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Can anyone find a hole in my proof or tell me how to make it stronger or simpler? The "lemma above" referenced in the fourth paragraph is a small but solid proof showing that $r$ is strictly less than $b$ in the division algorithm if it's written $a=bq+r$.

$\mathbf {Theorem}$ If $m$ and $n$ are positive integers that are relatively prime, then there are integers $a$ and $b$ such that $am+bn=1$.

$\mathbf {Proof}$ Let $D = \{d \in \Bbb Z : am+bn>0\}$. Then there is a least element $l$ in D. Since $m$ and $n$ are relatively prime, we need only show that $l$ divides $m$ and $n$.

Suppose $l$ does not divide $m$. Then there are integers $q$ and $r$ such that $m=ql+r$. Rewritten, we have that $r=ql-m=q(am+bn)-m=qam-m+qbn=m(qa-1)+n(qb)$.

Thus, $r \in D$. By the lemma above, $r<l$, contradicting that $l$ is the least element of $D$. We conclude then that $l$ divides $m$. Switching $m$ for $n$ gives the same result for $n$.

Since $m$ and $n$ are relatively prime, $l=1$ and it follows that there are integers $a$ and $b$ such that $am+bn=1$.

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marked as duplicate by rtybase, Siong Thye Goh, Xander Henderson, Namaste, Moya Sep 10 '17 at 20:31

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  • $\begingroup$ Have a look at this post : math.stackexchange.com/questions/2215445/… $\endgroup$ – zwim Sep 10 '17 at 1:04
  • $\begingroup$ @zwim Okay, I see. Are you implying that I still need to prove that any other common divisor of $m$ and $n$ divides $l$? I figured I didn't need to since we already know that they are relatively prime. $\endgroup$ – Austin Sep 10 '17 at 1:11
  • $\begingroup$ I didn't proofread your post in details, seemed ok, I just remembered having done it in details for any gcd (not just relatively prime numbers), so I linked it for you to have a comparison point. $\endgroup$ – zwim Sep 10 '17 at 1:18
  • $\begingroup$ Okay, thanks! I actually really appreciate that, it was just that link made think oh crap did I screw up? Thanks for your feedback, it's appreciated! $\endgroup$ – Austin Sep 10 '17 at 1:24
  • $\begingroup$ See this answer for a more conceptual presentation of the standard proof of Bezout's GCD identity. $\endgroup$ – Bill Dubuque Sep 10 '17 at 15:24
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proof-verification:

The "lemma above" referenced in the fourth paragraph is a small but solid proof showing that $r$ is strictly less than $b$ in the division algorithm if it's written $a=bq+r$. ("the lemma" should be clearly stated before or after the proof.)

Proof (Suppose m and n are two positive integers that are relatively prime.) Let $D = \{d \in \Bbb Z : am+bn>0\}$. (This set is ill-defined: what is the constraint for d?) Then there is a least element $l$ in D. (This is nontrivial. Can you explain?) Since $m$ and $n$ are relatively prime, we need only show that $l$ divides $m$ and $n$. (so that l must be 1, which is what we want to prove.)

Suppose $l$ does not divide $m$. Then there are integers $q$ and $r$ such that $m=ql+r$. (There should be some properties about q and r that you need later.) RewrittenRewriting it, we have that $r=ql-m=q(am+bn)-m=qam-m+qbn=m(qa-1)+n(qb)$. (Where is this rewritten formula used in the later argument?)

Thus, $r \in D$. By the lemma above ("the lemma" should be clearly stated before or after the proof.), $r<l$, contradicting that $l$ is the least element of $D$. We conclude then that $l$ divides $m$. Switching $m$ for $n$ gives the same result for $n$.

Since $m$ and $n$ are relatively prime, $l=1$ and it follows that there are integers $a$ and $b$ such that $am+bn=1$. (This sentence is redundant.)

The result you prove is called Bézout's identity. See also a proof in the linked page.

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  • $\begingroup$ Thank you for all the feedback! It's not too often that I get responses like this so I want to thank you for the effort you put into this. I have a couple of questions for you. The set $D$ is defined in that the constraint on $d$ is that it must be a positive integer such that $am+bn>0$. I'm wondering how I can define it better because that seems pretty clear cut to me.. as a beginner. The formula $m(qa-1)+n(qb)$ is used to show that $r \in D$. Perhaps the line break is confusing there... $\endgroup$ – Austin Sep 10 '17 at 1:49
  • $\begingroup$ The notation written in OP does not have the meaning you want. You could use (for fixed $m$ and $n$) $$D=\{d\in{\bf Z}: d=am+bn\, \hbox{ for some } a,b\in{\bf Z} \hbox{ and } d>0\}.$$ $\endgroup$ – Jack Sep 10 '17 at 12:08
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suppose $\exists a,b\in Z;am+bn=1$

and we have $(m,n)=1$ (because $m$ and $n$ are positive integers that are relatively prime)

$d=(m,n)$ so $d|m$ and $d|n$

but $d|am+bn$ (according to $Theorem$:$\forall x,y\in Z; c|a \wedge c|b \Rightarrow c|(ax+by)$

so $d|1$, but $d\in Z^*$ ($m$ and $n$ are positive integers)

so $d=1$

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