> I'm surprised there is no such operator for averaging. You can't have an operator for combining portions of groups with fractions alone, because 1/3 = 2/6. C

> I'm surprised there is no such operator for averaging. You can't have an operator for combining portions of groups with fractions alone, because 1/3 = 2/6. Combining groups of B boys and N people total, you get B1+B2 boys and N1+N2 people. Let's use @ for that operator, just to not distract from the usual addition. a/b @ c/d = (a+c)/(b+d). Let's combine a group of 1/3 boys with a group of 2/3 boys. 1/3 @ 2/3 = 3/6. But 1/3 = 2/6, so that should be the same as 2/6 @ 2/3 = 4/9. But 3/6 isn't 4/9. You end up with this issue of a/b @ c/d = p a / p b @ q c / q d = (p a + q c) / (p b + q d), which can be anything. So we have the conclusion that if this operator makes sense, then all numbers are equal. You end up with a notion of numbers that is useless for the original problem of combining groups of people of different genders. What we should do here is define it on pairs of numbers rather than fractions. A fraction and a total (p, N), or the number of boys and number of girls (b, g). The latter is super straightforward: (paul, jenny) + (bob, alice) = (paul and bob, jenny and alice), so numerically, it's (b1, g1) @ (b2, g2) = (b1+b2, g1 + g2), but (1, 3) is not (2, 6) here, unlike with fractions. Real simple. If we want to connect this back to the world of fractions, (p1, N1) @ (p2, N2) = (number of boys / total, total) = ( (p1 N1 + p2 N2)/(N1+N2), N1+N2 ). It's just a weighted average, so you need to keep track of the weights.