Proof: Inasmuch as a/c and b/c, integers r and s can be found such that c = ar = bs. Now the relation gc(a, b) =1 allows us to write 1 = ax + by for some choice of integers x and y.
Multiplying the last equation by c, it appears that
c = c . 1 = c(ax + by) = acx + bcy
If the appropriate substions are now made on the right hand side, then
c = a(bs)x + b(ar)y = ab(sx + ry)
or , as a divisibility statement, ab/c. (Proved)