Question 864447
<pre>
We will use the fact that the minimal operator 
'min' is associative, that is,

min(x,min(y,z)) = min(min(x,y),z)

which is easy to prove since both sides can only be the 
minimal of the three numbers x, y and z.

Suppose the prime factorizations of a, b and c are

a = <font size=2>p<sub>1</sub></font><font size=2 face="symbol"><sup>a<sub>1</sub></font><font size=2>p<sub>2</sub></font><font size=2 face="symbol"><sup>a<sub>2</sub></font>···<font size=2>p<sub>k</sub></font><font size=2 face="symbol"><sup>a<sub>k</sub></font>

b = <font size=2>p<sub>1</sub></font><font size=2 face="symbol"><sup>b<sub>1</sub></font><font size=2>p<sub>2</sub></font><font size=2 face="symbol"><sup>b<sub>2</sub></font>···<font size=2>p<sub>k</sub></font><font size=2 face="symbol"><sup>b<sub>k</sub></font>

c = <font size=2>p<sub>1</sub></font><font size=2 face="symbol"><sup>g<sub>1</sub></font><font size=2>p<sub>2</sub></font><font size=2 face="symbol"><sup>g<sub>2</sub></font>···<font size=2>p<sub>k</sub></font><font size=2 face="symbol"><sup>g<sub>k</sub></font>

Then

gcd(b,c) = <font size=2>p<sub>1</sub></font><sup>min(<font size=2 face="symbol">b<sub>1</sub>,g<sub>1</sub></font>)</sup><font size=2><font size=2>p<sub>2</sub></font><sup>min(<font size=2 face="symbol">b<sub>2</sub>,g<sub>2</sub></font>)</sup><font size=2>···<font size=2>p<sub>k</sub></font><sup>min(<font size=2 face="symbol">b<sub>k</sub>,g<sub>k</sub></font>)</sup><font size=2>

gcd(a,b) = <font size=2>p<sub>1</sub></font><sup>min(<font size=2 face="symbol">a<sub>1</sub>,b<sub>1</sub></font>)</sup><font size=2><font size=2>p<sub>2</sub></font><sup>min(<font size=2 face="symbol">a<sub>2</sub>,b<sub>2</sub></font>)</sup><font size=2>···<font size=2>p<sub>k</sub></font><sup>min(<font size=2 face="symbol">a<sub>k</sub>,b<sub>k</sub></font>)</sup><font size=2>

gcd(a,gcd(b,c)) = <font size=2>p<sub>1</sub></font><sup>min(<font face="symbol">a<sub>1</sub></font>,min(<font size=2 face="symbol">b<sub>1</sub>,g<sub>1</sub></font>))</sup><font size=2>p<sub>2</sub></font><sup>min(<font face="symbol">a<sub>2</sub></font>,min(<font size=2 face="symbol">b<sub>2</sub>,g<sub>2</sub></font>))</sup>···<font size=2>p<sub>k</sub></font><sup>min(<font face="symbol">a<sub>k</sub></font>,min(<font size=2 face="symbol">b<sub>k</sub>,g<sub>k</sub></font>))</sup>

Now since the minimal operator 'min' is associative, 
the above is equal to

<font size=2>p<sub>1</sub></font><sup>min(min(<font size=2 face="symbol">a<sub>1</sub>,b<sub>1</sub></font>),<font face="symbol">g<sub>1</sub></font>)</sup><font size=2>p<sub>2</sub></font><sup>min(min(<font size=2 face="symbol">a<sub>2</sub>,b<sub>2</sub></font>),<font face="symbol">g<sub>2</sub></font>)</sup>···<font size=2>p<sub>k</sub></font><sup>min(min(<font size=2 face="symbol">a<sub>k</sub>,b<sub>k</sub></font>),<font face="symbol">g<sub>k</sub></font>)</sup>

which is equal to

gcd(gcd(a,b),c)

and the associativity of gcd is proved.

Edwin</pre>