Summer Math Series:  Week 1

Notes by David Kosbie

1. Area of a Triangle = bh/2 2. Pythagorean Theorem:  Euclid’s Windmill Proof 3. Pythagorean Theorem:  Chinese Proof (or perhaps the Indian mathematician Bhaskara’s) 4. Pythagorean Theorem:  President Garfield’s Trapezoid Proof 1. The Distance Formula:  derived from Pythagorean Theorem!

1. Fermat’s Last Theorem:  xn + yn = zn has no positive integral solutions for n>2.
Proven recently by Andrew Wiles (omitted here for lack of room in the margin).

2. Hypotenuses in a “square root spiral” are of length sqrt 2, sqrt 3, sqrt 4, sqrt 5,…
(Inductive proof) 3. The square root of 2 is irrational.
1. (p/q) 2 = 2 è p2 = 2q2, then apply Fundamental Theorem of Arithmetic  è lhs has even # of prime factors, rhs has odd #, QED.
2. (p/q) 2 = 2 è p2 = 2q2 è p is even  è  è q is even, QED.

4. “Nearly all” real numbers are irrational!
1. The integers are countable (as are evens, primes, powers of 10, …)
2. Integer pairs – Z2 – are countable (dovetailing!)
3. Integer triplets, etc – Z3 , Z4,… – are countable.
4. Rationals are countable.
5. Algebraics are countable.
6. Reals are not countable (diagonalization!)
7. Thus, “nearly all” reals are irrational (even non-algebraic, hence transcendental!)

Summer Math Series:  Week 2

5. More on cardinality
1. From last week:  the following sets are all countable (“denumerable”):  Natural numbers (N), Integers (Z), Evens (E), Primes (P), Powers of 10, …, integer pairs (Z2), integer triplets, etc (Z3 , Z4,…), rationals (Q), and algebraics (A).

2. We say that |N| = |E| = |P| = |Z| = |Zk| = |Q| = |A| = א0

3. Reals (R) – actually, just the Reals in (0,1) – are not countable (diagonalization!).

4. |R| = |reals in (0,1)| = c (where c > א0)

Use y = (2x – 1) / (x – x2)

1. |R| = |R2| = |Rk| = c (shuffling!)

2. Cantor’s Theorem:  |P[A]| > |A|  (diagonalization!  See p.277)
(the power set of any infinite set A – written as P[A] or 2A – has greater cardinality than the original set)

3. Thus, |N|= א0 < |P[N]|  <  |P[P[N]]|  < |P[P[P[R]]]|  < ….

4. Cantor’s Paradox:  There is no Universal Set U = Set of All Sets (as |P[U]| > |U|)

5. |R| = c = |P[N]|

i.      |P[N]|  <= |R|
Given a subset of integers, construct a real number by placing a 3 in the ith decimal digit if the number i appears in the subset, otherwise insert a 7.  The result is a unique real in (0,1).

ii.      |R| <= |P[N]|
First, map the reals into (0,1), as above.  Then, given a real in (0,1), write the number in binary and include the integer i in the corresponding subset iff the ith digit is 1.

1. Cantor’s Continuum Hypothesis:  No set S exists where  א0 < |S| < c

i.      Godel proved the continuum hypothesis cannot be disproved!

ii.      Cohen proved the continuum hypothesis cannot be proved!

iii.      In this sense, it is like Euclid’s Parallel Postulate

1. We define א1 as the smallest cardinality greater than א0.

2. So we know |R| = c > א0, but:  Does |R| = c = א1?  This is undecidable!

1. Derive the Quadratic Formula by Completing the Square (coming soon: cubics, quartics!)

2. The Locker Problem  (coming soon:  Fundamental Thm of Arithmetic, number theory)

3. Prove the Binomial Theorem (by induction!) (coming soon:  more number theory!)
1. Hint:  Using construction method of Pascal’s Triangle, find recursive defn of nCk
2. Application:  prove:  nC0 + nC1 + … +  nCn= 2n

Summer Math Series:  Week 3 1. Pascal’s Triangle and Pascal’s Binomial Theorem
1. nCk = kth value in nth row of Pascal’s Triangle! (Proof by induction)
2. Rows of Pascal’s Triangle == Coefficients in (x + a)n.  That is: 2. The Circle Problem and Pascal’s Triangle 1. How many intersections of chords connecting N vertices?
2. How does this relate to Pascal’s Triangle?

3. Patterns in Pascal’s Triangle (see http://www.kosbie.net/lessonPlans/pascalsTriangle/)
1. Simple Patterns

i.      Natural Numbers (1,2,3,4…)

ii.      Triangular Numbers (1,3,6,10,…)

iii.      Binomial Coefficients (nCk)                    ß Pascal’s Binomial Theorem

iv.      Tetrahedral Numbers (1,4,10,20,…)

v.      Pentatope Numbers (1,5,15,35,70…)

1. More Challenging Patterns

i.      Powers of 2 (2,4,8,16,…)

ii.      Hexagonal Numbers (1,6,15,28,…)

iii.      Fibonacci Numbers (1,1,2,3,5,8,…)         ß Prove This!

iv.      Sierpinski’s Triangle

v.      Catalan Numbers (1,2,5,14,42,…)           ß Prove This!

vi.      Powers of 11 (11, 121, 1331, 14641,…)

1. Applications of the Binomial Theorem
1. Find the coefficient of x3 in (x + 5) 3
2. Prove:  nC0 + nC1 + … +  nCn= 2n  (Hint:  2 = 1+1, so what does 2n = ?)

Summer Math Series:  Week 4

1. Newton’s Binomial Theorem
1. Generalization to negative integer powers: 2. (P + PQ)m/n = P m/n + (m/n)AQ + (m-n)/(2n) BQ + (m-2n)/(3n) CQ + …
where A,B,C,… represent the immediately preceding terms
so B = (m/n)AQ,  C = (m-n)/(2n) BQ, …

3. After some algebra:
(1 + Q) m/n = 1 + (m/n)Q + (m/n)(m/n - 1)/2 Q2 + (m/n)(m/n - 1)(m/n - 2)/(3*2) Q3 + ...

4. That is: 2. Applications of Newton’s Binomial Theorem
1. 1 / (1 + x)3 = 1 – 3x + 6x2 – 10x3 + 15x4 – …
2. Sqrt(1 – x) = 1 – (1/2)x – (1/8)x2 – (1/16)x3 – (5/128)x4 – …
3. So, sqrt(7) = 3 sqrt(1 – 2/9) è fast approximation for square roots!
4. Also cube roots, etc, since (1 – x)1/3 can be expanded this way, too…

Glossary

Binomial coefficients (N choose K): The number of ways in which you can choose K elements from a set of N elements. This equals n! / ( k! (n-k)! ).

Catalan numbers (1, 2, 5, 14, 42, ...): The number of ways you can divide a polygon with N sides into triangles, using non-intersecting diagonals (a triangle has 1 way, a rectangle has 2 ways, a pentagon has 5 ways, a hexagon has 14 ways, and so on). The Catalan numbers can be computed using the formula: Fibonacci numbers (1, 1, 2, 3, 5, 8, ...): A series in which the first two numbers are 1 and each subsequent number is the sum of the preceding two numbers.

Hexagonal numbers (1, 6, 15, 28, ...): Numbers that can be represented as the number of points on the perimeter of a hexagon with a constant number of points on each edge. These are given by the formula N * (2N-1), and can be seen in the following figure: Pentatope numbers (1, 5, 15, 35, 70, ...) A figurate number (a number that can be represented by a regular geometric arrangement of equally spaced points) given by:
Ptopn = (1/4)Tn(n+3) = (1/24) n (n+1) (n+2) (n+3)
for tetrahedral number Tn. Note: pentatopes are 4-dimensional analogs of tetrahedra.

Sierpinski's triangle: a famous fractal formed by connecting triangle midpoints as such: Tetrahedral numbers (1, 4, 10, 20, ...): a figurate number formed by placing discrete points in a tetrahedron (triangular base pyramid). The formula is given by: n(n+1)(n+2)/6.

Triangular numbers (1, 3, 6, 10, ...): The number of dots you need to form a triangle: 