离散数学P4 Number Theory I

【P4】 -Mathematics for Computer Science_哔哩哔哩_bilibili

Lec 4 | MIT 6.042J Mathematics for Computer Science, Fall 2010

Number theory: Study of the integers

Def:m|a(m divides a)

• m∣a 是一个谓词，结果是真（true）或假（false）

• 通常要求 m≠0，因为除数不能为零

• “$|$” 是“整除符号”，不能与绝对值符号混淆

“A除B”时，实际计算为 B ÷ A。例如：“6除2”即2 ÷ 6=1/3。除数在前，被除数在后（B ÷ A）

“A除以B”时，计算为 A ÷ B。例如：“6除以2”即6 ÷ 2=3。被除数在前，除数在后（A ÷ B）

所以m|a可以读作m除a，即a除以m， a/m

m|a

exists k, whihc make a=km

suppose a is equal to 0,这里是假设被除数 $a$ 等于 0。 

any integer will define a, will define 0,

这句话的意思是，任何非零整数都能“整除”0。 根据整除的定义，如果 $m$ 整除 0，那么存在一个整数 $k$，使得 $0=k\cdot m$。

because I can take k to be equal to 0, so this is equal to 0 times any integer m,

这里明确了 $0=0\cdot m$。 这意味着，对于任何一个非零整数 $m$，我们都可以找到一个整数 $k=0$，使得 $0=k\cdot m$ 成立。

so m defines  0 for all integers,

对于所有的非零整数 $m$（通常在讨论整除时，除数不为0），我们都可以说 $m$ 整除 0。 换句话说，任何非零整数都能够整除 0。

so this is kind of the exception

这被称为一个“例外”或者“特殊情况”，因为通常我们讨论整除关系时，被除数 $a$ 不会是 0，或者说 0 作为被除数时具有这种普遍性，即任何非零整数都能整除 0。

 

Suppose we have a gallon jug , also have b gallon jug, a=3 b=5, and a<=b

00:11:15提出的theorem

Theorem: m|a and m|b then m|any result.

(这里的 "any result" 指的是水壶中的水量 $x$ 和 $y,通过一系列操作,\; emptying,\; filling,\; pouring来得到不同的x和y)$.

a=3,b=5, m=1

 

State Machine

States pairs(x,y)

x=# gallon in a jug

y=# gallon in b jug

start state:(0,0)

 

Transitions:

emptying

(x,y)->(0,y)

(x,y)->(x,0)

filling:

(x,y)->(a,y)

(x,y)>(x,b)

pouring:

pour form jug a to jug b

(x,y)->(0,x+y)  x+y<=b

(x,y)->(x-(b-y),b)->(x+y-b,b) x+y>=b

pour from jug b to jug a

(x,y)->(x+y,0) x+y<=a

(x,y)->(a,y-(a-x))->(a,x+y-a)  x+y>=a

Example a=3, b=5 
(0,0)->(3,0)->(0,3)->(3,3)->(1,5)->(1,0)->(0,1)->(3,1)->(0,4)

(0,0)->(0,5)->(3,2)->(0,2)->(2,0)->(2,5)->(3,4)

 

Theorem: m|a and m|b then m|any result

(这里的 "any result" 指的是水壶中的水量 $x$ 和 $y,通过一系列操作,\; emptying,\; filling,\; pouring来得到不同的x和y)$.

00:21:06开始证明， 前面10分钟在讲状态机

Proof by induction: Assume m|a and m|b

Invariant: P(n)=If (x,y) is the state after n transitions, then m|x and m|y

Base Case: (0,0) m|0 -> P(0)

Inductive Step: Assume P(n)

Suppose that (x,y) state after n transitions

P(n)-> m|x and m|y

After another transition, each of the jugs are filled with

0, a, b, x, y, x+y, x+y-a, x+y-b gallons

m|0 , m|a, m|b, m|x, m|y =>m divides any of above.=> P(n+1)   

这里用到了整除的封闭性如果 m∣p 且 m∣q，则 m∣(p+q)、m∣(p−q)

 

来这个证明是与倒水问题联系在一起的！这一下子就说通了，因为倒水问题恰好是一个非常适合用整除和裴蜀定理（Bézout's identity） 来分析的经典数学问题，而裴蜀定理的证明确实会涉及到归纳法或其等价形式。

倒水问题与 $ax+by$

我们来回顾一下经典的倒水问题：你有两个没有刻度的水壶，容量分别为 $A$ 升和 $B$ 升，目标是得到恰好 $C$ 升水。每次操作可以是：

1. 灌满一个水壶。
2. 倒空一个水壶。
3. 从一个水壶向另一个水壶倒水，直到其中一个满或另一个空。

在这个问题中，每次操作，水壶中水的总量变化总是 $A$ 的倍数或 $B$ 的倍数。最终留在水壶中的水量，总可以表示成 $Ax+By$ 的形式，其中 $x$ 和 $y$ 是整数（可以是正数、负数或零）。

例如，对于 3 升和 5 升的水壶，目标是得到 4 升水：

• 灌满 5 升水壶 (得到 5 升，即 $0\cdot 3+1\cdot 5$)
• 将 5 升水壶的水倒入 3 升水壶 (5 升水壶剩余 2 升，即 $0\cdot 3+1\cdot 5-1\cdot 3$)
• 倒空 3 升水壶
• 将 5 升水壶中剩余的 2 升水倒入 3 升水壶
• 灌满 5 升水壶 (得到 5 升，3 升水壶中有 2 升)
• 从 5 升水壶向 3 升水壶倒水（倒 1 升，使 3 升水壶满，5 升水壶剩余 4 升，即 $5-1=4$ 升）。
  • 在这里，4 升可以看作是 $3x+5y$ 的形式：$4=3\times (-2)+5\times 2=-6+10$。

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啊，原来这个证明是与倒水问题联系在一起的！这一下子就说通了，因为倒水问题恰好是一个非常适合用整除和裴蜀定理（Bézout's identity） 来分析的经典数学问题，而裴蜀定理的证明确实会涉及到归纳法或其等价形式。

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为什么牵扯到 $m|a$ 和 $m|b⟹m|(ax+by)$

这个倒水问题与 最大公约数 (GCD) 密切相关。Greatest Common Divisor

关键性质： 在只有容量为 $A$ 和 $B$ 的水壶的情况下，你能够量出的任何水量 $C$，都必然是 $A$ 和 $B$ 的最大公约数 $d=\text{gcd}(A,B)$ 的倍数。换句话说，如果 $A$ 和 $B$ 的最大公约数是 $d$，那么你只能得到 $d$ 的倍数的水量。

理由： 每一步操作，水壶中的水量都始终是 $A$ 和 $B$ 的整数线性组合（例如，灌满 5 升水壶是 $0\cdot 3+1\cdot 5=5$；将水倒掉 $3$ 升是从 $5$ 升中减去 $1\cdot 3$，等等）。因此，最终任何能测量出的水量，都形如 $Ax+By$。

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裴蜀定理 (Bézout's Identity)

这里就引入了非常重要的裴蜀定理：

定理： 对于任意不全为零的整数 $a$ 和 $b$，存在整数 $x$ 和 $y$，使得 $ax+by=\text{gcd}(a,b)$。 更一般地，$ax+by$ 的所有可能值都是 $\text{gcd}(a,b)$ 的倍数。反之，所有 $\text{gcd}(a,b)$ 的倍数也都可以表示成 $ax+by$ 的形式。

因此，如果 $m|a$ 且 $m|b$，那么 $m$ 必然整除 $\text{gcd}(a,b)$。进而，由于 $\text{gcd}(a,b)$ 整除 $ax+by$ 的所有可能值，那么 $m$ 也必然整除 $ax+by$ 的所有可能值。

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归纳法与裴蜀定理的证明

您在 MIT 6.042 中可能看到的归纳法证明，很可能是针对裴蜀定理本身的证明，或者是在证明扩展欧几里得算法时使用的归纳法。

扩展欧几里得算法就是一种能够找到 $x$ 和 $y$ 的方法，使得 $ax+by=\text{gcd}(a,b)$。这个算法通常用归纳法来证明其正确性。

在倒水问题中，目标水量 $C$ 能否达到，取决于 $C$ 是否是 $\text{gcd}(A,B)$ 的倍数。

• 对于 3 升和 5 升的水壶：$\text{gcd}(3,5)=1$。
• 因为 $\text{gcd}(3,5)=1$，这意味着你可以通过 3 升和 5 升水壶的组合得到任何 1 的倍数的水量。所以，得到 4 升水是可能的。

所以，当您说 "m|a and m|b then m|any result" 时，这里的 "any result" 指的就是 $ax+by$ 这种形式，而这个结论的根本支撑是裴蜀定理和整除的传递性，而裴蜀定理的证明往往会用到类似欧几里得算法的递推关系，这正是归纳法大显身手的地方。

 

🎯 定理要证：

如果 $m\mid a$ 且 $m\mid b$，那么在只允许用容量为 $a$ 和 $b$ 的两个水壶进行倒水操作的整个过程中，任意时刻两个壶中水量 $x,y$ 都满足：

m \mid x \text{ 且 } m \mid y
]

换句话说：

初始整除结构 + 合法操作 = 永远保持整除结构（整除关系作为不变量）

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🧠 证明策略：数学归纳法（Induction）+ 不变量法（Invariant）

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🔧 一、不变量（Invariant）设定

定义：

• 状态 $(x,y)$ 表示当前两个水壶中的水量

• 命题 $P(n)$ 表示：第 $n$ 步之后的状态 $(x,y)$ 满足 $m\mid x$ 且 $m\mid y$

我们的目标是证明：

$P(0)\Rightarrow P(1)\Rightarrow \cdots \Rightarrow P(n)\forall n\in \mathbb{N}$

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🧱 二、基础情况（Base Case）

初始状态是空壶：

• $(x,y)=(0,0)$

• 显然 $m\mid 0$，所以 $P(0)$ 成立

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🔁 三、归纳步骤（Inductive Step）

假设 $P(n)$ 成立，即：

在第 $n$ 步后，状态是 $(x,y)$，并且 $m\mid x$ 且 $m\mid y$

我们要证明下一步执行任意合法操作后，新状态 (x′,y′)(x', y')(x′,y′) 仍满足 m∣x′m \mid x'm∣x′ 且 m∣y′m \mid y'm∣y′

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⚙️ 四、所有合法操作对应的新状态

你列出的这些操作是典型的水壶倒水操作，包括：

操作	说明	对应表达式
倒空任意壶	$x=0$ 或 $y=0$	$0$
装满壶	$x=a$ 或 $y=b$	$a,b$
相加	把一个壶的水倒进另一个壶直到满或清空	$x+y$
减去壶容量	如 $x+y-a$，表示合并后减去 a 容量	$x+y-a$
其他中间状态	都是这几个的变形

 

重点是：

你所有能达到的新水量，都是在 $x,y,a,b$ 的线性组合范围内

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✅ 五、核心论点：整除的封闭性

归纳假设告诉我们：

• $m\mid x$，$m\mid y$

• 前提又告诉我们 $m\mid a$，$m\mid b$

• 又知道 $m\mid 0$

而整除的定理告诉我们：

如果 $m\mid p$ 且 $m\mid q$，则 $m\mid (p+q)$、$m\mid (p-q)$

所以无论你用 $x,y,a,b$ 做怎样的加减操作，只要是合法操作，就不会破坏“被 $m$ 整除”这个性质

因此：

下一状态中水壶水量仍然被 $m$ 整除，$P(n+1)$ 成立

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🧾 六、结论

归纳证明完成，得出结论：

在任意步操作之后，只要最初 $m\mid a$ 且 $m\mid b$，那么任意状态 $(x,y)$ 中都有 $m\mid x$ 且 $m\mid y$

00:32:38给出gcd的定义

Def: gcd(a,b)=the greatest common divisor of a and b

a=3, b=5, gcd(3,5)=1

00:34:24给出下面的定义

Def:a and b are relatively prime, if gcd(a,b)=1

 

for the jugs game

00:35:20给出下面的推论

Corollary: gcd(a,b) | any reuslt

 

00:36:43提出下面的定理

Theorem: (a<=b)Any linear combination L(L=sa+tb) of a nd b with 0=<L<=b can be reached.

4=-2*3+2*5 等式1

0= 5*3-3*5  等式2

等式1+等式2得到

4=3*3-1*5

Proof: Notice L=sa+tb=(s+mb)a+(t-ma)b

so eiist t',s' L=s'a+t'b with s'>0

Assume 0<L<b

Algorithm:

To obtain L gallon, repeat s' times

Fill a jug

Pour into b jug, when it becomes full, empty it out and continue pouring until a jug is empty

 

First loop:(0,0)->(3,0)->(0,3) 每次操作完, jug a是空的

Second loop:(0,3)-->(3,3)-->(1,5)-->(1,0)-->(0,1)  每次操作完, jug a是空的

Third loop:(0,1)-->(3,1)-->(0,4)

 

Filled the a jug s' times

Suppose b jug emptied u times

Let r be the remainder in b jug, 0=<r<=b

r=s'a-ub  L=s'a+t'b 这里要证明r=L

r=s'a+t'b-t'b-ub=L-(t'+u)b

if t'+u<>0, r<0 or r>b, contradiction here, so t'+u=0

so r=L

 

a=5 b=7 c=6

(0,0)->(5,0)->(0,5)->(5,5)->(3,7)->(3,0)->(0,3)  这里是2*5-7=3

(0,3)->(5,3)->(1,7)->(1,0)->(0,1)   这里是3*5-7*2=1

(0,1)->(5,1)->(0,6) 这里是4*5-7*2=6

 

01:00:10提到下面这个描述

There exists a unique q(quotient) and r(remainder) r=rem(b,a), such that b=qa+r with 0<=r<=a 这是一个描述，但是不打算证明

Lemma: gcd(a,b)=gcd(rem(b,a),a)

 

Euclid's Algorithm欧几里得算法

224=2*105+14

gcd(105,224)=gcd(rem(224,105),105)=gcd(14,105)

105=7*14+7

gcd(14,105)=gcd(rem(105,14),14)=gcd(7,14)

gcd(7,14)=gcd(rem(14,7),7)=gcd(0,7)=7

欧几里得算法（辗转相除法）

原理：

gcd⁡(a,b)=gcd⁡(b,a mod b)

不断用较小数除较大数，直到余数为 0。最后一个非零余数就是最大公约数。

 

Lemma: gcd(a,b)=gcd(rem(b,a),a)

01:07:40开始证明的

Proof: Suppose 【m|a and m|b】  =>  【m| b-qa=rem(b,a)  and m|a】 因为m|a m|b 所以m|rem(b,a) 因为rem(b,a)其实是a和b的线性组合，根据除法的封闭性可知，m|b-qa成立

If rem(b,a)<>0 then 【m| b-qa=rem(b,a) and m|a 】 => m|a and m|b

这一块听不懂，不知道在说什么，参考下面Gemini通过

整除的封闭性，如果m|a and m|b 那么m| (xa+yb)

 

• 双向不等式法：用于证明两个具体的数值或代数表达式相等。它直接得出 $X=Y$ 的结论。
• 夹逼定理：用于证明函数的极限相等。它通过另外两个函数的极限来推导出中间函数的极限。

所以，虽然两者都涉及到不等式，但它们的应用场景和目的完全不同。

When a proof like $\text{gcd}(a,b)=\text{gcd}(\text{rem}(b,a),a)$ is structured by showing two inequalities, it's a very common and robust technique. Here's why it's done that way:

Why Prove with Two Inequalities ($\le$ and $\ge$)?

To prove that two quantities, let's call them $X$ and $Y$, are equal ($X=Y$), a common strategy is to demonstrate two separate inequalities:

1. $X\le Y$: This shows that $X$ is "no larger than" $Y$.
2. $X\ge Y$: This shows that $X$ is "no smaller than" $Y$.

If $X$ is simultaneously less than or equal to $Y$, AND greater than or equal to $Y$, the only possible conclusion is that $X$ must be exactly equal to $Y$.

How it Applies to $\text{gcd}(a,b)=\text{gcd}(\text{rem}(b,a),a)$

Let's call:

• $X=\text{gcd}(a,b)$
• $Y=\text{gcd}(\text{rem}(b,a),a)$

The proof you likely saw in the MIT course (or a similar one) would break down as follows:

You're asking a great question that gets to the heart of how mathematical proofs often work, especially when showing two quantities are equal!

When a proof like $\text{gcd}(a,b)=\text{gcd}(\text{rem}(b,a),a)$ is structured by showing two inequalities, it's a very common and robust technique. Here's why it's done that way:

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Why Prove with Two Inequalities ($\le$ and $\ge$)?

To prove that two quantities, let's call them $X$ and $Y$, are equal ($X=Y$), a common strategy is to demonstrate two separate inequalities:

1. $X\le Y$: This shows that $X$ is "no larger than" $Y$.
2. $X\ge Y$: This shows that $X$ is "no smaller than" $Y$.

If $X$ is simultaneously less than or equal to $Y$, AND greater than or equal to $Y$, the only possible conclusion is that $X$ must be exactly equal to $Y$.

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How it Applies to $\text{gcd}(a,b)=\text{gcd}(\text{rem}(b,a),a)$

Let's call:

• $X=\text{gcd}(a,b)$
• $Y=\text{gcd}(\text{rem}(b,a),a)$

The proof you likely saw in the MIT course (or a similar one) would break down as follows:

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Step 1: Prove $X\le Y$ (i.e., $\text{gcd}(a,b)\le \text{gcd}(\text{rem}(b,a),a)$) 这个证明认为最大的gcd是右侧的

This part of the proof usually involves showing that any common divisor of a and b is also a common divisor of rem(b,a) and a.

Let d=gcd(a,b).
Since d divides a and d divides b, and we know r=b−q⋅a (where r=rem(b,a)), it follows that d must also divide r. (Because b is a multiple of d, and q⋅a is a multiple of d, so their difference r must also be a multiple of d).
So, d is a common divisor of r and a.
Since Y=gcd(r,a) is the greatest common divisor of r and a, and d is a common divisor of r and a, it must be that d≤Y.
Therefore, gcd(a,b)≤gcd(rem(b,a),a). 也就是X≤Y

Step 2: Prove $X\ge Y$ (i.e., $\text{gcd}(a,b)\ge \text{gcd}(\text{rem}(b,a),a)$) 这个证明认为最大的gcd是右侧的

This part of the proof involves showing that any common divisor of $\text{rem}(b,a)$ and $a$ is also a common divisor of $a$ and $b$.

Let d′=gcd(rem(b,a),a).

Since d′ divides $r=\text{rem}(b,a)$ and d′ divides $a$, and we know $b=q\cdot a+r$, it follows that d′ must also divide $b$. (Because $q\cdot a$ is a multiple of d′, and $r$ is a multiple of d′, so their sum $b$ must also be a multiple of d′).

So, d′ is a common divisor of $a$ and $b$.

Since $X=\text{gcd}(a,b)$ is the greatest common divisor of $a$ and $b$, and d′ is a common divisor of $a$ and $b$, it must be that d′≤X.

Therefore, $\text{gcd}(a,b)\ge \text{gcd}(\text{rem}(b,a),a)$. 也就是X$\ge Y$

(Note: This is often stated as X≥d′, which is the same as d′≤X. Since Y=d′, we get $Y\le X$, or $X\ge Y$).

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Conclusion

By proving both $\text{gcd}(a,b)\le \text{gcd}(\text{rem}(b,a),a)$ and $\text{gcd}(a,b)\ge \text{gcd}(\text{rem}(b,a),a)$, you conclusively establish that the two quantities must be equal. This two-sided inequality approach is a powerful and common method for proving equality in mathematics, especially when dealing with concepts like greatest common divisors.

 

The proof structure using $X\ge Y$ and $X\le Y$ implicitly handles the case where $\text{rem}(b,a)=0$, but it's good to explicitly confirm it.

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What Happens When $\text{rem}(b,a)=0$?

If $\text{rem}(b,a)=0$, it means that $b$ is perfectly divisible by $a$. In other words, $a$ is a divisor of $b$.

In this scenario:

• $\text{gcd}(a,b)$ would be $a$ (since $a$ divides both $a$ and $b$, and $a$ is the largest number that divides $a$).
• The expression $\text{gcd}(\text{rem}(b,a),a)$ becomes $\text{gcd}(0,a)$.

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Understanding $\text{gcd}(0,a)$

By definition, the greatest common divisor of $0$ and any non-zero integer $a$ is $|a|$. This is because:

• Any integer divides $0$ (since $0=k\cdot x$ for any $x$, where $k=0$).
• The largest divisor of $a$ is $|a|$.
• So, the greatest common divisor of $0$ and $a$ is $|a|$.

For the context of the Euclidean algorithm, we usually deal with positive integers for $a$ and $b$, so $\text{gcd}(0,a)=a$.

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Checking the Proof with $\text{rem}(b,a)=0$

Let's re-examine our two inequalities when $\text{rem}(b,a)=0$:

We want to prove $\text{gcd}(a,b)=\text{gcd}(0,a)$.

1. Prove $X\le Y$ (i.e., $\text{gcd}(a,b)\le \text{gcd}(0,a)$)

• As established, if $\text{rem}(b,a)=0$, then $\text{gcd}(a,b)=a$.
• Also, $\text{gcd}(0,a)=a$.
• So, $a\le a$, which is true. This inequality holds.

2. Prove $X\ge Y$ (i.e., $\text{gcd}(a,b)\ge \text{gcd}(0,a)$)

• Again, $\text{gcd}(a,b)=a$.
• And $\text{gcd}(0,a)=a$.
• So, $a\ge a$, which is also true. This inequality holds.

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Conclusion

The proof by showing $X\le Y$ and $X\ge Y$ does indeed hold even when $\text{rem}(b,a)=0$. The definition of $\text{gcd}(0,a)=a$ is crucial here and seamlessly integrates into the logic.

In fact, the termination condition of the Euclidean algorithm itself relies on this: when you reach a remainder of 0, the GCD is the last non-zero remainder, which corresponds to the 'a' in $\text{gcd}(0,a)$.

 

Theorem: gcd(a,b) is a linear combination of a and b 也被称为Bézout's Identity (or Bézout's Lemma)

Proof by Induction:

Invariant:

P(n)=If Euclid's algorithm reaches gcd(x,y) after n steps, then both x and y are linear combinations of a and b. and we also know gcd(a,b)=gcd(x,y)

Base Case: 0 steps, we do nothing, x=a, y=b, P(0) is true

Inductive Step: Assume P(n) is true

Notice that exists q which make rem(y,x)=y-qx =>因为y是a和b的线性组合, x也是。->rem(y,x) is linear combinations of a and b

=>P(n+1) 得证？

gcd(0,y)=y

You've accurately captured the essence of an inductive proof for Bézout's Identity in the context of the Euclidean Algorithm, as it might be presented in a course like MIT's 6.042J! It's a clever way to frame it. Let's break down why this inductive approach works and clarify the parts that might be unclear.

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Understanding the Inductive Proof for Bézout's Identity

The goal of Bézout's Identity is to show that for any two integers $a$ and $b$, their greatest common divisor $\text{gcd}(a,b)$ can be written as $sa+tb$ for some integers $s$ and $t$.

gcd(a,b)=sa+tb

The inductive proof you've outlined uses an invariant (a property that stays true throughout the algorithm's execution) and focuses on the intermediate values generated by the Euclidean Algorithm.

The Invariant $P(n)$

Your statement for $P(n)$ is a good start, but let's refine it for clarity and precision, especially in the context of the proof. A more precise $P(n)$ would be:

$P(n)$: "If the Euclidean algorithm has taken $n$ steps to transform the pair $(a,b)$ into (xn​,yn​) (where yn​ is the current dividend and xn​ is the current divisor), then both xn​ and yn​ are linear combinations of $a$ and $b$, and gcd(a,b)=gcd(xn​,yn​)."

Let's use the standard notation for the algorithm steps:

Start with (a0​,b0​)=(a,b).

Step 1: (a1​,b1​)=(b0​,rem(a0​,b0​))

Step 2: (a2​,b2​)=(b1​,rem(a1​,b1​))

...

Step $n$: (an​,bn​)=(bn−1​,rem(an−1​,bn−1​))

So, for your $P(n)$, xn​=bn​ and yn​=an​. (Or, if you prefer to keep $x$ as the divisor and $y$ as the dividend, let xn​ be the divisor at step $n$, and yn​ be the dividend at step $n$).

Base Case: $P(0)$

• $n=0$ steps: The algorithm hasn't started yet. The current pair is $(a,b)$.
• We need to show that $a$ and $b$ are linear combinations of $a$ and $b$.
  • $a=1\cdot a+0\cdot b$
  • $b=0\cdot a+1\cdot b$
  • This is clearly true.
• We also need to show $\text{gcd}(a,b)=\text{gcd}(a,b)$. This is trivially true.

Therefore, $P(0)$ is true.

Inductive Step: Assume $P(k)$ is true, Prove $P(k+1)$ is true

Assume that after $k$ steps, we have reached the pair (xk​,yk​) (where xk​ is the divisor and yk​ is the dividend in the current step), and our inductive hypothesis $P(k)$ states:

1. xk​=sk​a+tk​b for some integers sk​,tk​.
2. yk​=uk​a+vk​b for some integers uk​,vk​.
3. gcd(a,b)=gcd(xk​,yk​).

Now we perform one more step of the Euclidean algorithm to get to (xk+1​,yk+1​). The next step involves dividing yk​ by xk​: yk​=q⋅xk​+r, where r=rem(yk​,xk​).

The new pair in the algorithm will be (xk+1​,yk+1​)=(r,xk​). (Or, if you follow $(b,\text{rem}(a,b))$ structure, it's (xk​,r)). Let's stick with the general form $(x,y)\to (\text{rem}(y,x),x)$. So the new pair is (r,xk​).

We need to prove $P(k+1)$ for the pair (r,xk​):

1. Is $r$ a linear combination of $a$ and $b$?

   • We know r=yk​−q⋅xk​.
   • From our assumption $P(k)$, we know:
     • yk​=uk​a+vk​b
     • xk​=sk​a+tk​b
   • Substitute these into the expression for $r$: r=(uk​a+vk​b)−q(sk​a+tk​b) r=(uk​−qsk​)a+(vk​−qtk​)b
   • Since (uk​−qsk​) and (vk​−qtk​) are integers, $r$ is indeed a linear combination of $a$ and $b$.
   • We also know xk​ is already a linear combination of $a$ and $b$ by the inductive hypothesis.

2. Is gcd(a,b)=gcd(r,xk​)?

   • We know from our core Euclidean Algorithm property that gcd(yk​,xk​)=gcd(rem(yk​,xk​),xk​), which means gcd(yk​,xk​)=gcd(r,xk​).
   • By our inductive hypothesis $P(k)$, we have gcd(a,b)=gcd(yk​,xk​).
   • Therefore, by transitivity, gcd(a,b)=gcd(r,xk​).

So, we've shown that if $P(k)$ is true, then $P(k+1)$ is true.

Conclusion

By the principle of mathematical induction, $P(n)$ is true for all $n\ge 0$.

Now, let's consider the final step of the Euclidean Algorithm. The algorithm stops when we reach a remainder of 0. Let's say this happens at step $N$, where rN​=0. The last non-zero remainder, which is the $\text{gcd}(a,b)$, is rN−1​ (or xN​ in some notations).

So, the algorithm eventually yields a pair $(\text{gcd}(a,b),0)$. According to $P(N-1)$ (or $P(N)$ depending on how you index your steps), the first element of this pair, $\text{gcd}(a,b)$, must be a linear combination of $a$ and $b$. Thus, $\text{gcd}(a,b)=sa+tb$ for some integers $s$ and $t$.

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Why this might have seemed confusing:

The confusion often arises because:

• Defining $P(n)$ precisely: Getting the exact formulation of $P(n)$ and what (xn​,yn​) represents at step $n$ can be tricky.
• Connection to the final result: It's important to explicitly state how the truth of $P(n)$ for all $n$ (specifically, for the final step of the algorithm) proves Bézout's Identity.

This inductive proof is a robust and elegant way to demonstrate Bézout's Identity, showing that the property of being a linear combination of $a$ and $b$ is maintained throughout the Euclidean Algorithm's execution, right up to the point where the GCD is isolated.

 

01:18:35给出这个定理，但是没写证明

Theorem: gcd(a,b) is the smallest positive linear combinations of a and b.

Theorem: $\text{gcd}(a,b)$ is the smallest positive linear combination of $a$ and $b$.

Proof Strategy: We will prove this in two parts:

1. Show that $\text{gcd}(a,b)$ is a positive linear combination of $a$ and $b$. (This is what Bézout's Identity already tells us).
2. Show that $\text{gcd}(a,b)$ is smaller than or equal to any other positive linear combination of $a$ and $b$.

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Part 1: $\text{gcd}(a,b)$ is a positive linear combination of $a$ and $b$.

This part is covered by Bézout's Identity itself. As discussed, the Extended Euclidean Algorithm (or the inductive proof you mentioned) guarantees that for any non-zero integers $a$ and $b$, there exist integers $s$ and $t$ such that:

$\text{gcd}(a,b)=sa+tb$

Since $\text{gcd}(a,b)$ is defined as a positive integer, it satisfies the "positive linear combination" requirement.

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Part 2: $\text{gcd}(a,b)$ is the smallest positive linear combination of $a$ and $b$.

Let $d=\text{gcd}(a,b)$. Let $L$ be any positive linear combination of $a$ and $b$. This means $L=xa+yb$ for some integers $x$ and $y$, and $L>0$.

We need to show that $d\le L$.

Since $d=\text{gcd}(a,b)$, by definition:

• $d$ divides $a$ (i.e., $a=kd$ for some integer $k$).
• $d$ divides $b$ (i.e., $b=md$ for some integer $m$).

Now, substitute these into the expression for $L$: $L=xa+yb$ $L=x(kd)+y(md)$ $L=d(xk+ym)$

Since $x,k,y,m$ are all integers, $(xk+ym)$ is also an integer. Let $C=(xk+ym)$. So, $L=dC$.

This means that $d$ divides $L$.

Since $L$ is a positive integer ($L>0$) and $d$ divides $L$, it must be that $L$ is a positive multiple of $d$. If a positive integer $L$ is a multiple of another positive integer $d$, then $L$ must be greater than or equal to $d$. Therefore, $d\le L$.

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Conclusion

From Part 1, we know that $\text{gcd}(a,b)$ itself is a positive linear combination of $a$ and $b$.

From Part 2, we showed that $\text{gcd}(a,b)$ is less than or equal to any other positive linear combination of $a$ and $b$.

Combining these two facts, $\text{gcd}(a,b)$ must be the smallest positive linear combination of $a$ and $b$.

This proof relies on the fundamental properties of divisibility and the prior knowledge that $\text{gcd}(a,b)$ can indeed be expressed as a linear combination (Bézout's Identity).