Prove that if $\gcd( a, b ) = 1$ then $\gcd( ac, b ) = \gcd( c, b ) $

Question

I know it might be too easy for you guys here. I'm practicing some problems in the textbook, but this one really drove me crazy.
From $\gcd( a, b ) = 1$, I have $ax + by = 1$, where should I go from here? The extra $ac$ is so annoying. Any hint?

Thanks,
Chan

Answer 1

Let $d_1 = \gcd(c,b)$ and $d_2 = \gcd(ac,b)$.

So we have $cx_1 + by_1 = d_1$ and $acx_2 + by_2 = d_2$.

Further we have $ax + by = 1$.

Now multiply $ax+by = 1$ by $d_1$, write $d_1 = cx_1 + by_1$ and rearrange to show that $d_2|d_1$ i.e.

$d_1 \times (ax+by = 1) \Rightarrow ax(cx_1 + by_1) + bd_1y = d_1 \Rightarrow a c (x x_1) + b(a x y_1 + d_1 y) = d_1$. $d_2 = \gcd(ac,b)$ divides any integer linear combination of $ac$ and $b$ and hence $d_2 | d_1$

And by a similar argument multiply $ax+by = 1$ by $d_2$, write $d_2 = acx_2 + by_2$ and rearrange to show that $d_1|d_2$ i.e.

$d_2 \times (ax+by = 1) \Rightarrow ax(acx_2 + by_2) + bd_2y = d_2 \Rightarrow c (a^2 x x_2) + b(a x y_2 + d_2 y) = d_2$. $d_1 = \gcd(c,b)$ divides any integer linear combination of $c$ and $b$ and hence $d_1 | d_2$

So we have $d_1 | d_2$ and $d_2 | d_1$ and $d_1$,$d_2$ are non-negative since they are the gcd of two integers and hence $d_1 = d_2$

Answer 2

My attempt at a Bill Dubuquesque argument: \begin{align*} r|ac,\;b &\Longleftrightarrow r|ac,\;bc,\;b\\ &\Longleftrightarrow r|\gcd(ac,bc),\;b\\ &\Longleftrightarrow r|\gcd(a,b)c,\;b &\quad&\mbox{(since $\gcd(ac,bc)=\gcd(a,b)c$)}\\ &\Longleftrightarrow r|c,\;b \end{align*}

Answer 3

This follows easily from basic laws of GCD arithmetic. First I will present the proof using the standard notation $\rm\: (a,b)\:$ for $\rm\: gcd(a,b),\: $ immediately followed by a proof employing a more suggestive arithmetical notation, namely denoting $\rm\:\gcd(a,b)\:$ by $\rm\ a \dot+ b\:.\:$ Because the arithmetic of GCDs shares many of the same basic laws of the arithmetic of integers, the proof becomes more intuitive using a notation that highlights this common arithmetical structure. The proof employs only said basic GCD laws, namely the associative law, and the commutative law, combined with the fundamental distributive law $\rm\, \ (ac,bc)\ =\ (a,b)\:c\,\ $ and the GCD-specific law $\rm\: \color{#C00}{(c,1) = 1}.\: $

$\begin{eqnarray} \qquad\ &\rm (ac,\:b)\: &=&\rm\: (ac,b\:\color{#C00}{(c,1)})\: &=&\rm\: (ac,\:bc,\:b) &=&\rm\ ((a,b)\:c,b)\: &[\rm\: &=&\rm\: (c,\:b)\ &\rm if\ \ \ (a,b) &=&\rm 1\:] \\ &\rm ac \dot+ b &=&\rm\ ac \dot+ b\:\color{#C00}{(c \dot+ 1)} &=&\rm\,\ ac\dot+bc\dot+b &=&\rm\ (a\dot+b)\:c \dot+ b\ &[ &=&\rm\ \: c \dot+ b\ \:\ &\rm if\rm\quad a\dot+b &=&\rm 1\:] \end{eqnarray}$

Notice how the suggestive notation in the second proof invites us to exploit our well-honed arithmetical intuition regarding the associative, commutative and distributive laws of integer arithmetic during analogous GCD arithmetic proofs. $\ $ For a less trivial example see my similar proof of the Freshman's Dream $\rm\ (A,B)^n\, =\ (A^n,B^n)\ $ for GCDs and cancellable ideals.

The motivation behind this powerful abstract axiomatic approach becomes much clearer should you go on to study ideal theory and/or divisor theory (and you should, for there lies much beauty).

For further discussion and generalizations see this post and see my menagerie of posts on GCDs.

Answer 4

Not sure if I can prove it algebraically but can try to explain logically. There is no common denominator between a and b (this is what gcd(a,b) = 1 means).

When you multiply a * c you introduce c's prime factors to the new number "ac".

So, if any new common factors (between ac and b) are introduced, we know that they came solely from c. We know this because prior to multiplying by c there were no common denominators. So, the only commonality that could originate here is from c and b.

Answer 5

(a,c)=1, so there are integers x,y such that ax+cy=1. (b,c)=1, so there are integers z,w such that bz+cw=1.

multiplying the two equations we get, axbz+axcw+cybz+cycw=1

simplifying our answer we have, ab(xz)+c(axw+yzb+ycw)=1.

it follows that (ab,c)=1.

Answer 6

We could prove the contrapositive: $\gcd(ac,b) \neq \gcd(c,b) \Longrightarrow \gcd(a,b) \neq 1$. Let $d = \gcd(ac,b)$ and $d' = \gcd(c,b)$. Now $d' < d$. So $d$ does not divide $c$ and $b$ and $d$ divides $ac$ and $b$. So $\gcd(a,b) \neq 1$.