MAS114 Revison Cards

Pregunta	Respuesta
Russell's Paradox: Prove that the set of all sets which are not elements of themselves is self-contradictory:	If S∈S then by definition of S S∉S. If S∉S, then by definition of S S∈S.
Two functions are equal if:	-Same domain and co-domain -For all a∈A, we have f(a)=g(a)
-Negation of A: -Contra-positive of A⇒B:	¬A ¬A ⇒ ¬B
Equivalence of: ¬(∀x ∈ X, P(x)) ¬(∃x ∈ X s.t P(x))	∃x ∈ X s.t ¬P(x) ∀x ∈ X, ¬P(x)
Euclid's Theorem	There are infinitely many prime numbers.
Definitions: -Greatest common divisor: -Least common multiple: -Co-prime:	-gcd(a,b) is the largest positive integer which is a divisor of both a and b. -lcm(a,b) is the smallest positive integer which is a multiple of both a and b. -Two integers a and b are said to be co-prime if the gcd(a,b) = 1.
gcd properties:	gcd(a,b) = gcd(b,a) gcd(a,0) = gcd(a,a) = a gcd(a,1) = 1 gcd(a,b) = gcd(a,-b) gcd(a,b) = gcd(a + kb, b)
a ≡ b (mod m)	m \| (a-b)
Congruence properties:	- a ≡ a (mod m) - a ≡ b (mod m) ⇒ b ≡ a (mod m) - a ≡ b & b ≡ c ⇒ a ≡ c (mod m) - a ≡ b & c ≡ d ⇒ a+c ≡ b+d & a-c ≡ b-d & ac ≡ bd (mod m)
Fermat's little theorem	a^(p-1)≡1 (mod m)
Fermat-Euler Theorem	Let a, n be integers w/ gcd(a,n)=1 a^φ(n)≡1 (mod n). Where φ(n) is the number of integers between 1 to n which are co-prime to n
Wilson's Theorem	(n-1)! ≡1 (mod m) iff n is prime
x is irrational y is rational	Then: x+y is irrational xy is irrational (if y≠0)
Convergence definition:	A sequence a_i converges to x if: ∀ε>0, ∃N∈N s.t. ∀n>N, \|a_n - x\|<ε
The four group axioms:	G1 (Closure): ab∈G G2 (Associativity): (ab)c = a(bc) G3 (Neutral Element): ∃e s.t eg = g = ge G4 (Inverse): gh = e = hg
Albanian Group	ab = ba
(gh)^-1	h^-1g^-1
Homomorphism: Isomorphism:	f(xy) = f(x)f(y) Isomorphism is a bijective Homomorphism
Subgroup Criteria	SG1: H ≠ ∅ SG2: gh∈H SG3: h^-1∈H
Group Actions	GA1: e*x =x GA2: g(hx) = (gh)x
Orbit Counting Theorem	n = 1/\|G\|*∑ \|fix(g)\|
Equivalence relations	Reflexivity: a~a for all a Symmetry: if a~b then b~a for all a,b Transitivity: a~b and b~c then a~c

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