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

Discrete Mathematics2026年3月4日发布 芮和
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Elementary Number Theory

  • Divisibility
    primes, division algorithm, greatest common divisor, fundamemtal theorem of arithmetic
  • Congruences
    congruence, residue classes, Euler’s theorem
  • Application (Public-key encryption)
    RSA, modular arithmetics, square and multiply, Euclidean algorithm, prime number generation, CRT
  • Application (Key exchange)
    groups, subgroups, cyclic groups, Discrete logarithm, Diffie-Hellman key exchange

Divisibility

Notation:ℕ={0,1,2,…};ℤ={0,±1,…};ℚ={ba:a,b∈ℤ,a≠0}\mathbb{N}=\left\{0,1,2,\dots\right\};\,\mathbb{Z}=\left\{0,\pm 1,\dots\right\};\,\mathbb{Q}=\left\{\frac{b}{a}:\, a,b\in\mathbb{Z},\, a\neq 0\right\}(rational); ℝ={a.b1b2…,a,b1,b2,⋯∈ℤ}\mathbb{R}=\left\{a.b_1b_2\dots,\,a,b_1,b_2,\dots\in\mathbb{Z}\right\}(real)

Definition: Let a∈ℤ∖{0}a\in\mathbb{Z}\setminus\left\{0\right\} and let b∈ℤb\in\mathbb{Z}.

  • aadividesbb: there is an integer c∈ℤc\in\mathbb{Z} such that b=acb=ac
  • aa is a divisor of bb; bb is a multiple of aa
  • a|ba\mid b: aa divides bb; a∤ba\nmid b: aa does not divide bb
  • n∈{2,3,…}n\in\left\{2,3,\dots\right\} is a prime if the only positive divisors of nn are 11 and nn
  • If n∈{2,3,…}n\in\left\{2,3,\dots\right\} is not a prime, then nn is called a composite
  • Example: nn is composite iff ∃a,b∈(1,n)∩ℤ\exist a,b\in\left(1,n\right)\cap\mathbb{Z} such that n=abn=ab

Theorem (Fundamental Theorem of Arithmetic):
Every integer n>1n>1 can be uniquely written as n=p1e1…prern=p_1^{e_1}\dots p_r^{e_r}, where p1<⋯<prp_1<\dots<p_r are primes and e1,…,er≥1e1,\dots,e_r\geq 1.


FTA Proof

Existence: by mathematical induction on the integer nn

  • n=2n=2: 2=212=2^1 is a product of prime powers
  • Induction hypothesis: suppose there is an integer k>2k>2 such that the theorem is true for all integers nn such that 2≤n<k2\leq n<k
  • Prove the theorem is true for n=kn = k
    • n=kn=k is a prime
      • n=kn=k is a product of prime powers
    • n=kn=k is composite
      • There are integers n1,n2n_1,n_2 such that 1<n1,n2<n1<n_1,n_2<n and n=n1n2n=n_1n_2
      • By induction hypothesis, n1=p1α1…prαrn_1=p_1^{\alpha_1}\dots p_r^{\alpha_r} and n2=q1β1…qsβsn_2=q_1^{\beta_1}\dots q_s^{\beta_s}
        • p1,…,pr,q1,…,qsp_1,\dots,p_r,q_1,\dots,q_s are primes; α1,…,αr,β1,…,βs≥1\alpha_1,\dots,\alpha_r,\beta_1,\dots,\beta_s\geq 1
      • n=n1n2=p1α1…prαr⋅q1β…qsβn=n_1n_2=p_1^{\alpha_1}\dots p_r^{\alpha_r}\cdot q_1^{\beta}\dots q_s^{\beta} is a product of prime powers

Division Algorithm

r=a−bx,{a=bq+r0≤r<br=a-bx,\,\begin{cases}a=bq+r\\0\leq r<b\end{cases}

The Well-Ordering Property: Every non-empty subset of ℕ\mathbb{N} has a least element.

THEOREM (Division Algorithm):
Let a,b∈ℤa,b\in\mathbb{Z} and b>0b>0. Then there are unique q,r∈ℤq,r\in\mathbb{Z} such that 0≤r<b0\leq r<b and a=bq+ra=bq+r.

  • Existence: Let S={a−bx:x∈ℤ and a−bx≥0}S=\left\{a-bx:\,x\in\mathbb{Z}\text{ and }a-bx\geq 0\right\}. Then
    • S≠∅S\neq\emptyset and S⊆ℕS\subseteq\mathbb{N}
      • SS has a least element, say r=a−bq≥0r=a-bq\geq 0
      • If r≥br\geq b, then r−b=a−b(q+1)∈Sr-b=a-b(q+1)\in S and r−b<rr-b<r.
        • The contradiction shows that 0≤r<b0\leq r<b.
  • Uniqueness: Suppose that q′,r′∈ℤ,0≤r′<bq’,r’\in\mathbb{Z},\,0\leq r'<b and a=bq′+r′a=bq’+r’
    • Recall that a=bq+r,0≤r<ba=bq+r,\,0\leq r<b
      • Then b(q−q′)=r′−r∈(−b,b)b\left(q-q’\right)=r’-r\in(-b,b)
        • It must be the case that q=q′q=q’ and thus r=r′r=r’

Ideal

{a=bq′+r′0≤r′<b\begin{cases}a=bq’+r’\\0\leq r'<b\end{cases}

bq′+r′=bq+rbq’+r’=bq+r
b(q′−q)=r−r′∈(−b,b)b\left(q’-q\right)=r-r’\in(-b,b)

DEFINITION: Let I⊆ℤI \subseteq \mathbb{Z} be nonempty. II is called an ideal of ℤ\mathbb{Z} if:

  • 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.
    • Example: dℤ=0,±d,±2d,…d\mathbb{Z} = {0, \pm d, \pm 2d, \dots} is an ideal of ℤ\mathbb{Z} for all d∈ℤd \in \mathbb{Z}.

THEOREM: Let II be an ideal of ℤ\mathbb{Z}. Then ∃d∈ℤ\exists d \in \mathbb{Z} such that I=dℤI = d\mathbb{Z}.

  • If I={0}I = \{0\}, then d=0d = 0.
  • Otherwise, let S={a∈I:a>0}S = \left\{a \in I :\,a > 0\right\}.
    • S⊆ℕS \subseteq \mathbb{N} and S≠∅S \neq \emptyset.
    • due to the well-ordering property, SS has a least element, say d∈Sd \in S.
      • dℤ⊆Id\mathbb{Z} \subseteq I:
        • d∈I⟹dr∈Id \in I \implies dr \in I for any r∈ℤr \in \mathbb{Z}.
      • I⊆dℤI \subseteq d\mathbb{Z}:
        • ∀x∈I,x=dq+r,0≤r<d\forall x \in I,\,x = dq + r,\,0 \le r < d.
        • r=x−dq∈I,0≤r<dr = x – dq \in I,\,0 \le r < d.
        • r=0r = 0 (otherwise, there is a contradiction)
        • x=dq∈dℤx = dq \in d\mathbb{Z}.

DEFINITION: Let I1,I2I_1, I_2 be ideals of ℤ\mathbb{Z}. Then the sum of I1I_1 and I2I_2 is defined as I1+I2={x+y:x∈I1,y∈I2}I_1 + I_2 = \left\{x + y :\, x \in I_1, y \in I_2\right\}

THEOREM: If I1,I2I_1, I_2 are ideals of ℤ\mathbb{Z}, then I1+I2I_1 + I_2 is an ideal of ℤ\mathbb{Z}.

  • ∀a,b∈I1+I2,a+b∈I1+I2\forall a, b \in I_1 + I_2,\, a + b \in I_1 + I_2
    • ∃x1,x2∈I1,y1,y2∈I2\exists x_1, x_2 \in I_1,\,y_1, y_2 \in I_2 such that a=x1+y1;b=x2+y2a = x_1 + y_1;\,b = x_2 + y_2
    • a+b=(x1+x2)+(y1+y2)∈I1+I2a + b = (x_1 + x_2) + (y_1 + y_2) \in I_1 + I_2
  • ∀a∈I1+I2,r∈ℤ,ra∈I1+I2\forall a \in I_1 + I_2,\,r \in \mathbb{Z},\,ra \in I_1 + I_2
    • ∃x∈I1,y∈I2\exists x \in I_1,\,y \in I_2 such that a=x+ya = x + y
    • ra=(rx)+(ry)∈I1+I2ra = (rx) + (ry) \in I_1 + I_2

EXAMPLE:3ℤ+5ℤ=ℤ;4ℤ+6ℤ=2ℤ3\mathbb{Z} + 5\mathbb{Z} = \mathbb{Z};\,4\mathbb{Z} + 6\mathbb{Z} = 2\mathbb{Z}

  • 3ℤ+5ℤ⊆ℤ3\mathbb{Z} + 5\mathbb{Z} \subseteq \mathbb{Z}: this is obvious
  • ℤ⊆3ℤ+5ℤ\mathbb{Z} \subseteq 3\mathbb{Z} + 5\mathbb{Z}:
    • For every n∈ℤ,n=3⋅(2n)+5⋅(−n)∈3ℤ+5ℤn \in \mathbb{Z},\,n = 3 \cdot (2n) + 5 \cdot (-n) \in 3\mathbb{Z} + 5\mathbb{Z}
aℤ+bℤ=gcd⁡(a,b)⋅ℤa\mathbb{Z} + b\mathbb{Z} =\gcd(a,b)\cdot\mathbb{Z}

Greatest Common Divisor

DEFINITION: Let a,b∈ℤa, b \in \mathbb{Z} and at least one of them is nonzero.

  • common divisor: an integer dd such that d|a,d|bd\mid a, d\mid b
  • greatest common divisorgcd⁡(a,b)\gcd(a, b): the largest common divisor
    • relatively prime:gcd⁡(a,b)=1\gcd(a, b) = 1

THEOREM: Let a,b∈ℤa, b \in \mathbb{Z} and at least one of them is nonzero. Then aℤ+bℤ=gcd⁡(a,b)ℤa\mathbb{Z} + b\mathbb{Z} = \gcd(a, b)\mathbb{Z}.

  • a,b≠0⟹aℤ+bℤ≠0{a, b} \neq {0} \implies a\mathbb{Z} + b\mathbb{Z} \neq {0}
  • There exists d∈ℤ∖0d \in \mathbb{Z} \setminus {0} such that aℤ+bℤ=dℤa\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z}. W.l.o.g., d>0d > 0.
    • dd is a common divisor of a,b:a⋅1+b⋅0∈dℤa, b:\,a \cdot 1 + b \cdot 0 \in d\mathbb{Z}
    • dd is greatest: Suppose that d′d’ is a common divisor of a,ba, b
      • d′|a,d′|bd’|a,\,d’|b
      • aℤ+bℤ=dℤ⟹d=as+bta\mathbb{Z} + b\mathbb{Z} = d\mathbb{Z} \implies d = as + bt for some integers s,ts, t
      • d′|dd’\mid d and thus d′≤dd’ \leq d

THEOREM (Bézout): There exist s,t∈ℤs,t\in\mathbb{Z} such that gcd⁡(a,b)=as+bt\gcd(a,b)=as+bt.

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93 条评论

  • 冥界幽歌
    冥界幽歌 读者

    质因数分解唯一性看着简单,写证明时手抖😂

    印度
    回复
  • 铜制钥匙
    铜制钥匙 游客

    密码学全靠这些数学撑着,头皮发麻又佩服

    中国江苏
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  • Ellie
    Ellie 读者

    看到理想那段直接懵了,dℤ和gcd怎么扯上的?

    中国内蒙古
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  • 暗物质研究员
    暗物质研究员 游客

    b要是负数余数咋算?感觉定义会崩

    中国山东
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  • 幽影咒术师
    幽影咒术师 游客

    排版真的劝退,公式挤成一坨谁顶得住啊

    中国北京
    回复
  • 神经猫
    神经猫 读者

    当年考试就栽在扩展欧几里得上,求个步骤图解

    中国陕西
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  • 不说话的小树
    不说话的小树 读者

    RSA那块实际应用挺酷的

    中国湖北
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