Theorem:
If $a$ and $b$ are any integers not both zero, and if $q$ and $r$ are any integers such that $a = bq + r$, then GCD($a, b$) $=$ GCD($b, r$).
Proof:
Suppose $a$ and $b$ are integers, with $a≠0$ and $b≠0$, such that $a=bq+r$ for some integers $q$ and $r$.
Step 1:
Let $d$ be a common divisor of $a$ and $b$ such that $d|a$ and $d|b$. By definition of divisibility, $a=dn$, and $b=dm$ for some integers n and m.
Then, $a=bq+r ⟹ dn=dmq+r ⟹ dn-dmq=r ⟹ d(n-mq)=r$.
Let $n-mq=k$. Note that $k$ is an integer since it's the product and difference of other integers. Hence, by definition of divisibility, $r=dk$ for some integer $k$; thus, $d|r$. Therefore, $d$ is a common divisor of $b$ and $r$.
Step 2:
Let $d$ be a common divisor of $b$ and $r$ such that $d|b$ and $d|r$. By definition of divisibility, $b=dx$ and $r=dy$ for some integers $x$ and $y$.
Then, $a=bq+r ⟹ a=dxq+dy ⟹$ $a=d(xq+y)$.
Let $xq+y=h$. Note that $h$ is an integer. Hence, by definition of divisibility, $a=dh$ for some integer $h$; thus, $d|a$. Therefore, $d$ is a common divisor of $a$ and $b$.
Since the set of common divisors for $a$ and $b$ is the same for $b$ and $r$, it then follows that GCD($a,b$) $=$ GCD($b,r$).
• Do I use different symbols for both steps?
"By definition of divisibility, a=dn, and b=dm for some integer n and m."
Let n-mq=k.
By definition of divisibility, b=dx and r=dy for some integer r and s.
Let xq+y=h.
• Is splitting the proof into two steps acceptable?
I am unsure whether or not some definitions are missing in this proof.
I would appreciate some feedback.
