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Very confusing, not really sure how I'm supposed to deduce what $\gcd (x,y)$ is and how $$\gcd \left(\frac{x}{\gcd(x,y)} , \frac{y}{\gcd(x,y)}\right)$$ can be $1$?

Daniel Buck
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user122661
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  • They have no common factors, except $1$. – gaoxinge Jan 27 '14 at 05:16
  • the greatest common divisor of x, y also has the property that any other divisor of x, y divides it. Some would say that this is in fact the definition of the greatest common divisor. – ZKe Jan 27 '14 at 05:47

4 Answers4

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HINT: Given two positive numbers and then dividing both of them by their largest common factor, now what is the greatest common factor of the resulting two numbers?

Building on my hint,

Say $x$ has a prime factorization $p_1^{e_1}p_2^{e_2}...p_n^{e_n}$ and $y$ has a prime factorization of $q_1^{e'_1}q_2^{e'_2}...q_m^{e'_m}$, with $e_i,e'_j \ge 1$, both prime factorizations are guaranteed to be unique by the Unique Factorization Theorem, then $gcd(x,y)$ will be found by multiplying all the common primes (and also taking the highest power of that prime found in the two numbers).

Now removing all these common primes and their highest common powers, what can we say about the remaining prime factorization of $\frac{x}{gcd(x,y)}$ and $\frac{y}{gcd(x,y)}$?

ireallydonknow
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One of the basic algebraic facts about $\gcd$ is

$$ \gcd(ab, ac) = a \gcd(b,c) $$

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It is an immediate consequence of the gcd distributive law $\rm(\color{#c00}{DL})$ $\rm\ \ (a,b)c\, =\, (ac,bc).$

Indeed $\rm\,\ d = (x,y) = \left(d\dfrac{x}d, d\dfrac{y}d\right)\,\overset{\rm\color{#c00}{DL}}=\, d\left(\dfrac{x}d,\dfrac{y}d\right)$
$\smash[b]{\rm\overset{\large cancel\ d}\Rightarrow\ 1 = \left(\dfrac{x}d,\dfrac{y}d\right) }\quad $ QED

Below are $\,3\,$ proofs of DL using Bezout's identity, universal gcd laws, and unique factorization.

First we show that the gcd distributive law follows immediately from the fact that, by Bezout, the gcd may be specified by linear equations. Distributivity follows because such linear equations are preserved by scalings. Namely, for naturals $\rm\:a,b,c,x \ne 0$

$\rm\qquad\qquad \phantom{ \iff }\ \ \ \:\! c = (a,b) $

$\rm\qquad\qquad \iff\ \: c\:\ |\ \:a,\:b\ \ \ \ \ \ \&\ \ \ \ c\ =\ na\: +\: kb,\ \ \ $ some $\rm\:n,k\in \mathbb Z$

$\rm\qquad\qquad \iff\ cx\ |\ ax,bx\ \ \ \&\ \ \ cx = nax + kbx,\ \,$ some $\rm\:n,k\in \mathbb Z$

$\rm\qquad\qquad { \iff }\ \ cx = (ax,bx) $

The reader familiar with ideals will note that these equivalences are captured more concisely in the distributive law for ideal multiplication $\rm\:(a,b)(x) = (ax,bx),\:$ when interpreted in a PID or Bezout domain, where the ideal $\rm\:(a,b) = (c)\iff c = gcd(a,b)$

Alternatively, more generally, in any integral domain $\rm\:D\:$ we have

Theorem $\rm\ \ (a,b)\ =\ (ax,bx)/x\ \ $ if $\rm\ (ax,bx)\ $ exists in $\rm\:D.$

Proof $\rm\quad\: c\ |\ a,b \iff cx\ |\ ax,bx \iff cx\ |\ (ax,bx) \iff c\ |\ (ax,bx)/x\ \ \ $ QED

The above proof uses the universal definitions of GCD, LCM, which often served to simplify proofs, e.g. see this proof of the GCD * LCM law.

Alternatively, comparing powers of primes in unique factorizations, it reduces to the following $$ \min(a+c,\,b+c)\ =\ \min(a,b) + c$$

The proof is precisely the same as the prior proof, replacing gcd by min, and divides by $\le$, and

$$\begin{eqnarray} {\rm employing}\quad\ c\le a,b&\iff& c\le \min(a,b),\\ \rm the\ analog\ of\quad\ c\ \, |\, \ a,b&\iff&\rm c\ \,|\,\ (a,b) \end{eqnarray}$$

Community
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Bill Dubuque
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GCD is greatest common divisor,for example $gcd(4,12)=4$

because $4$ is greatest integer on which both number is divisible,now

if $gcd(x,y)=1$ then clearly $gcd(x/gcd(x,y),y/gcd(x,y)=gcd(x,y)=1$

if $gcd(x,y)=x$ then clearly $gcd(x/gcd(x,y),y/gcd(x,y)=gcd(1,y/x)=1$

this can be done if $gcd(x,y)=y$

we almost done,now consider that $gcd(x,y)=m$,could you continue?

for instance $gcd(10,12)=2$

what is

$gcd(10/2,12/2)=gcd(5,6)$

?

dato datuashvili
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