Prove all other common divisors divide $\gcd$

Question

In the book of Silverman, the below proof is given of the above :

$$a= q_1b + r_1$$ $$b =q_2 r_1+ r_2$$ $$r_1 =q_3r_2 + r_3$$ $$\vdots$$ $$r_{n-3} = q_{n-1}r_{n-2} + r_{n-1}$$ $$r_{n-2} = q_n r_{n-1} + r_n$$ $$r_{n-1} = q_{n+1}r_n + 0$$

But why is $r_n$ the greatest common divisor of $a$ and $b$? Suppose that $d$ is any common divisor of $a$ and $b$. We will work our way back down the list of equations. So from the first equation $a = q_1b + r_1$ and the fact that $d$ divides both $a$ and $b$, we see that $d$ also divides $r_1$.
Then the second equation $b = q_2r_1 + r_2$ shows us $d$ must divide $r_2$.
Continuing down line by line, at each of the i-th step, we will know $d$ divides the previous two remainders $r_{i-1}$ and $r_i;$ and the i-th step $r_{i-1} = q_{i+1}r_i+ r_{i +1}$ will tell us that $d$ also divides the next remainder $r_{i + 1}$.
Eventually, we reach the penultimate line $r_{n-2} = q_nr_{n-1} + r_n$, at which point we conclude that $d$ divides $r_n$. So we have shown that if $d$ is any common divisor of $a$ and $b$, then $d$ will divide $r_n$. Therefore, $r_n$ must be the greatest common divisor of $a$ and $b$.

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Say, $b = 32, a=32*8=256, r_n = b = \gcd(a,b)= 32, r_1=0, d= 4$.
$a=32*8 + 0$, with $b=32, q_1=8.$
Get, $d=4\mid r_n(=32)$.

Or, $b=32, a=33*8.$ Get, $a = b*8(=q_1) + 8(=r_1).$ Then, $b = 4(=q_2)*8(=r_1) + 0(=r_2 = r_{n+1}).$
Get, $d = 4\mid r_n(=8).$

But, it appears unconvincing to me from the very start.
May be a contradiction based proof would have worked better, by taking the prime factorization of $a,b$, and showing that a non-common divisor would not divide at any step, at least two of the three terms.

Or, may this approach is not proper.

Answer 1

Here is a slightly different proof that $r_n$ is the greatest common divisor, based more on what @dssknj remarked (still not by contradiction though).

Lemma: $\gcd(x,y)=gcd(x,y+rx)$ for any integers $x,y,r$.

Proof:
If $d|x$ and $d|y$ then clearly $d|y+rx$. So every common divisor of $x$ and $y$ is also a common divisor of $x$ and $y+rx$.
The converse is also true:
If $d|x$ and $d|y+rx$ then $d|y$ because $y=(y+rx)-rx$. So every common divisor of $x$ and $y+rx$ is also a common divisor of $x$ and $y$.

This means that $\{x,y\}$ have exactly the same common divisors as $\{x,y+rx\}$, and therefore will also have the same greatest common divisor. $\square$

You can now traverse the sequence of equations once in either direction to show that $$ \gcd(a,b) = \gcd(b,r_1) = \gcd(r_1,r_2) = \gcd(r_2,r_3) = ...\\ ... = \gcd(r_{n-2},r_{n-1}) = \gcd(r_{n-1},r_n) = \gcd(r_n,0) = r_n$$

Basically, instead of traversing the sequence of equations once in each direction, this proof builds a two-way connection between adjacent entries, and then you only have to go through the sequence once in either direction to fully prove the connection we want between the first and last.

Answer 2

If you want it by contradiction then here it is...

1) $a=bq+r$ then any common divisor of $a$ and $b$ is common divisor of $b$ and $r$.

Using 1) it is easy to show that $r_n$ is a common divisor of $a$ and $b$. Let assume that $r_n\neq \gcd(a, b)$ then $\gcd(a, b)=d, d\gt r_n$ $\quad d$ is a common factor of $a$ and $b$ and using 1) we have $d\mid r_n \implies d\le r_n$ Hence contradiction.

Answer 3

DEFINITION: Let $a,b \in \mathbb Z$. Then $g \in \mathbb Z^+$ is the greatest common divisor of $a$ and $b$ if

$\qquad \text{$(1.) \quad g \mid a$ and $g \mid b$}$.

$\qquad \text{$(2.) \quad \forall x \in \mathbb Z, \ x\mid a$ and $x \mid b$ implies $x \mid g$}$.

Now, suppose $x \mid a$ and $x \mid b$. Then

\begin{array}{rcl} a= q_1b + r_1 &\implies &x \mid r_1 \\ b =q_2 r_1+ r_2 &\implies &x \mid r_2 \\ r_1 =q_3r_2 + r_3 &\implies &x \mid r_3 \\ &\vdots \\ r_{n-3} = q_{n-1}r_{n-2} + r_{n-1} &\implies &x \mid r_{n-1} \\ r_{n-2} = q_n r_{n-1} + r_n &\implies &x \mid r_n \\ \end{array}

So $r_n$ satisfies condition $(2.)$.

In the other direction.

\begin{array}{rcl} r_{n-1} = q_{n+1}r_n + 0 &\implies &r_n \mid r_{n-1} \\ r_{n-2} = q_n r_{n-1} + r_n &\implies &r_n \mid r_{n-2} \\ r_{n-3} = q_{n-1}r_{n-2} + r_{n-1} &\implies &r_n \mid r_{n-3} \\ &\vdots \\ r_1 =q_3r_2 + r_3 &\implies &r_n \mid r_1 \\ b =q_2 r_1+ r_2 &\implies &r_n \mid b \\ a= q_1b + r_1 &\implies &r_n \mid a \\ \end{array}

So $r_n$ satisfies conditions $(1.)$. Hence $r_n=\gcd(a,b)$