is a given expression an irreducible fraction The following statement is pretty obvious:
For six integers a, b, c, A, B, C

if 

    the six integers are pairwise co-prime, and
    the absolute value of each is greater than 1

then

    a/A is an irreducible fraction
    b/B is an irreducible fraction
    c/C is an irreducible fraction
    (a/A)*(b/B)*(c/C) is an irreducible fraction

But what about this product:
    (1 - a/A)*(1 - b/B)*(1 - c/C)

Is this product always an irreducible fraction?
In other words, can the product never be an integer?
If the product can never be an integer, how do I prove it?
Does it make any difference if A,B,C are squares?
 A: The product can be an integer. Take $(a,b,c,A,B,C)=(3,11,37,5,7,2)$. Note these are pairwise coprime (indeed they are each in fact prime).
Then
$$
\begin{align}
\left(1-\frac{a}{A}\right)\left(1-\frac{b}{B}\right)\left(1-\frac{c}{C}\right)
&=\left(1-\frac{3}{5}\right)\left(1-\frac{11}{7}\right)\left(1-\frac{37}{2}\right)
\\&=\left(\frac{2}{5}\right)\left(\frac{-4}{7}\right)\left(\frac{-35}{2}\right)
\\&=4
\end{align}$$
The idea is that while $a,b,c,A,B,C$ may be pairwise coprime, it does not follow that $A-a,B-b,C-c,A,B,C$ are pairwise coprime (Note that $1-\dfrac{x}{X}=\dfrac{X-x}{X}$).
A: Considering that the product can be written as $\displaystyle\frac{(A-a)(B-b)(C-c)}{ABC}$: 


*

*The product is not always an irreducible fraction: for example, if $A, a, B, b, C, c$ are equal to $5, 3, 2, 17, 13, 11$, respectively, then $A-a=2$ or $C-c=2$ in the numerator can be simplified with $B=2$ in the denominator. 

*The product can be an integer: for example, if $A, a, B, b, C, c$ are equal to $3, 7, 5, 13, 4, 19$, we have $\displaystyle\frac{(3-7)(5-13)(4-19)}{3\cdot5\cdot4}=-8$.

*If $A, B, C$ are squares, the product can still be an integer: for example, if $A, a, B, b, C, c$ are equal to $9, 19, 25, 41, 16, 61$, we have $\displaystyle\frac{(9-19)(25-41)(16-61)}{9\cdot25\cdot16}=-2$.
