Instructor: Dr. Daniel Saracino
Date: From 2024-08-29 to 2024-12-20
For a printable pdf file, please click here
For all integers a, b, c, d:
- If a∣b and b∣c, then a∣c.
- If a∣b and a∣c, then for all x,y∈Z, a∣(xb+yc).
- If a∣b and c∣d, then ac∣bd.
Suppose a and b are integers that are not both 0, so that (a,b) exists.
Then for every n∈Z,
(a,b)=(a+nb,b)=(a,b+na)Every integer n≥2 is the product of one or more primes.
There exist infinitely many primes.
For every integer n≥2, the factorization of n into primes is unique, except for the order in which the factors are written.
Suppose m∈Z+ and a,b∈Z.
Then the following are equivalent:
- a≡b (mod m), i.e., m∣(a−b).
- a=b+mℓ for some ℓ∈Z.
- am=bm. 1
Thus, the remainders are equal, and any one of (i), (ii), or (iii) proves the other two.
Basic Properties of Congruence Modulo m:
Suppose m∈Z+, then the following all hold:
Reflexivity: For every a∈Z,
a≡a (mod m).
Symmetry: For all a,b∈Z,
if a≡b (mod m), then b≡a (mod m).
Transitivity: For all a,b,c∈Z,
if a≡b (mod m) and b≡c (mod m), then a≡c (mod m).
When a relationship between elements of a set S is reflexive, symmetric, and transitive, the relationship is called an equivalence relation on S.
Thus, Theorem 7 says that "congruence modulo m" is an equivalence relation on the set Z.
Suppose m∈Z+, and let a,b,c,d∈Z.
Then, if a≡b (mod m) and c≡d (mod m), we have:
- a+c≡b+d (mod m)
- ac≡bd (mod m)
- Let m∈Z and a,b,c∈Z.
Then
ca≡cb(modm)if and only if
a≡b(mod(c,m)m)The congruence ax≡b(modm) has solutions if and only if (a,m)∣b.
If (a,m)∣b then there is a unique solutions x0 such that 0≤x0<(a,m)m, and a complete set of solutions (modm) is
{x0+(a,m)mt∣0≤t<(a,m)}Chinese Remainder Theorem/CRT
Suppose b1,⋯,bn∈Z and m1,⋯,mn∈Z+.
Consider the system
xx⋮x≡b1(modm1)≡b2(modm2)≡bn(modmn)If m1,⋯,mn are pairwise relatively prime, meaning (mi,mj)=1 where i=j, then the system has a solution.
And all the solutions are ≡(modm1∗m2∗⋯∗mn)
Let p be prime then:
If a∈Z and (a,p)=1, then ap−1=1(modp)
For all integers a, ap≡a(modp)
For every prime p, (p−1)!≡−1(modp)
If a∈Z and (a,n)=1 then
aφn=1(modn)Notice how a can be any number coprime to n
If n≥2 and the prime power decomposition of n is n=p1e1×⋯×pkek then
φ(n)=p1e1−1(p1−1)×p2e2−1(p2−1)×⋯×pkek−1(pk−1)e.g,
(3,100)3φ(100)100φ(100)340=1=1(mod100)=22⋅52=22−1(2−1)×52−1(5−1)=40=1(mod100)If n is composite, then nth Mersenne number is composite.
If p is an odd prime, then every prime factor q of Mp has the form q=2kp+1 for some k∈Z+
n is perfect if and only if σ(n)=2
For every prime p,
σ(p)=p+1σ(pk)=1+p+p2+⋯+pk=p−1pk+1−1σ is multiplicative
If n is an even perfect number, then there exists a Mersenne prime 2p−1 such that
n=2p−1(2p−1)i.e., all even perfect numbers comes from n=2p−1(2p−1)
Suppose f:S→T and g:T→U Then:
- If f and g are both one-to-one, then so is g∘f
- If f and g are both onto, then so is g∘f
Suppose a<b where a,b∈R Then:
- (a,b)∼(c,d) for all c<d in R
- (a,b)∼(c,∞) for all c∈R
- (a,b)∼R
The set of real numbers in the interval (0,1) is uncountable(equivalently, R is uncountable)
If B is countable, and f:A→B is an injection, then A is countable.
If A1,⋯,Ak are finitely many countable sets, then A1×A2×⋯×Ak is countable.
the union of countably many countable sets are countable.
i.e., the union of countable collection of countable sets is countable.
note: the ®sets are allowed to overlap.
If m and n are integers that are not both zero, such that (m,n) exists,
then there exist x,y∈Z such that xm+yn=(m,n).
Suppose a, b, c are integers such that a∣bc and (a,b)=1.
Then a∣c.
If a and m are integers and m is positive, then there exist integers q and r such that a=qm+r and 0≤r≤m
The congruence ax≡b(modm) is solvable if and only if
(a,m)∣bThen there is a unique solution x0 such that
0≤x0<(a,m)mThen the complete set of solutions is
{x0+(a,m)mt∣0≤t<(a,m)}There are exactly (a,m) incongruent solutions modm
If S and T are finite sets with the same number of elements, and f:S→T, then f is one-to-one if and only if f is onto.
∣S∣=∣T∣&f:S→T⇒onto⇔one-to-oneIf x is a nonempty countable set, then there exists an injection f:x→Z+
If S⊆Z+, then S is countable.
We say integers a and b are relatively prime (or "coprime") if (a,b)=1.
An integer p is prime if p>1 and the only positive integers that divide p are 1 and p.
φ(n)=the number of elements in {0,1,⋯,n−1} that are coprime to ne.g,
φ(1)=1Since {(1-1)}={0} and 0 is coprime to 1If n∈Z+, the nth Mersenne number is 2n−1
If n∈Z+, the proper factors of n are the positive integers that divide n but are not n.
A positive integer is called perfect if it equals the sum of its proper factors.
If n∈Z+, then σ(n) denotes the sum of all positive integers that divide n.
We say f is a function(or mapping) from S to T and we write
f:S→Tif f associates to each s∈S a uniquely determined element f(s)∈T
If f:S→T we say f is one-to-one(or "1-1")
- if whenever s1=s2 in S, then f(s1)=f(s2) in T.
- if and only if when s1,s2∈S and f(s1)=f(s2),thens1=s2.
If f:S→T we say f is onto if for every t∈T there exists some s∈S such that f(s)=t.
f is a bijection if f is both one-to-ton and onto. When a function is a bijection, there is an inverse function and the domain eqals codomain.
Suppose f:S→T is a bijection. Then for each t∈T there is a unique s∈S such that f(s)=t. So we can define a function f−1:T→S by letting.
f−1(t)=(the unique s∈S such that f(s)=t)f−1 is the inverse function of f.
If x is a set, then the powerset x is the set whose elements are the subsets of x. We denote the powerset of x by P(x)
So P(x)={A∣A≤x}
e.g., If x={1,2}, P(x)={∅,{1},{2},{1,2}}
If A and B are sets, their cartesian product is the set A×B={(a,b)∣a∈Aandb∈B}
Note that if A and B are finite sets, then
∣A×B∣=∣A∣×∣B∣If f:S→T and g:T→U, we define the composite function g∘f:S→U by letting
g∘f(s)=g(f(s))for all s∈S
Every integer n≥2 has a unique representation in the form:
n=p∗1e∗1×p∗2e∗2×p∗3e∗3×⋯×p∗ke∗kwhere the p's are distinct primes and each ei≥1. This is called the prime-power decomposition of n.
Suppose n≥2 and the prime-power decomposition (ppd) of n is:
n=p∗1e∗1×p∗2e∗2×p∗3e∗3×⋯×p∗ke∗kThen if m is any positive integer that divides n,
m=p∗1f∗1×p∗2f∗2×p∗3f∗3×⋯×p∗kf∗kfor some integers f1,⋯,fk such that 0≤fi≤ei.
If
n=p∗1c∗1×p∗2c∗2×p∗3c∗3×⋯×p∗kc∗kand
m=p∗1d∗1×p∗2d∗2×p∗3d∗3×⋯×p∗kd∗k,where p1,⋯,pk are distinct primes, and the ci's and di's are nonnegative integers, then
(m,n)=p1min(c1,d1)×p2min(c2,d2)×⋯×pkmin(ck,dk)If n is an integer such that n≥2, then there exists a positive integer m such that n=m2 if and only if all the exponents in the ppd of n are even.
Suppose a≡b (mod m). Then:
- a≡b (mod m)
- a2≡b2 (mod m)
- a3≡b3 (mod m)
- ...
- ak≡bk (mod m) for every k∈Z+.
Thus, using part (ii) of Theorem 8 repeatedly, we get:
If a≡b (mod m), then ak≡bk (mod m) for every k∈Z+.
If p(x) is a polynomial with integer coefficients and a≡b (mod m), then p(a)≡p(b) (mod m).
If n≥3, then φ(n) is even.
For n=1 or 2, φ(n)=1
If p∣n, then (p−1)∣φ(n)
If the prime power decomposition of n is
n=p1e1×⋯×pkekthen
σ(n)=σ(p1e1)×⋯×σ(pkek)=(p1−1p1e1+1−1)×⋯×(pk−1pkek+1−1)Subsets of countable sets are countable.
Given ax≡b(modm), since a−(a−m)=m,
a≡a−m(modm)In modular arithmetic, any integer is equivalent to itself plus or minus any multiple of the modulus.
so
ax≡(a−m)x(modm).Thus
(a−m)x≡b(modm)Every integer is congruent(mod9) to one of {0,1,2,⋯,8}