gcd and divisibility in ℤ

Question

I'm studying the divisibility properties in ℤ, there's a passage in my manual I find difficult to grasp. I don't understand why it is consequent that $d_1$ divides $d_2$ in the following statement:

Let $d_1 = \gcd(b, c)$ and $d_2 = \gcd(ab, c)$. Now $d_1 = \gcd(b, c)$ ⇒[($d_1$ divides $b$) and ($d_1$ divides $c$)]. Consequently, $d_1$ divides $ab$, and so it follows that $d_1$ divides $d_2$.

Couldn't be $d_1$ greater than $d_2$, so that it cannot divide $d_2$?

Answer 1

$d_2$ is GCD of $(ab,c)$ which means it is the greatest among all the numbers that divide both ab and c. Since $d_1$ is one such number dividing both ab and c, hence, $d_1\leq d_2$.

Answer 2

By definition, $\gcd(a,b)$ is an integer number such that

• $\gcd(a,b)$ divides $a$ and $\gcd(a,b)$ divides $b$
• if an integer $c$ divides both $a$ and $b$ then $c$ divides $\gcd(a,b)$

Now by assumption you have that $d_1$ divides $b$. This means that there exists an integer $x$ such that $b=d_1x$. Therefore, $ab=d_1ax$. Hence $d_1$ divides both $ab$ and $c$. By the second property of $\gcd(ab, c)$, you have that if a number divides both $ab$ and $c$ then it must divide $\gcd(ab, c)$. Hence $d_1$ divides $\gcd(ab, c)$.