L02 – Equivalence, Congruence, and Residue Class

Discrete Mathematics2026年3月6日发布 芮和
47.5K 490
9,943字
42–63 分钟

Summary of Lecture 1

L01 – Fundamental theorem of arithmetic, the well-ordering property, division algorithm, ideal,greatest common divisor, and Bézout’s theorem

Divide, Divisor, Multiple, Prime, Composite

  • Fundamental Theorem of Arithmetic:n=p1e1…prern=p_1^{e_1}\dots p_r^{e_r}
  • The Well-Ordering Property:∅≠S⊆ℕ⟹min⁡S∈S\emptyset\neq S\subseteq\mathbb{N}\implies\min{S}\in S
  • Division Algorithm:a∈ℤ,b∈ℤ+,∃!p,q, s.t. a=bq+r,0≤r<ba\in\mathbb{Z},\,b\in\mathbb{Z}^+,\,\exist!p,q,\text{ s.t. }a=bq+r,\,0\leq r<b
  • Ideal of ℤ\mathbb{Z}: A nonempty set I⊆ℤI\subseteq\mathbb{Z} such that a,b∈I⟹a+b∈Ia,b\in I\implies a+b\in I and a∈I,r∈ℤ⟹ra∈Ia\in I,\,r\in\mathbb{Z}\implies ra\in I
    • THEOREM:II is an ideal of ℤ⟺I=dℤ\mathbb{Z}\iff I=d\mathbb{Z}
  • Sum of Ideals:I1+I2={x+y:x∈I1,y∈I2}I_1+I_2=\set{x+y:\,x\in I_1,\,y\in I_2}
    • THEOREM:I1,I2I_1,I_2 are ideals of ℤ⟹I1+I2\mathbb{Z}\implies I_1+I_2 is an ideal of ℤ\mathbb{Z}
  • Sum of ideals:aℤ+bℤ=gcd⁡(a,b)ℤ.{a,b}≠0a\mathbb{Z}+b\mathbb{Z}=\gcd(a,b)\mathbb{Z}.\,\set{a,b}\neq 0
  • Greatest common divisor:gcd⁡(a,b)=as+bt.\gcd(a,b)=as+bt.

FTA Proof

THEOREM: If a,b,c∈ℤ,c|aba,b,c\in\mathbb{Z},\,c\mid ab and gcd⁡(c,a)=1\gcd(c,a)=1, then c|bc\mid b.

  • ∃s,t∈ℤ\exists s,t\in\mathbb{Z} such that c⋅s+a⋅t=1=gcd⁡(c,a)⟹b⋅(c⋅s+a⋅t)=b⋅1c\cdot s+a\cdot t=1=\gcd(c,a)\implies b\cdot(c\cdot s+a\cdot t)=b\cdot 1
    • bcs+abt=bbcs+abt=b
    • c|(bcs),c|(abt)⟹c|bc\mid(bcs),\,c\mid(abt)\implies c\mid b

THEOREM: If pp is a prime and p|abp\mid ab, then p|ap\mid a or p|bp\mid b.

  • p|ap\mid a: done
  • p∤a⟹gcd⁡(p,a)=1p\nmid a\implies\gcd(p,a)=1
    • gcd⁡(p,a)=1∧p|ab⟹p|b\gcd(p,a)=1\wedge p\mid ab\implies p\mid b
p|(a1a2…an)⟹p|a1,p|a2,…,p|an 至少一个成立p\mid\left(a_1a_2\dots a_n\right)\implies p\mid a_1,\,p\mid a_2,\,\dots,\,p\mid a_n\text{ 至少一个成立}

FTA: proof of uniqueness

n=p1e1…prern=p_1^{e_1}\dots p_r^{e_r}
  • Suppose that n=p1…pr=q1…qsn=p_1\dots p_r=q_1\dots q_s, where pi,qjp_i,q_j are all primes
    • p1|n⟹p1|q1…qs⟹p1|qjp_1\mid n\implies p_1\mid q_1\dots q_s\implies p_1\mid q_j for some j⟹p1=qjj\implies p_1=q_j
    • W.l.o.g., we suppose that j=1j=1. Then p2…pr=q2…qsp_2\dots p_r=q_2\dots q_s
    • The theorem is true by induction.

FTA Applications

THEOREM: Suppose that a=p1α1…prαr,b=p1β…prβa=p_1^{\alpha_1}\dots p_r^{\alpha_r},\,b=p_1^{\beta}\dots p_r^{\beta}. Then gcd⁡(a,b)=d\gcd(a,b)=d, where

d:=p1min⁡({α1,β1})…prmin⁡({αr,βr}).d:=p_1^{\min\left(\set{\alpha_1,\beta_1}\right)}\dots p_r^{\min\left(\set{\alpha_r,\beta_r}\right)}.
  • dd is a common divisor of a,ba,b
  • dd is largest among the common divisors
    • Suppose that d′d’ is a common divisor of a,ba,b
    • d′=p1e1…prerd’=p_1^{e_1}\dots p_r^{e_r}
      • d′|a⟹ei≤αid’\mid a\implies e_i\leq \alpha_i for all i∈[r]i\in[r];
        d′|b⟹ei≤βid’\mid b\implies e_i\leq \beta_i for all i∈[r]i\in[r];
        • ei≤min⁡{αi,βi}e_i\leq\min\set{\alpha_i,\beta_i} for all i∈[r]i\in[r]

THEOREM (Euclid) There are infinitely many primes.

  • Supposethereare only nn primes: p1,…,pnp_1,\dots,p_n
  • By FTA, N=p1…pn+1N=p_1\dots p_n+1 must be a product of primes
  • ∃i∈[n]\exists i\in[n] such that pi|Np_i\mid N
  • But pi∤Np_i\nmid N

Equivalence Relation

DEFINITION: Let A,BA,B be two sets. A binary relation from AA to BB is a subset R⊆A×BR\subseteq A\times B. ∀a∈A,b∈B,aRb⟺(a,b)∈R\forall a\in A,\,b\in B,\,aRb\iff(a,b)\in R

EXAMPLE:R={(a,a):a∈ℤ}R=\set{(a,a):\,a\in\mathbb{Z}} is a binary relation from ℤ\mathbb{Z} to ℤ\mathbb{Z}

  • A=ℤ,B=ℤ,R={(a,a):a∈ℤ}⊆A×B,∀a,b∈ℤ,aRb⟺(a,b)∈R⟺a=bA=\mathbb{Z},\,B=\mathbb{Z},\,R=\set{(a,a):\,a\in\mathbb{Z}}\subseteq A\times B,\,\forall a,b\in\mathbb{Z},\,aRb\iff(a,b)\in R\iff a=b

DEFINITION: Let AA be a set. An equivalence relationR⊆A×AR\subseteq A\times A on AA is a binary relation RR from AA to AA such that

  • Reflective:∀a∈A,aRa\forall a\in A,\,aRa
  • Symmetric:∀a,b∈A,aRb⟹bRa\forall a,b\in A,\,aRb\implies bRa
  • Transitive:∀a,b,c∈A,aRb,bRc⟺aRc\forall a,b,c\in A,\,aRb,\,bRc\iff aRc

DEFINITION: The equivalence class of a∈Aa\in A is the set

[a]R={b∈A:aRb}[a]_R=\set{b\in A:\,aRb}

Equivalence Class

THEOREM: Let RR be an equivalence relation on AA. For any a,b∈A,[a]R=[b]Ra,b\in A,\,[a]_R=[b]_R if and only if aRbaRb.

  • ⟹:[a]R=[b]R⟹{a∈[b]Rb∈[a]R⟹aRb\implies:\,[a]_R=[b]_R\implies\begin{cases}a\in[b]_R\\b\in[a]_R\end{cases}\implies aRb
  • ⟸:aRb\impliedby:\,aRb
    • ∀x∈[b]R,bRx\forall x\in[b]_R,\,bRx;
      ∀x∈[a]R,aRx⟹xRb⟹x∈[b]R\forall x\in[a]_R,\,aRx\implies xRb\implies x\in[b]_R
    • [b]R⊆[a]R[b]_R\subseteq[a]_R
    • Similarly, [a]R⊆[b]R[a]_R\subseteq[b]_R

THEOREM: Let RR be an equivalence relation on AA. For any a,b∈Aa,b\in A, either [a]R∩[b]R=∅[a]_R\cap[b]_R=\emptyset or [a]R=[b]R[a]_R=[b]_R.

  • [a]R∩[b]R=∅:x∈[a]R∩[b]R[a]_R\cap[b]_R=\emptyset:\,x\in[a]_R\cap[b]_R, done
  • [a]R∩[b]R≠∅[a]_R\cap[b]_R\neq\emptyset
    • ∃x∈[a]R∩[b]R,{aRxbRx⟹aRb\exists x\in[a]_R\cap[b]_R,\,\begin{cases}aRx\\bRx\end{cases}\implies aRb, i.e., [a]R=[b]R[a]_R=[b]_R

The equivalence classes under RR form a partition of AA: a set {A1,A2,…,An}\set{A_1,A_2,\dots,A_n} of nonempty subsets of AA such that

  • Ai∩Aj=∅,∀i≠jA_i\cap A_j=\emptyset,\,\forall i\neq j
  • ⋃i=1nAi=A\bigcup_{i=1}^n A_i=A

Congruence

THEOREM: Let n∈ℤ+n\in\mathbb{Z}^+. Then R={(a,b)∈ℤ2:n|(a−b)}⊆ℤ×ℤR=\set{(a,b)\in\mathbb{Z}^2:\,n\mid(a-b)}\subseteq\mathbb{Z}\times\mathbb{Z} is an equivalence relation on ℤ\mathbb{Z} (from ℤ\mathbb{Z} to ℤ\mathbb{Z}).

  • RR is a binary relation from ℤ\mathbb{Z} to ℤ\mathbb{Z}
    • Reflective: ∀a∈ℤ,n|(a−a)⟹aRa\forall a\in\mathbb{Z},\,n\mid(a-a)\implies aRa
    • Symmetric: ∀a,b∈ℤ,aRb⟹n|(a−b)⟹b|(b−a)⟹bRa\forall a,b\in\mathbb{Z},\,aRb\implies n\mid(a-b)\implies b\mid(b-a)\implies bRa
    • Transitive: ∀a,b,c∈ℤ,{aRbbRc⟹{n|(a−b)n|(b−c)⟹n|(a−c)⟹aRc\forall a,b,c\in\mathbb{Z},\,\begin{cases}aRb\\bRc\end{cases}\implies\begin{cases}n\mid(a-b)\\n\mid(b-c)\end{cases}\implies n\mid(a-c)\implies aRc

DEFINITION: Let n∈ℤ+n\in\mathbb{Z}^+ and R={(a,b)∈ℤ2:n|(a−b)}R=\set{(a,b)\in\mathbb{Z}^2:\,n\mid(a-b)}.

  • The notation a≡b(modn)a\equiv b\pmod{n} means that aRbaRb.
    • a≡bmodna\equiv b\mod n is called a congruence
      • Read as: aa is congruent to bb modulo nn
      • nn is called the modulus of the congruence
    • a≢b(modn):(a,b)∉Ra\not\equiv b\pmod{n}:\,(a,b)\notin R, or equivalently n∤(a−b)n\nmid(a-b)
      • Read as: aa is not congruent to bb modulo nn

Residue Class

DEFINITION: Let a∈ℤ,n∈ℤ+a\in\mathbb{Z},\,n\in\mathbb{Z}^+. We denote the equivalence class of aa under the equivalence relation modn\mod n with [a]n[a]_n and call it the residue class of amodna\bmod n.

  • [a]n={b∈ℤ:a≡bmodn}={a+nx:x∈ℤ}=a+nℤ[a]_n=\set{b\in\mathbb{Z}:\,a\equiv b\mod n}=\set{a+nx:\,x\in\mathbb{Z}}=a+n\mathbb{Z}
    • any element of [a]n[a]_n is a representative 代表元 of [a]n[a]_n

EXAMPLE:[0]6={0,±6,±12,…}={0+6x:x∈ℤ}=6ℤ;[1]6={…,−11,−5,1,7,13,…};…[0]_6=\set{0,\pm 6,\pm 12,\dots}=\set{0+6x:\,x\in\mathbb{Z}}=6\mathbb{Z};\,[1]_6=\set{\dots,-11,-5,1,7,13,\dots};\,\dots

THEOREM: Let n∈ℤ+n \in \mathbb{Z}^+. For any a∈ℤa \in \mathbb{Z}, there is a unique integer rr such that 0≤r<n0 \leq r < n and a≡r(modn)a \equiv r \pmod{n}. [a]n=[r]n[a]_n=[r]_n

  • Existence: by division algorithm, ∃q,r∈ℤ\exists q, r \in \mathbb{Z} such that 0≤r<n,a=qn+r0 \leq r < n,\,a = qn + r
    • a≡r(modn)a \equiv r \pmod{n}
  • Uniqueness: suppose that 0≤r′<n0 \leq r’ < n and a≡r′(modn)a \equiv r’ \pmod{n}
  • |r−r′|<n\left|r – r’\right| < n and r≡r′(modn)r \equiv r’ \pmod{n}
    • |r−r′|<n\left|r – r’\right| < n and n|(r−r′)n \mid\left(r – r’\right)
      • r=r′r = r’

COROLLARY: For n∈ℤ+n \in \mathbb{Z}^+, there are nn residue classes, i.e., {[0]n,[1]n,…,[n−1]n}\set{[0]_n, [1]_n, \dots, [n-1]_n}.


mod\bmod

DEFINITION: Let a,n∈ℤa, n \in \mathbb{Z} and n>0n > 0. Then there are unique integers q,rq, r such that 0≤r<n0 \leq r < n and a=nq+ra = nq + r.

  • We define amodna \bmod n as rr.

DEFINITION: Let α∈ℝ\alpha \in \mathbb{R}.

  • ⌊α⌋\lfloor \alpha \rfloor: floor of α\alpha, the largest integer ≤α\leq\alpha
  • ⌈α⌉\lceil \alpha \rceil: ceiling of α\alpha, the smallest integer ≥α\geq \alpha
    • If a=nq+ra = nq + r, then q=⌊an⌋q = \left\lfloor \frac{a}{n} \right\rfloor and r=a−nqr = a – nq
α=−1.2⟹{⌊α⌋=−2⌈α⌉=−1\alpha=-1.2\implies\begin{cases}\lfloor \alpha \rfloor=-2\\\lceil \alpha \rceil=-1\end{cases}

ℤn\mathbb{Z}_n

DEFINITION: Let nn be any positive integer. We define ℤn\mathbb{Z}_n to be the set of all residue classes modulo nn.

  • ℤn={[0]n,[1]n,…,[n−1]n}\mathbb{Z}_n = \set{[0]_n, [1]_n, \dots, [n-1]_n} 完全剩余系
    • ℤn={0,1,…,n−1}\mathbb{Z}_n = \set{0, 1, \dots, n-1};
  • ℤn={[1]n,[2]n,…,[n]n}\mathbb{Z}_n = \set{[1]_n, [2]_n, \dots, [n]_n}
    • ℤn={1,2,…,n}\mathbb{Z}_n = \set{1, 2, \dots, n}

EXAMPLE: Two representations of the set ℤ6\mathbb{Z}_6

  • ℤ6={[0]6,[1]6,[2]6,[3]6,[4]6,[5]6}={0,1,2,3,4,5}\mathbb{Z}_6 = \set{[0]_6, [1]_6, [2]_6, [3]_6, [4]_6, [5]_6}=\set{0, 1, 2, 3, 4, 5}
  • ℤ6={[−3]6,[−2]6,[−1]6,[0]6,[1]6,[2]6}={−3,−2,−1,0,1,2}\mathbb{Z}_6 = \set{[-3]_6, [-2]_6, [-1]_6, [0]_6, [1]_6, [2]_6}= \set{-3, -2, -1, 0, 1, 2}

DEFINITION: Let n∈ℤ+n \in \mathbb{Z}^+. For all [a]n,[b]n∈ℤn[a]_n, [b]_n \in \mathbb{Z}_n, define

  • addition:[a]n+[b]n=[a+b]n[a]_n + [b]_n = [a + b]_n
  • subtraction:[a]n−[b]n=[a−b]n[a]_n – [b]_n = [a – b]_n
  • multiplication:[a]n⋅[b]n=[a⋅b]n[a]_n \cdot [b]_n = [a \cdot b]_n

Well-defined?[a]n=[a′]n,[b]n=[b′]n⟹[a]n⋆[b]n=[a′]n⋆[b′]n,⋆∈{+,−,⋅}[a]_n = \left[a’\right]_n,\,[b]_n = \left[b’\right]_n \implies [a]_n \star [b]_n = \left[a’\right]_n \star \left[b’\right]_n,\,\star \in \set{+, -, \cdot}?

  • Yes.
  • [a]n=[a′]n⟹a≡a′(modn)⟹n|(a−a′)⟹∃x[a]_n = \left[a’\right]_n \implies a \equiv a’ \pmod{n} \implies n \mid \left(a – a’\right) \implies \exists x such that a−a′=nxa – a’ = nx
  • [b]n=[b′]n⟹b≡b′(modn)⟹n|(b−b′)⟹∃y[b]_n = \left[b’\right]_n \implies b \equiv b’ \pmod{n} \implies n \mid \left(b – b’\right) \implies \exists y such that b−b′=nyb – b’ = ny
    • (a+b)−(a′+b′)=nx+ny(a + b) – \left(a’ + b’\right) = nx + ny
    • (a−b)−(a′−b′)=nx−ny(a – b) – \left(a’ – b’\right) = nx – ny
    • ab−a′b′=a(b−b′)+b′(a−a′)=any+b′nxab – a’b’ = a\left(b – b’\right) + b’\left(a – a’\right) = any + b’nx
  • [a]n⋆[b]n=[a′]n⋆[b′]n[a]_n \star [b]_n = \left[a’\right]_n \star \left[b’\right]_n

ℤn∗\mathbb{Z}_n^*

{[bs]n=[b]n⋅[s]n=[1]n[bt]n=[b]n⋅[t]n=[1]n⟹[bs]n=[bt]n,bs≡bt(modn)\begin{cases}[bs]_n=[b]_n\cdot[s]_n=[1]_n\\[bt]_n=[b]_n\cdot[t]_n=[1]_n\end{cases}\implies[bs]_n=[bt]_n,\,bs\equiv bt\pmod{n}

DEFINITION: Let n∈ℤ+n \in \mathbb{Z}^+ and [b]n∈ℤn[b]_n \in \mathbb{Z}_n. The residue class [s]n∈ℤn[s]_n \in \mathbb{Z}_n is called an inverse of [b]n[b]_n if [b]n[s]n=[1]n[b]_n [s]_n = [1]_n.

  • The inverse of a residue class is unique (if exist). We denote ([b]n)−1=[s]n\left([b]_n\right)^{-1} = [s]_n.
  • division: If [b]n[s]n=[1]n[b]_n [s]_n = [1]_n, define [a]n[b]n=[a]n⋅([b]n)−1=[a]n⋅[s]n\frac{[a]_n}{[b]_n} = [a]_n \cdot \left([b]_n\right)^{-1} = [a]_n \cdot [s]_n

THEOREM: Let n∈ℤ+n \in \mathbb{Z}^+. [b]n∈ℤn[b]_n \in \mathbb{Z}_n has an inverse if and only if gcd⁡(b,n)=1\gcd(b, n) = 1.

  • Only if: ∃s\exists s such that [b]n[s]n≡[1]n;∃t,bs−1=nt;gcd⁡(b,n)=1[b]_n [s]_n \equiv [1]_n;\,\exists t, bs – 1 = nt;\,\gcd(b, n) = 1
  • If: ∃s,t\exists s, t such that bs+nt=1;bs≡1(modn)bs + nt = 1;\,bs \equiv 1 \pmod{n}

DEFINITION: Let n∈ℤ+n \in \mathbb{Z}^+. Define ℤn∗={[b]n∈ℤn:gcd⁡(b,n)=1}\mathbb{Z}_n^* = \set{[b]_n \in \mathbb{Z}_n :\,\gcd(b, n) = 1}.

  • If nn is prime, then ℤn∗={1,2,…,n−1}\mathbb{Z}_n^* = \set{1, 2, \dots, n-1}
  • If nn is composite, then ℤn∗⊂ℤn\mathbb{Z}_n^* \subset \mathbb{Z}_n

EXAMPLE:ℤ5∗={1,2,3,4};ℤ6∗={1,5};ℤ8∗={1,3,5,7}\mathbb{Z}_5^* = \set{1, 2, 3, 4};\,\mathbb{Z}_6^* = \set{1, 5};\,\mathbb{Z}_8^* = \set{1, 3, 5, 7}

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  • 绝影无踪
    绝影无踪 读者

    mod 6的剩余类举的例子挺直观

    日本
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  • SoloSquad
    SoloSquad 读者

    那个划分(partition)的定理怎么理解?

    中国重庆
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  • 冰霜圣歌者
    冰霜圣歌者 读者

    除法的定义这里没看懂,为啥要反过来乘逆元啊

    印度
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  • 月光秘法师
    月光秘法师 读者

    代表元选负数那部分有点绕

    中国浙江
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  • Wandering Monk
    Wandering Monk 读者

    ℤₙ*的定义终于看懂了,gcd(b,n)=1这个条件

    美国
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  • 东皇
    东皇 读者

    同余的传递性证明蛮清晰的

    中国山东
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    • 竹幽
      竹幽 读者

      是啊,这块讲得明白

      中国浙江@ 东皇
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  • 夜风吟游
    夜风吟游 读者

    等价关系那三个性质确实容易搞混

    中国江西
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  • 松鼠饼干
    松鼠饼干 读者

    模运算的逆元存在条件挺关键的

    美国
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  • 桃子味
    桃子味 游客

    剩余类的代表元可以随便选吗?

    中国上海
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  • 奶油小甜心
    奶油小甜心 读者

    ideal那块定义挺抽象的

    美国
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  • 影畔独行
    影畔独行 读者

    这同余类定义绕晕我了

    韩国
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  • 暗影猎
    暗影猎 游客

    同余关系对称性那里卡住了

    中国山东
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  • 兰皋
    兰皋 游客

    这些定理证明过程跳得有点快

    日本
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  • 绯月吟游者
    绯月吟游者 游客

    模运算的逆元存在条件记不住

    日本
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  • 机智的鹦鹉
    机智的鹦鹉 游客

    离散数学果然是我的噩梦 😭

    中国湖北
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  • 孤城夜雨
    孤城夜雨 游客

    有人能举个具体的同余反例吗?

    日本
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  • 暴走恐龙崽
    暴走恐龙崽 游客

    整除性和等价类之间的联系挺有意思

    印度
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  • 金陵烟雨
    金陵烟雨 游客

    讲义里例子太少了,多来点实际计算就好了

    中国浙江
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  • 镖师张猛
    镖师张猛 游客

    贝祖定理在密码学里是不是很重要?

    中国浙江
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  • 宇宙第一呆
    宇宙第一呆 游客

    理想和最大公约数的关系没太懂

    中国四川
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  • SweetiePie
    SweetiePie 读者

    完全剩余系的概念好抽象

    中国湖南
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  • 话痨雷达
    话痨雷达 游客

    欧几里得那个素数无穷的证明真巧妙

    中国广东
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  • 帆布包
    帆布包 读者

    模n运算到底有啥实际用途啊?

    中国
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  • 天涯漫游
    天涯漫游 游客

    同余关系这块以前考试老错

    中国香港
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  • 社恐但话多
    社恐但话多 游客

    这数学符号看得我眼花缭乱 🤯

    日本
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  • 郑十二
    郑十二 游客

    之前搞密码学作业就栽在 residue class 上,现在看还是有点懵

    中国湖北
    回复
  • 小绒绒糖糖
    小绒绒糖糖 读者

    mod n 的逆元那块卡了好久,有人能说说为啥 gcd(b,n)=1 就一定有逆吗?

    中国北京
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  • 韵凝霜雪
    韵凝霜雪 读者

    模运算的应用场景多吗?

    奥地利
    回复
    • 糖豆子
      糖豆子 读者

      很多,密码学里常用

      中国北京@ 韵凝霜雪
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  • 乐呵呵
    乐呵呵 读者

    这些定理证明写得好详细啊

    日本
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    • 暗夜独航
      暗夜独航 读者

      证明过程确实清晰

      印度@ 乐呵呵
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  • 社交催化剂
    社交催化剂 读者

    同余类乘法那块儿有点懵

    中国山东
    回复
  • 宇宙第一帅的猪
    宇宙第一帅的猪 读者

    模运算这块儿挺绕的

    中国浙江
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  • 夜色如墨
    夜色如墨 读者

    等价关系的定义还是得背熟

    德国
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  • 糖豆妹妹
    糖豆妹妹 读者

    同余那块儿老记混,得再瞅瞅

    爱尔兰
    回复
  • Classy Clyde
    Classy Clyde 读者

    数学系的吧,看着就眼熟

    阿根廷
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  • 米粒粒
    米粒粒 读者

    等价类的划分原来这么来的

    巴西
    回复
    • 铁背苍龙
      铁背苍龙 读者

      同感,一下就明白了

      印度@ 米粒粒
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  • value29
    value29 读者

    同余关系这块得画个图才清楚

    墨西哥
    回复
    • PinkyPinky
      PinkyPinky 读者

      我也画了图才明白

      中国湖南@ value29
      回复
  • Silent Thunder
    Silent Thunder 读者

    模运算这块儿有点绕啊

    美国
    回复
  • 谦和
    谦和 读者

    这讲义写得也太干了吧,看得我头大 😵

    中国山东
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