Artin Field Extension

Emile Artin's very unique book "Galois Theory" (1971) on "Finite Field Extension" interpreted by Vector Space.

Let H a Field with subfield G
F is G's subfield:
H ⊃ G ⊃ F

Example:
Let
F = Q = Rational Field
G = Q(√2) = Larger Extended Field Q with irrational root √2
H = Q(√2, √3) = Largest Extended Field Q with irrational roots (√2 & 3)

{1, √2} forms basis of Q(√2) over Q

{1, √3} basis of Q(√2, √3) over Q(√2)
[since √3 ≠ p+ q√2 , ∀p,q ∈ Q]

=> {1,√2, √3, √6} basis of Q(√2, √3) over Q
=> Q(√2, √3) is a 4-dimensional Vector Space over Q.

Isomorphism (≌)

Q(√2) ≌ Q[x] / {x² - 2}

Read as:

Q(√2) isomorphic to the quotient of the Polynomial ring Q[x] modulo the Principal Ideal {x² - 2}

Q[x] the Polynomial Ring

{x² - 2} is the Principal Ideal

Complex Number (C)

C = R[x] / {x² + 1}

R[x] the Polynomial Ring with coefficients in the Field R

{x² + 1} is the Principal Ideal

Questions:

Since R[x] / {x² + 1} is the Field C

Why below are not Fields ?
R[x] / {x³ + 1}
R[x] / {x^4 + 1}
C[x] / {x² + 1}

Hint: they are not irreducible in that particular Field, not a Principal Ideal.

Note: C[x] the Polynomial Ring with coefficients in the Field C

Google Symmetries

Google = 10 ^ 100

1000 = 10³ = 2³ x 5³
10^google = (10³)³³. 10
= (2³)³³ x (5³)³³. (2x5)
= 2¹ºº x 5¹ºº

=> 100 flips (2) , 100 rotations of pentagons (5) = Google symmetries.

Irrational 'e'

In secondary school, we know how to prove √2 is irrational, how about e ?

e= 1 + 1/1!  +1/2!  + 1/3!  + 1/4!  +...

Prove by Reductio ad absurdum (contradiction):

Assume e= p/q as rational
multiply both sides by q!
LHS: e. q!= (p/q) .q! = p.(q-1)!  => integer

RHS:  q!+q! + (3.4...q)+ (4.5...q) +...1 + 1/(q+1) +.... => fraction

Contradiction !

Therefore e is irrational.

Tie shoe lace in 1 sec

How to tie shoelace in 1 second:

http://www.youtube.com/watch?v=wMuNjnNyaiA

Origamatic: Fold T-Shirt

Origami + Mathematics = Origamatic
http://www.langorigami.com/science/math/math.php

Watch this video on useful trick to fold T-shirt:

http://www.youtube.com/watch?v=BAxhr0j0thY

Abelian Group

This interesting example is like solving simultaneous equations in Group, using only one Group property tool (Inverse => cancellation law)

Let Group G, ∀a,b ∈G,
for any 3 consecutive integers i,
$latex (a.b)^{i}= a^{i}.b^{i} $
Prove：G is abelian?
[Herstein: i, i+1,i+2]

Proof:
(a.b)ⁿ= aⁿ.bⁿ ...(1)
(a.b)ⁿ⁺¹= aⁿ⁺¹.bⁿ⁺¹ ...(2)
(a.b)ⁿ⁺² = aⁿ⁺².bⁿ⁺² ...(3)
Take inverse (1):
(a.b)⁻ⁿ = (aⁿ.bⁿ)⁻¹ = b⁻ⁿ.a⁻ⁿ ...(4)

Left * (4) to (2)
(a.b)ⁿ⁺¹(a.b)⁻ⁿ =ab

{Right *} (4) to (2):

ab = (aⁿ⁺¹.bⁿ⁺¹).(b⁻ⁿ.a⁻ⁿ) = aⁿ⁺¹.b.a⁻ⁿ
{Left *} a⁻¹
=> a⁻¹(ab) = b = (a⁻¹aⁿ⁺¹).b.a⁻ⁿ = aⁿ.ba⁻ⁿ
Right x aⁿ
=> b.aⁿ = aⁿ.b(a⁻ⁿaⁿ) = aⁿ.b ...(5)
Take inverse of (2):
(a.b)⁻ⁿ⁻¹= b⁻ⁿ⁻¹.a⁻ⁿ⁻¹ ...(6)
Right x (6) to (3)
(a.b) = aⁿ⁺².bⁿ⁺².(b⁻ⁿ⁻¹.a⁻ⁿ⁻¹)
ab = aⁿ⁺².b.a⁻ⁿ⁻¹ ...(7)
Right x aⁿ⁺¹
abaⁿ⁺¹ = aⁿ⁺².b
Left x a⁻¹
baⁿ⁺¹ = aⁿ⁺¹.b ...(8)
(b.aⁿ).a = (aⁿ.b).a from (5) b.aⁿ = aⁿ.b
aⁿ.b.a = aⁿ⁺¹.b
cancellation law:
=> ba= ab
=> G is abelian [QED]

Note:

Trick is inverse (1) then x to (2):

b.aⁿ = aⁿ.b ...(5)
Similarly inverse (2) then x (3):
baⁿ⁺¹ = aⁿ⁺¹.b ...(8)
Solve (5) & (8): by cancellation law
aⁿ.b.a = aⁿ⁺¹.b
=> ba= ab

Get Rich Rule 72

Rule 72: Compound Interest

Let P= Principal at interest rate i after n years:

Compound Interest Formula:

$latex P'=P.(1+i)^{n}$

$latex (1+i)^{n} = 1 + ni + \frac {1}{2} (ni)^{2} + \dots $

if ni=0.72 =72%

$latex (1+i)^{n} = 2$

=> P'=2 P

ie ni=0.72, P double

so if i = ROI (Buffett recommended) = 15% => n=72/15= 4.8 ~ 5 years

=> every 5 yrs x2

=> every 10 yrs x4

If you wish to double many times to get 1 million after 10 years, reverse the calculation:

Invest $ 1 m / 2 / 2 = $250,000 now

Buffett advised to put a  MoS (Margin of Safety=50%) to buffer mistake buys and have bigger ROI,

=> invest 250K - 50% = $125 K (now)

Spread it into a portfolio of max. 5 stocks at any time => S$25 K/ stock.

If the companies are good one, they will increase equity, or ultimately acquired, and you will be paid handsomely with higher stocks. eg. Google, DELL 10 years ago.