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Let $ a \in \mathbb{N}$, $ b \in \mathbb{N}_0 $, $ r \in \mathbb{N}_0 $, $ q\in \mathbb{N} $ and $ b \leq a$ so that $$a = bq + r $$with $$ 0 \leq r \lt b$$

Proof:

Let $ D = \{d \in \mathbb{Z} \vert ~ d \mid a \text{ and } d \mid b \}$ and Let $ D' = \{d \in \mathbb{Z} \vert ~ d \mid b \text{ and } d \mid r \}$

Let $ x \in D $, so that $x \mid a$ and $x \mid b$:

$$ a = bq +r $$ $$ a - bq = r $$

So $x \mid a-bq=r$ so $x \mid r$

Let $$ x \in D', $$ so that $x \mid b$ and $x \mid r$:

$$ a = bq +r $$

So $x \mid b$ and $x \mid r$ but also $x \mid a$ and since $$ a \in D$$

$$ \gcd(a,b) = \gcd(b,r)$$

Any critique? Did i proved it somehow in a correct way? If not, could you show me what i did wrong or how to do it properly? Thanks in advance!

Bill Dubuque
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JohnGam
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  • Use $x$ for in-line notation $x$. Use $$x$$ for centred $x$, like so: $$x$$ – Shaun May 06 '21 at 20:02
  • @Shaun updated. Thanks for letting me know! – JohnGam May 06 '21 at 20:06
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    The proof looks correct to me. You proved that the set of common divisors of both pairs is identical, from which it follows that the smallest element of those sets also must be the same. – Robert Shore May 06 '21 at 20:19
  • @RobertShore the only thing that bothers me is the part with 0 ≤ r < b and the definition b≤a. It feels to me like i have to include that somewhere, but i don't know really where. – JohnGam May 06 '21 at 20:22
  • Did you mean to write $a \in D$ or did you mean $x \in D$? They are both true I don't see how $a\in D$ leads to your conclusion. ..... "It feels to me like i have to include that somewhere, but i don't know really where" It's not required. If $a= bn + s$ then $\gcd(a,b) = \gcd(b,s)$ regardless of how big $a$ and $s$ are. The only reason those are specified is to show it is true for the unique remainder $r$ (it's also true of the $s= r + Mb$ for any $r$ plus some multiple of $b$)... Basically an number $s$ and any $s+Mb$ will have the exact same factors in common with $b$. – fleablood May 06 '21 at 20:51
  • The hypotheses $0 \le r < b$ and $b < $ are not relevant here. They become useful when you want to iterate the construction to calculate $\gcd(a, b)$ (Euclid's algorithm). – Rob Arthan May 06 '21 at 21:30
  • The proof shows that if $,a = bq+r,$ for $,q,r\in\Bbb Z,,$ then $,x\mid a,b\Rightarrow x\mid r,,$ $\rm \color{#c00}{thus}$ $,D\subseteq D',,$ i.e. $,{\rm CD}(a,b)\subseteq {\rm CD}(b,r),,$ for $,\rm CD := $ set of common divisors (you forgot to state the conclusion after $\rm \color{#c00}{thus}),,$ and similarly $,{\rm CD}(b,r)\subseteq {\rm CD}(a,b),,$ so the CD sets are equal, so they have equal greatest elements, i.e. equal GCDs. This is a very common proof, e.g. see here the linked dupe for a concise proof. – Bill Dubuque May 06 '21 at 22:11

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