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Here's what I have so far, I let $d_1$ divide $a$ and $b$ so I could write $a$ and $b$ as $a = d_1k$ and $b = d_1j$. After manipulation, I was able to achieve that $d_1 \mid r$ after substituting values for $a$ and $b$ into $a - bq = r$. Ultimately, I want to show that $d_1 \mid d_2$ and vice-versa to show that my $\gcd$ is unique but how do I go about establishing that it is the greatest. Thank you for your time.

Mr. Brooks
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ponderingdev
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2 Answers2

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Hint $\ $ If $\ d\mid \color{#c00}b\ $ then $\, d\mid \color{#c00}bq\!+\!r\iff d\mid r,\,$ therefore $\, b,bq\!+\!r\,$ and $\,b,r\,$ have the same set $\cal D$ of common divisors $\,d,\,$ hence they have same greatest common divisor $(= \max\cal D).$

Bill Dubuque
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Let $d = \gcd(a,b)$ and $e = \gcd(b,r)$. Since $d | a $ and $d | b$, $d | (a - bq)$, i.e., $d | r$. Since $d | b$ and $d | r$, $d | e$. Argue similarly to obtain $e | d$. Then $e = d$.

kobe
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  • can d ever be bigger than e? – ponderingdev Mar 25 '15 at 16:32
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    By definition of $\gcd(b,r)$, any common divisor of $b$ and $r$ also divides $\gcd(b,r)$. So since $d$ is a common divisor of $b$ and $r$, $d | \gcd(b,r)$, i.e., $d | e$. – kobe Mar 25 '15 at 16:33