Summer Math Series: Week 1
Notes by David
- Area of a Triangle = bh/2
- Pythagorean Theorem: Euclid’s
- Pythagorean Theorem: Chinese Proof (or perhaps the Indian mathematician Bhaskara’s)
- Pythagorean Theorem: President Garfield’s Trapezoid Proof
- The Distance Formula: derived from Pythagorean Theorem!
- Fermat’s Last Theorem: xn
+ yn = zn has no positive integral solutions
Proven recently by Andrew Wiles (omitted here for lack of room in the
- Hypotenuses in a “square root spiral” are of
length sqrt 2, sqrt 3,
sqrt 4, sqrt 5,…
- The square root of 2 is irrational.
- (p/q) 2 = 2 è p2 = 2q2, then apply
Fundamental Theorem of Arithmetic è lhs has even # of prime factors, rhs has odd #, QED.
- (p/q) 2 = 2 è p2 = 2q2 è p is even
è … è q is even, QED.
- “Nearly all” real numbers are irrational!
- The integers are countable (as are evens,
primes, powers of 10, …)
- Integer pairs – Z2 – are countable
- Integer triplets, etc – Z3
, Z4,… – are countable.
- Rationals are countable.
- Algebraics are countable.
- Reals are not
- Thus, “nearly all” reals are irrational (even non-algebraic, hence
Summer Math Series: Week 2
- More on cardinality
- 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
- We say that |N| = |E| = |P| = |Z| = |Zk| = |Q| = |A| = א0
- Reals (R) – actually, just the Reals
in (0,1) – are not countable (diagonalization!).
- |R| = |reals in (0,1)|
= c (where c > א0)
y = (2x – 1) / (x – x2)
- |R| = |R2| = |Rk|
= c (shuffling!)
- 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)
- Thus, |N|= א0 < |P[N]| <
|P[P[P[R]]]| < ….
- Cantor’s Paradox: There is no Universal Set U = Set of
All Sets (as |P[U]| > |U|)
- |R| = c = |P[N]|
|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).
|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.
- Cantor’s Continuum Hypothesis: No set S exists where א0 < |S| < c
Godel proved the continuum hypothesis cannot be disproved!
Cohen proved the
continuum hypothesis cannot be proved!
In this sense, it
is like Euclid’s
- We define א1 as the smallest cardinality greater than א0.
- So we know |R| = c > א0, but:
Does |R| = c = א1? This is
- Derive the Quadratic Formula by Completing the
Square (coming soon: cubics, quartics!)
- The Locker Problem (coming soon: Fundamental Thm
of Arithmetic, number theory)
- Prove the Binomial Theorem (by induction!) (coming soon:
more number theory!)
Using construction method of Pascal’s Triangle, find recursive defn of nCk
+ nC1 + … + nCn= 2n
Summer Math Series: Week 3
Triangle and Pascal’s Binomial Theorem
- nCk = kth value in nth
row of Pascal’s Triangle! (Proof by induction)
- Rows of Pascal’s Triangle == Coefficients in (x
+ a)n. That is:
Circle Problem and Pascal’s Triangle
- How many intersections of chords connecting N
- How does this relate to Pascal’s Triangle?
in Pascal’s Triangle (see
- Simple Patterns
Coefficients (nCk) ß Pascal’s Binomial Theorem
- More Challenging Patterns
Powers of 2 (2,4,8,16,…)
(1,1,2,3,5,8,…) ß Prove This!
(1,2,5,14,42,…) ß Prove This!
Powers of 11 (11,
121, 1331, 14641,…)
of the Binomial Theorem
- Find the coefficient of x3 in (x + 5)
- Prove: nC0
+ nC1 + … + nCn=
2n (Hint: 2 = 1+1, so what does 2n =
Summer Math Series: Week 4
- Generalization to negative integer powers:
- (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,
- 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 + ...
- That is:
of Newton’s Binomial Theorem
- 1 / (1 + x)3 = 1 – 3x + 6x2
– 10x3 + 15x4 – …
- Sqrt(1 – x) = 1 – (1/2)x – (1/8)x2 –
(1/16)x3 – (5/128)x4 – …
- So, sqrt(7) = 3 sqrt(1 – 2/9) è fast approximation for square roots!
- Also cube roots, etc, since (1 – x)1/3 can be expanded this way,
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: