BC} \), then \( \mAngle{ACB} < \mAngle{BAC} \).

Let \( \triangle{ABC} \) be given. If \( \mAngle{ACB} < \mAngle{BAC} \), then \( \mSegment{AB} < \mSegment{BC} \).

Let \( \triangle{ABC} \) be given. Then the sum of the segments \( \segment{AB} \) and \( \segment{BC} \) is greater in measure than the segment \( \segment{BC} \).

Let \( \triangle{ABC} \) be given. If \( D \) is in the interior of \( \triangle{ABC} \), then \( \mSegment{BD} < \mSegment{AB} \) and \( \mAngle{BAC} < \mAngle{BDC} \).

Let \( \alpha \), \( \beta \), and \( \gamma \) be segments. If the measure of the pairwise sums of these segments always exceed the measure of the remaining segment, then there is a triangle with sides equal in measure to \( \alpha \), \( \beta \), and \( \gamma \).

Let \( A \), \( B \), \( C \), \( D \), and \( E \) be points with \( C \), \( D \), and \( E \) non-collinear. There is a point \( F \) with \( \mAngle{FAB} = \mAngle{DCE} \).

Turn in the preliminary maths report for your term paper.

Let \( \triangle{ABC} \) and \( \triangle{DEF} \) be given. If \( \mSegment{AB} = \mSegment{DE} \), \( \mSegment{AC} = \mSegment{DF} \), and \( \mAngle{BAC} > \mAngle{EDF} \), then \( \mSegment{BC} > \mSegment{EF} \).

Let \( \triangle{ABC} \) and \( \triangle{DEF} \) be given. Assume \( \mAngle{ABC} = \mAngle{DEF} \) and \( \mAngle{BCA} = \mAngle{EFD} \). If either \( \mSegment{BC} = \mSegment{EF} \) or \( \mSegment{AB} = \mSegment{DE} \), then \( \triangle{ABC} \cong \triangle{DEF} \).

Let \( \triangle{ABC} \) and \( \triangle{DEF} \) be given. If \( \mSegment{AB} = \mSegment{DE} \), \( \mSegment{AC} = \mSegment{DF} \), and \( \mSegment{BC} > \mSegment{EF} \), then \( \mAngle{BAC} > \mAngle{EDF} \).

Let lines \( AB \), \( CD \), and \( EF \) be given with \( E \in \segment{AB} \) and \( F \in \segment{CD} \). If \( \mAngle{AEF} = \mAngle{EFD} \), then \( AB \parallel CD \).

Let distinct lines \( AB \), \( CD \), and \( EF \) be given so that no three of these points are collinear. Assume there are points \( G \in \segment{AB} \cap \segment{EF} \) and \( H \in \segment{CD} \cap \segment{EF} \). If either \( \mAngle{EGB} = \mAngle{GHD} \) or the measure of the sum of \( \angle{BGH} \) and \( \angle{DHG} \) is equal to that of two right angles, then \( AB \parallel CD \).

Let distinct lines \( AB \), \( CD \), and \( EF \) be given so that no three of these points are collinear. Assume there are points \( G \in \segment{AB} \cap \segment{EF} \) and \( H \in \segment{CD} \cap \segment{EF} \). If \( AB \parallel CD \), then the sum of the angles \( \angle{BGH} \) and \( \angle{GHD} \) is equal in measure with the sum of two right angles.

If lines \( \alpha \parallel \beta \) and \( \beta \parallel \gamma \), then \( \alpha \parallel \gamma \). That is, parallelism is a transitive relation on lines.

Let \( A \), \( B \), and \( C \) be non-collinear points. There is a line \( \ell \) with \( A \in \ell \) and \( \ell \parallel BC \).

We briefly sketched the structure of the remainder of Book 1. Mostly it's about…

• areas of parallelograms,
• areas of triangles, and
• constructions of rectangles and squares.

The Pythagorean Theorem!

We looked briefly at Euclid's proof, to appreciate the complexity and delicacy of his argument. However, we gave a complete argument using Reverend Wheeler's proof.

• Born in Syracuse, Sicily. Killed during a Roman invasion. Lived 287–212 BCE.
• Studied in Alexandria, Egypt at the great Library.
• Genius inventor.
• All around cool dude (if a bit smelly).

Determined a formula for the area of the circle—the formula given is the . The key ingredients to his argument (see your lecture notes) were:

• The ratio of the circumference to the diameter is a constant (in Euclidean geometry). Today we call that constant \( \pi \).
• The area and perimeter of all regular \( n \)-gons.
• Every circle has an inscribed square (Euclid 4.6).
• Eudoxus' Method of Exhaustion (this can be understood as a rudimentary version of limits).
• Areas are always comparable (i.e., exactly one of \( A < B \) or \( A = B \) or \( A > B \) always holds).
• Proof by contradiction.

Archimedes' argument recovers Euclid 12.2. His argument is "non-constructive" in an interesting way: it compares the area of the circle with that of a triangle whose existence is merely postulated, rather than constructed by Euclidean means. This has the advantage that it works—the specific triangle he used cannot be constructed with Euclid's axioms alone!

• Invented the calculus, though not in the form we currently see it.
• Invented the binomial series!
• Used calculus to study optics, planetary motion, projectile motion, and much more.

We saw in lecture Newton's approach to approximating \( \pi \). This involved several key components.

1. The binomial series yields an approximation of \( \sqrt{3} \) to arbitrary (finite) precision.
2. The semicircle \( 0 \leq y \leq \sqrt{x(1-x)} \) has area approximation from \( x = 0 \) to \( x = \tfrac{1}{4} \) using the "Method of Inverse Fluxions" (i.e., Newton's integral).
3. The same portion of the semicircle has a geometric area approximation via sectors!
4. Finally, comparing the true value and using our approximated values (to only \( 9 \) terms!) we can approximate \( \pi \) to \( 7 \) decimal places.