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I tried to prove it with linear combination.

Also other information,

  • $h*\gcd(m, n)=h*am+h*bn$
  • At each iteration $i$, $y_i=x_iq+r$
  • Last iteration $t$, so $x_t=\gcd(m, n)$.

Base case: Iteration t. $x_t=h_t*\gcd(m, n)=h_t*am+h_t*bn$. Last iteration, $x_t=\gcd(m, n)$, so $h_t = 1$. Hence, $\gcd(m, n)=am+bn$

I don't know how to prove that m, n are positive integers and a, b are integer.

Assume iteration k, $x_k=h_k*\gcd(m, n)=h_k*am+h_k*bn$, for some integer h, $x_k$ is a multiple of gcd.

We need to prove $x_{k+1}=h_{k+1}*\gcd(m, n)=h_{k+1}*am+h_{k+1}*bn$.

I don't know is that correct, if it is wrong, please tell me

Also, I don't know is this prove correct.

Bernard
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Business Man
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  • This is called Bezout's theorem, you can look it up. – Ritam_Dasgupta May 11 '21 at 12:34
  • What exactly is the statement your are proving with induction. That is, you're proving some statement of the form $\forall k \in {\mathbb N}. P(k)$ by using induction; what exactly is $P(k)$ in your case? – Magdiragdag May 11 '21 at 12:59
  • My $P(k)=x_k=h_kgcd(m, n)=h_kam+h_k*bn$. $x_k$ is the $y_k=x_kq+r$ in iteration in Euclidian algorithm. – Business Man May 11 '21 at 13:03
  • I did say exactly for a reason. What is $h_k$? What are $a, b$? Where are the quantifiers? – Magdiragdag May 11 '21 at 13:04
  • Oh sorry, $h_k$ is just some integer that $x_k=gcd\times h_k$ so $x_k$ is some multiple of $gcd$. From the question, a, b are some integers and m, n are positive integers. What do you mean by "quantifiers"? sorry – Business Man May 11 '21 at 13:07
  • Are you allowed to use the well-ordering principle instead? There is a pretty direct proof using it, and the WOP is equivalent to the principle of induction. – Favst May 11 '21 at 13:10
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    No, $a$ and $b$ are not some integers, at least, they are not fixed integers. They appear quantified (with $\exists$) in $P(k)$! Once you write down really exactly what it is you're trying to prove, you'll see that the base case is totally trivial. – Magdiragdag May 11 '21 at 13:10
  • @Magdiragdag, I know how to do it now, thanks for directing me – Business Man May 11 '21 at 13:15
  • @Favst, we didn't learn about that, so we can't write that, but thanks for helping – Business Man May 11 '21 at 13:16
  • The inductive step is the same as here except use mod (remainder) vs. subtraction as the method of descent to a smaller element. i.e. $,a\Bbb Z+b\Bbb Z = \color{#0a0}{(a!-!qb)\Bbb Z+b\Bbb Z} = \gcd(a!-!qb,b)\Bbb Z = \gcd(a,b)\Bbb Z,$ where $,\color{#0a0}{a!-!qb} = a\bmod b,$ is the remainder left by $,a\div b.\ $ See the other dupes for more conceptual ways to do the induction (Euclidean descent) using a fundamental lemma about sets of integers closed under subtraction (or closed under mod or remainder). – Bill Dubuque May 11 '21 at 20:03

3 Answers3

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So we need to prove $ h_{k+1} * $ gcd $ (m,n) = h_{k+1} * am + h_{k+1} * bn $

Actually this is Bezout's Identity. It states $ax+by=$gcd$(a,b)$. So are you in a good position to complete the proof OP?

Edit: Just as Magdiragdag wrote in comments:

"Actually a and b are just random integers. Once write down really exactly what you are trying to prove and you'll see that the base case is totally trivial"

  • The help that I need is not this, I need help proving that for any positive integer m,n, there exists integer a, b. I need help proving base case. – Business Man May 11 '21 at 13:05
  • "Actually a and b are just random integers. Once write down really exactly what you are trying to prove and you'll see that the base case is totally trivial" –  May 11 '21 at 13:55
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Let $k = \gcd(m,n)$. This implies that $m=kc$ and $n=kd$ for some integers $c$ and $d$, such that $\gcd(c,d)=1$. Now, it is only necessary to prove that two integers exist such that $ac+bd=1$, as one can then multiply both sides by $k$ to solve the original problem.

$$ac+bd=1 \implies \frac {ac}{d}+b=\frac {1}{d} \implies ac \equiv 1 \pmod d$$

Further equivalent to proving that some $a$ exists such that, for at least one value of $n$,

$$ac=nd+1$$

Edit: my bad, I didn't notice the request for this to be in induction.

Bernard
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merelymyself
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The simplest is to prove the basis of the extended euclidean algorithm: at each step of the euclidean algorithm, the remainder is a linear combination of $m$ and $n$.

Let's fix some notations: we have a sequence of euclidean divisions \begin{alignat}{2} m&=q_1n&&+r_1 \\ n&=q_2 r_1 &&+r_2 \\ \vdots&&&\phantom{+}\!\!\vdots \\ r_{i-1}&=q_{i+1}r_i&&+r_{i+1}\\ \vdots&&&\phantom{+}\!\!\vdots \\ \end{alignat} and we rewrite the $(i+1)$th step as $$r_{i+1}=r_{i-1}-q_{i+1}r_i,\tag{1}$$ so that we can make an induction of order $2$:

  • Initialisation: conventionally, $r_{-1}=m,\: r_0=n$, and we can write $$r_{-1}=1\cdot m+0\cdot n,\qquad r_0=0\cdot m+1\cdot n.$$
  • Inductive step: if we suppose $r_{i-1}=a_{i-1}m+b_{i-1}n,\quad r_i=a_i m+b_in$, and using $(1)$, we deduce instantly that $$r_{i+1}=(a_{i-1}-q_{i+1}a_i)m+(b_{i-1}-q_{i+1}b_i)n.$$
Bernard
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