Prove $\overline{a} \mid \overline{b} \iff gcd(a,n) \mid gcd(b,n)$

Asked Apr 27 '22 at 19:00

Active Apr 27 '22 at 19:19

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Let $\overline{a},\overline{b} \in\mathbb{Z}_n$. If $\overline{a}\mid\overline{b}$, then $gcd(a,n) \mid gcd(b,n).$

Proof:

$(\Rightarrow)$ By definition, $\overline{a},\overline{b} \in\mathbb{Z}_n \Leftrightarrow a \equiv 0 \pmod n, b \equiv 0 \pmod n.$ Subtracting those, we get $a-b \equiv 0 \pmod n$, i.e., $a-b=nq \Leftrightarrow b=a-nq \,(1), \,q\in\mathbb{Z}.$

By definition, $gcd(a,n) \mid a$ and $gcd(a,n) \mid n \,(2)$. Then, by $(1)$ and $(2)$, $gcd(a,n) \mid b$.

Also by definition, $gcd(a,n)=ax+ny$ and $gcd(b,n)=bv+nw.$ We know $gcd(a,n) \mid n$ and $gcd(a,n) \mid b$. Therefore, $gcd(a,n) \mid gcd(b,n).$

$(\Leftarrow)$ I'm still thinking about it.

Is this proof correct?

Thanks in advance :)

edited Apr 27 '22 at 19:19

Bill Dubuque

272,048

asked Apr 27 '22 at 19:00

geep

Yes this is correct so far. – PTrivedi Apr 27 '22 at 19:13
1

By here in dupe $, a\mid b,$ in $\Bbb Z_n!!\iff (a,n)\mid b!!\overset{\rm \color{#c00}U!!}\iff (a,n)\mid (b,n),$ where $\rm\color{#c00}U$ = gcd universal propergty – Bill Dubuque Apr 27 '22 at 19:19
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The first sentence of the argument is already incorrect,. $\bar a \in \Bbb Z_n,$ denotes a coset $,a+n\Bbb Z,,$ which is the set of all integers $\equiv a\pmod{n},,$ but this is not equivalent to $,a\equiv 0\pmod{n},$ as you claim. – Bill Dubuque Apr 27 '22 at 19:27
The linked answer in the dupe (just now clarified) proves this relation between divisibility in $\Bbb Z$ and divisibility in $\Bbb Z_n = \Bbb Z\bmod n\ \ $ – Bill Dubuque Apr 27 '22 at 19:55

Prove $\overline{a} \mid \overline{b} \iff gcd(a,n) \mid gcd(b,n)$

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