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I'm trying to prove\show the associative property for the $\gcd(a,b)$ function.

$$\gcd( a ,\space \gcd(b,c) ) = \gcd(\space \gcd(a,b) , c )$$

Let:

$$e = \gcd( a ,\space \gcd(b,c) )\text{ and }f = \gcd(\space \gcd(a,b) , c )$$

From here you can see that

$$ e\mid a,\gcd(b,c) \Longleftrightarrow e\mid a,b,c $$

And similarly,

\begin{align*} f|\gcd(a,b),c &\Longleftrightarrow f\mid a,b,c\\ &\Longleftrightarrow f\mid e\\ &\Longleftrightarrow e\mid f \end{align*}

If $ e\mid f$ and $f\mid e \Longrightarrow |e| = |f| $ and since $e$ and $f$ are both non-negative we have the result, $e = f$, as required.

Is this proof sound?

Michael Hardy
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kvmu
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    How can you conclude that e|f and f|e. –  May 07 '13 at 02:24
  • I simply used the definition of GCD to come to that, is it a problem?

    e|a,b,c and since f|a,b,c this implies e|f (and vice-versa in this case, f|e).

     – kvmu May 07 '13 at 02:34
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    $\TeX$ tip: You can write "\gcd" and "\mid" to produce "$\gcd$" and "$\mid$" respectively. Both commands automatically format spacing quite nicely. – JavaMan May 07 '13 at 02:59

2 Answers2

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The idea is a good one. But the many logical symbols obscure the logic of the argument.

As you wrote, let $e=\gcd(a,\gcd(b,c))$ and $f=\gcd(\gcd(a,b),c)$.

We want to show that $e$ divides $f$ and $f$ divides $e$. Let us show that $e$ divides $f$. (Remark: it is not a good idea to try to prove two things at once.)

Because $e=\gcd(a,\gcd(b,c))$, it follows that $e$ divides $a$ and $e$ divides $\gcd(b,c)$. So $e$ divides $a$, $b$, and $c$.

It follows that $e$ divides $\gcd(a,b)$ and $e$ divides $c$. This implies that $e$ divides $\gcd(\gcd(a,b),c)$, that is, $e$ divides $f$.

The fact that $f$ divides $e$ is proved in the same way.

Remark: We have used without proof the fact that $x$ divides $y$ and $z$ if and only if $x$ divides $\gcd(y,z)$. One direction is obvious. But if we define $\gcd(y,z)$ as the greatest common divisor of $y$ and $z$, it is not obvious that if $x$ divides $y$ and $z$, then $x$ divides $\gcd(y,z)$. But we can get this by using Bezout's Theorem, and in other ways.

André Nicolas
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  • Thank you! This is exactly how I thought of it in my head, except my presentation is lacking in comparison to yours. – kvmu May 07 '13 at 04:35
  • @kvmu: I meant seriously the parenthetical remark about (except in really trivial cases) not trying to prove $A$ iff $B$ in one step. It is too easy to lose control of the logic. So waste some paper, or electrons, and remain in control. – André Nicolas May 07 '13 at 04:51
  • Point duly noted, thanks :) – kvmu May 07 '13 at 05:38
  • @AndréNicolas the non-obvious direction can also be easily shown using the theorem (see Cormen intro to algorithms 3ed, p. 930) that gcd(a, b) = smallest positive ax+by, where x,y are integers, one of which may be negative. – xdavidliu Dec 20 '17 at 21:38
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We can try the following way:

Let us denote $(x,y)$ as gcd$(x,y)$

Let the highest powers of prime $p$ in $a,b,c$ are $A,B,C$ respectively.

So, the highest power of $p$ that divides $(b,c)$ will be max $(B,C)$

Similarly, the highest power of $p$ that divides $(a,b)$ will be max $(A,B)$

So, the highest power of $p$ that divides $(a,(b,c))$ will be max $(A, $ max$(B,C))=$max$(A,B,C)$

Similarly, the highest power of $p$ that divides $((a,b),c)$ will be max $($ max$(A,B),C))=$max$(A,B,C)$

So, $(a,(b,c))=(a,(b,c))=(a,b,c)$

lab bhattacharjee
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