PART IA: Numbers and Sets
Let 2≤n. Then a is a unit modulo n if and only if hcf(a,n)=1.
- Use hcf to give linear combination. - ax≅1(mod n) for some integer x.
Bezout's Theorem - ax+by=c has integer solution <=> hcf(a,b) divides c.
=> Show hcf divides a, b, c <= Suppose h divides c, and create a linear combination of a&b.
Strong Principle of Induction is equivalent to the Well-Ordering Principle
(1) Suppose that P(n) isn't true and use WOP. (2) Suppose that there is no least n such that P(n) holds, then consider 'not P(n)'
The following are equivalent: (i) X is countable. (ii) there is an injection X->N (iii) X is empty or there is a surjection N->X.
(i)->(ii): If X is finite, obviously injects, and if X bijects with N, then it injects. (ii)->(i): If there's an injection f:X->N, then f is a bjiection between X and S=f(X). If S finite, so is X. If S infinite, then bijection g:S->N and thus g o f is a bijection. (i)->(iii) plainly (iii)->(ii): Suppose x non-empty and surjection f:N->X. Define g:X->N by g(a)=min(f^-1({a})), existing since f surjective, by WOP. g injective by construction.
A countable union of countable sets is countable. PROOF 1.
- Assume countable sets indexed by N. - For each iϵN, we can list elements as a(i,1), a(i,2),...(might terminate). - Define f: Union -> N and x->2^i3^j where x=a(i,j) for the least i s.t. xϵAi. This is an injection.
A set of size n has exactly 2^n subsets. DIRECT PROOF
- Assume that the set is {1,2,...,n}. - To specify a subset, each element of the set is either IN the subset or OUT of it, so it's 2*2*2*... for each element, so it's 2^n.
A set of size n has exactly 2^n subsets BY INDUCTION
- Clearly true for n=0. - Given n>0, and T⊆{1,2,...,n-1}, there are 2 sets S such that S∩{1,2,...,n-1}=T, namely T and T∩{n}. - Hence the number of subsets of {1,2,...,n} = 2 * number of subsets of {1,2,...,n-1} = 2 * 2^n-1 =2^n.
N x N is countable. PROOF 2
- Define f:NxX->N, (x,y)->2^x3^y - Then f is injective by the Fundamental Theorem of Arithmetic.
(Fact A) For any polynomial p, ∃ constant K s.t. |p(x)-p(y)|≤K|x-y| for all 0≤x,y≤1.
- Factor out (x-y) from p(x)-p(y), and take the absolute value of both sides.
There are uncountably many transcendental numbers.
If R\A were countable, then since A is countable, R=R\A U A is countable.
If p is prime and p divides ab, then p divides a or p divides b.
Suppose p divides ab but not a, then show p divides b using.
Fundamental Theorem of Arithmetic - Every natural number 2≤n is expressible as a product of primes, uniquely up to reordering.
- For uniqueness, use induction - Suppose n=p1p2..pk=q1q2..ql, and relabel so p1 divides q1, etc. - divide through by each factor and use inductive hypothesis.
For any set X, there is no bijection between X and P(X).
- Given any function f:X->P(X) - Let S={xϵX: x∉f(x)}. Then S does not belong to the image of f, since ∀xϵX, S and f(x) differ in the element x, and thus S≠f(x) for any x. - f is not a surjection.
If An tends to c and Bn tends to d, then An+Bn tends to c+d (as n tends to infinity).
- Given ε>0, ∃N∈N s.t. ∀n≥N, |An-c|<ε/2, and ∃M∈N s.t. ∀n≥M, |Bn-d|<ε/2. - ∀n≥Max(N,M), |An+Bn-(c+d)|≤|An-c|+|Bn-d|=ε.
The power set P(N) of the natural numbers is uncountable. PROOF BY CONTRADICTION.
- If P(N) were countable, we could list the subsets of N as S1,S2,S3,.... - Let S = {nϵN: n∉Sn}. Then S is not on our list since ∀nϵN, S≠Sn, a contradiction. - P(N) uncountable.
Any subset of the natural numbers is countable.
- If S⊆N and non-empty, by WOP, there's a least element s1ϵS. If S\{s1} is non-empty, then WOP. If S\{s1,s2},..... - If this ends, then S={s1,s2,...,sn} is finite. - If it goes on forever then map g(n)=sn is well-defined and injective. - If kϵS, then kϵN, and there are <k elements of S <k, so k=sn for some n<k, surjective.
Fermat's Little Theorem - Let p be prime. Then a^p≅a(mod p) for all integer a. Also a^(p-1)≅1(mod p) for non-zero a.
- If a≇0(mod p), then a is a unit mod p, so ax≅ay iff x≅y. - Hence a,2a,3a,...,(p-1)a are pairwise distinct mod p, and ≇0 mod p, so ≅1,2,3,..,p-1 in some order. - Multiply the lot together for (p-1)!a^p-1≅(p-1)!(mod p), but (p-1)! is a unit (product of units), so can be cancelled.
Fermat-Euler Theorem - Let hcf(a,m)=1. Then a^φ(m)≅1(mod m).
- Label each coprime integer u1,u2,...,uφ(m). - Since a is a unit, au1,au2,...,auφ(m) distinct and invertible, so ≅u1,u2,...,uφ(m) in some order. - Multiply all together for a^φ(m)z≅z(mod m) and cancel z.
A countable union of countable sets is countable. PROOF 2.
- Let I be a countable index set, and assume that for all iϵI, Ai countable. - I countable so injection f:I->N, and Ai countable so injection g(i):Ai->N. - Construct injection h: Union -> NxN: for each xϵU, pick Mx=min{jϵN:xϵAi,f(i)=j}, existing by WOP. - Let ixϵI be s.t. f(ix)=Mx (ix unique because f injective). -h(x)=(Mx,g(ix)(x)) defines an injection.
There exists x∈R with x^2=2.
- Let S be all x with x^2<2. Show S is non-empty, and bounded above, so must have a supremum c s.t 1<c<2. - Suppose that c^2<2, and show that (c+t)^2<2 for small enough t for contradiction, and likewise for c^2>2.
Inclusion-Exclusion Principle - Let S1,S2,...,Sn be finite sets. Then |S1∪S2∪...∪Sn| = (|a|=1)Σ|Sa| - (|a|=2)Σ|Sa| +(|a|=3)Σ|Sa|-....+(-1)^n+1(|a|=n)Σ|Sa| where Sa=∩Si.
- Let xϵS1∪S2∪...∪Sn, say xϵSi for k of the Si. We want x to be counted precisely once in the RHS. In general, #{A: |A|=r with xϵSa} = kCr for r≤k, and =0 for r>k. - The number of times x is counted on the RHS is k-kC2+kC3-....+(-1)^k+1kCk = 1 - (1+(-1))^k = 1 for 1≤k.
The power set P(N) is uncountable. PROOF 2
- Note that there is an injection from (0,1) into P(N): write xϵ(0,1) in binary 0.x1x2x3.... not ending in an infinite string of 1s, and set f(x)={nϵN:xn=1}. E.g. 0.11101000->{1,2,3,5}. - This is an injection
Let ~ be any equivalence relation on X. Then the equivalence classes form a partition of X.
- Since ~ reflexive, xϵ[x] for all xϵX, so X is the union of the equivalence classes of all xϵX. - Suppose [x]∩{[y] is non-empty and let z be in it. Then z~x, so by symmetry x~z and z~y, so x~y by transitivity. - Let now wϵ[y], so y~w. x~y and y~w, so x~w and wϵ[x], so [x]=[y] for non-empty intersection.
Cantor (1870s). R is uncountable.
- Suppose R countable and can be indexed as r(1),r(2),.... - Write each r(n) in decimal form in some way, i.e r(1)=n(1).d(1,1)d(1,2)d(1,3).... - Define r=0.d(1)d(2).... by d(n)=2 if d(n,n)=1, d(n)=1 if d(n,n)≠1. - This r has only on decimal expansion and is not on the list because r≠r(n) ∀nϵN, a contradiction.
Axiom of Archimedes - The Naturals, N, are not bounded above in the Reals, R.
- Suppose c is the supremum of N. - c-1 is not an upper bound, so ∃ n>c-1. - n+1∈N, with n+1>c. Contradiction.
e is irrational.
- Suppose e=p,q where q>1. Then q!e is natural, and = q!+q!/1!+...+q!/q! + q!/(q+1)!+.... - Show that q!/(q+1)!+.... <1, contradicting q!e natural.
Given non-empty sets A and B, ∃ injection f:A->B if and only if ∃ surjection g:B->A.
- Suppose f:A->B injective. Fix a0ϵA. Define g:B->A as b-> unique aϵA s.t. f(a)=b if it exists, and a0 otherwise. Then g surjective. - Suppose g:B->A is surjective. Let f:A->B be a-> some bϵB s.t. g(b)=a. Then f is injective.
Let p be an odd prime. Then -1 is a square modulo p if and only if p≅1(mod 4).
- Suppose p≅1(mod 4), then use Wilson's Theorem to show -1≅(-1)^(p-1)/2((p-1/2)!)^2 (mod p). Sub in p=4k+1. - Suppose that p=4k+3 and use Fermat's Little Theorem with z^2≅-1(mod p) to obtain contradiction.
N x N is countable. PROOF 1
- Take a(1)=(1,1) and a(n-1)=(p,q) - Define a(n) by a(n)=(p+q,1) if p=1 and a(n)= (p-1,q+1) otherwise. - This list includes every point (x,y)ϵNxN (by induction on x+y).
The set A of algebraic numbers is countable.
- The map Pd->Z^d+1 where p(x)->(a(d),a(d-1),....a(1),a(0)) is an injection. - Since Z^d+1 is countable, Pd is countable. - Set of all polynomials with integer coefficients is countable, so A is a countable union of countable sets.
Wilson's Theorem - Let p be prime. Then (p-1)!≅-1(mod p)
- True for p=2 so assume p>2. - Note that units modulo p come in pairs whose product ≅1, with some self-inverses where x^2≅1(mod p) [THESE ARE -1 and +1]. - (p-1)! is the product of (p-3)/2 pairs of inverses with +1 and -1, so (p-1)!≅-1(mod p)
Let p be prime. Then x^2≅1(mod p) if and only if x≅1(mod p) or x≅-1(mod p).
- Use D.O.T.S. - Either x+1 or x-1 ≅0(mod p)
Let p be prime. Then every a≇0(mod p) is a unit modulo p.
- Use hcf to give linear combination of a and p. - ax=1-py, so ax≅1(mod p) for some integer x.
1/Liouville,1851 - The number L= Σ1/10^(n!) is transcendental.
- Write Ln as Σ1/10^(k!) from k=1 to n, so Ln tends to L. Suppose there is polynomial p with root L, so |p(L)-p(Ln)|≤K|L-Ln|. - Note |L-Ln|≤2/10(n+1)! and Ln=s/10^n! for some natural s, so p(Ln)=t/10^dn! for some integer t. - Suppose p has degree d. For sufficiently large n, Ln is not a root of p. Hence |p(Ln)|≥1/10^dn!, so |p(Ln)-p(L)|≥1/10^dn! -Therefore 1/10^dn!≤|p(Ln)-p(L)|≤K|Ln-L|≤K2/10^(n+1)!
Z x Z is countable.
- Z countable, so injection Z->N, so because NxN countable, we have injection ZxZ->NxN->N.
The Rationals are Dense in the Reals, i.e for all a<b∈R,∃c∈Q with a<c<b.
- ∃n∈N with 1/n<b-a. Let T be the set of naturals k such that k/n>b. Show non-empty with AoA. - WOP on T gives least m, and set c=(m-1)/n. Since m-1 not in T, c<b. - If c<a, m/n=b, contradiction.
Schroder-Bernstein Theorem: If f:A->B and g:B->A are injections, then ∃ bijection h:A->B.
-For aϵA, write g^-1(a) for the bϵB (if it exists) s.t. g(b)=a. -Similarly, for bϵB, write f^-1(b) for the aϵA (if it exists) s.t. f(a)=b. We call g^-1(a),f^-1(g^-1(a)),g^-1(f^-1(g^-1(a))),... the ancestor sequence of aϵA. Similarly, we can define the Ancestor Sequence of bϵB. - Let A0 be the AS stopping in A, A1 stopping in B, and Ainf does not stop, similarly B0,B1,Binf. - f bijects A0 with B1 (since every bϵB1 is f(a) for some aϵA0), and similarly g bijects B0 with A1. And f or g biject Ainf with Binf. - Then h:A->B, a-> f(a) for aϵA0, g^-1(a) for aϵA1, and f(a) for aϵAinf is a bijection by construction.
Every bounded monotonic sequence converges.
-Suppose An is increasing. Then set {An:n>1} is non-empty and bounded above by supremum l. - Given ε>0, l-ε is not an upper bound, so ∃M∈N s.t. AM>l-ε. Thus l-ε<An<l ∀n>M.
If An≤d for all n, and An tends to c as n tends to infinity, then c≤d.
-Suppose c>d. Let ε=|c-d|>0. Then ∃N∈N s.t. ∀n≥N, |An-c|<ε. -This implies An>d, a contradiction.
Any natural number n can be expressed as n=qk+r, with 0≤r≤k-1
By induction on n, considering r<k-1 and r=k-1.
Condition for sequence tending to limit as n tends to infinity.
∀ε>0,∃M∈N s.t. ∀n>M, |an-l|<ε
Let {Ai : iϵI} be a family of open intervals of R that are pairwise disjoint. {Ai: iϵI} is countable. PROOF 1
Each interval Ai contains a rational, and Q is countable, so since the intervals are disjoint, we have an injection from I into Q. Hence I is countable.
We can write hcf(a,b) as a linear combination of a&b.
Either: -Run Euclid's algorithm to obtain hcf and work backwards. -Let h be the least positive linear combination of a&b, then show h=hcf.
Chinese Remainder Theorem - Let m,n be coprime. Then there is a unique solution to congruences x≅a(mod m) and x≅b(mod n). There is a solution to x≅a(mod m) and x≅b(mod m) and y is a solution if and only if x≅y(mod mn)
Existence: Use hcf for linear combination sm+tn=1. Then x=a(tn)+b(sm) is a solution. Uniqueness: - Suppose y is also a solution, so y≅x(mod n) and y≅x(mod m). - m and n are factors of (y-x), then show using mods.
(Fact B) A non-zero polynomial of degree d has at most d roots.
Given a polynomial p of degree d, we may assume by induction that it holds for all polynomials for degree <d, and that p has root a. - By long division we can write p(x)=(x-a)q(x) for polynomial q of degree d-1. So each root of p is either a or a root of q. But by inductive hypothesis, q has at most d-1 roots.
nCk = n!/(n-k)!k!
Given a set of size n, there are n(n-1)(n-2)....(n-k+1) ways to pick k elements, one by one, in order. But each subset of size k is picked in k(k-1)(k-2)...2*1 ways by this method. Hence the number of subsets of size k in {1,2,...,n} is n(n-1).....(n-k+1)/k(k-1)....2*1.
For any t>0, ∃n∈N with 1/n<t.
Given t>0, ∃n∈N s.t. n>1/t by AoA. Hence 1/n<t.
Any subset of a countable set is countable.
If Y⊆X and X countable, then take injection X->N restricted to Y.
Every natural number n≥2 can be written as a product of primes.
Induction on n, taking into account n as prime and composite.
The output of Euclid's algorithm on a&b gives the hcf(a,b).
Induction on: (i) the output divides both a and b (ii) any factor of both a&b divides the output
Euclid, 300BC - There are infinitely many primes.
Suppose there are finitely many, and find another.
Let {Ai : iϵI} be a family of open intervals of R that are pairwise disjoint. {Ai: iϵI} is countable. PROOF 2
The set {iϵI: Ai has length ≥1/2} is countable as it injects into 0.5Z. More generally, for each nϵN, {iϵI: Ai has length ≥ 1/n} is countable. So {Ai : iϵI} is countable as I is a countable union of countable sets.
Binomial Theorem - (a+b)^n= a^n + na^n-1b+....+nab^n-1+b^n.
aWhen we expand (a+b)^n=(a+b)(a+b)...(a+b), we obtain the number of terms of the form a^n-kb^k as we must specify k brackets from which to pick b. Hence (a+b)^n=ΣnCka^n-kb^k
