> If someone has a better explanation of why subtracting relatively prime numbers repeatedly will always result in one I'm not sure about better, but I think I

> If someone has a better explanation of why subtracting relatively prime numbers repeatedly will always result in one I'm not sure about better, but I think I can explain it more simply. Our subtraction is "a - b = c", where "a" and "b" have no factors in common. And the result is that "c" also has no factors in common with "b". Well, what if they did? What would it look like if "b" and "c" had a common factor? Let's use the example of "b = 6", that was used in the article. 6 has two factors: 2 and 3, so let's see what happens if 2 is a factor of the resulting "c": a - b = c ---------- 8 - 6 = 2 10 - 6 = 4 12 - 6 = 6 14 - 6 = 8 You can see what values of "a" are required to get a factor of "2" as the result, and at this point, I reckon most folk can see the flaw in this cunning plan. But let's rub the point in with 3: a - b = c ---------- 9 - 6 = 3 12 - 6 = 6 15 - 6 = 9 18 - 6 = 12 Again, you can see the result. In order for "b" to have a common factor with "c", "a" has to be some multiple more of that common factor. (slightly more mathy: we're looking at "a = b + c", where b = px and c = qx (x is the common factor), so we have "a = px + qx" yet we're claiming that "x" is not a factor of "a") Anyway, since we're subtracting and always get a smaller number, and it's always a relative prime, we clearly must end up with 1.