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I'm aware of the proof of this statement. However, the proof simply shows that some integer $d$ is a common divisor. But how does it being a common divisor of $a$ and $b$, $b$ and $r$ show that it is the greatest common divisor?

Bill Dubuque
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airwick
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1 Answers1

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The existence of the common divisor $d$ of any $2$ integers, say $a, b$ assume that they both be different from $0$ comes from the unique prime factorization of both $a,b$. If $a = p_1^{n_1}\cdot p_2^{n_2}\cdots p_m^{n_m}, b = q_1^{s_1}\cdot q_2^{s_2}\cdots q_k^{s_k}$ whereas the $p_i, q_j$ are primes and $n_i, s_j$ are naturals. Then if some $i, j$ such that $p_i = q_j$, then $d$ exists and $d = p_i^{\text{min}(n_i,s_j)}$ is a common divisor since $d \mid a$ and $d \mid b$. If not, then $d = 1$. Among all the common divisors of $a, b$, we can take the largest and call it $\text{gcd}(a,b)$. For the sake of notation, let $d = \text{gcd}(a,b)$, then $d \mid a$, $d \mid b$, so $d \mid (a - bq)$, and hence $d \mid r$. So $d \mid \text{gcd}(b,r)$. If $f = \text{gcd}(b,r)$, then $f \mid b$, $f \mid r$, then $f \mid (bq+r)$ or $f \mid a$. Thus $f \mid \text{gcd}(a,b)$. In brief, we have shown that $\text{gcd}(a,b) \mid \text{gcd}(b,r)$ and $\text{gcd}(b,r) \mid \text{gcd}(a,b) \implies \text{gcd}(a,b) = \text{gcd}(b,r)$.

Wang YeFei
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