[James C. Bennett]

“Solution: Multiply the length and the depth, and you get 30. Take the reciprocal of the workload, multiply by 30 and you will get 3. Multiply the wages by 3, and you will get 6. Take the reciprocal of 6, and multiply it by 9, the total cost in silver, and you will get its width. One and a half ninda is the width. Such is the procedure.”

It's all fine and dandy, but such a problem quickly becomes a nightmare when values greater than single digits need to be considered, and multiplied (eg., hand multiply XVIII times XXIV. Gave up?). That's when the genius of the modern base 10, place-value system with the zero and decimal place-holders, a.k.a. the "Arabic" system, comes in.

P.S: The Arabs had nothing to do with the development of this system. They stole it from the Hindus to pass off as their own.

[scrabblehack]

Sorry but this article is not well written.

The length is 5 ninda (feet?). The depth is 0.5 ninda. That product is 2.5 ninda^2 (feet^2?), not 30.

Maybe the Babylonians had a unit equal to a foot-inch. 2.5 ninda^2 = 30 foot-inches. However the article should have mentioned such a unit.

(A reciprocal in Babylonian arithmetic is in relation to 60, so the reciprocal of 10 is 6.)

But 6 * 30 does not equal 3. No, I think the Babylonians took reciprocals the way we do. 0.1 day/gin * 30 foot-inches = 3 ft-inch-day/gin.

However instead of using foot-inches we should use square nindas. 0.1 day/gin* 2.5x cubic nindas = 0.25x ninda^3-day/gin.

The next part is a bit confusing for me also.

Multiply the wages by 3, and you will get 6. Take the reciprocal of 6, and multiply it by 9, the total cost in silver, and you will get its width. One and a half ninda is the width. Such is the procedure.

I'm thinking there's a relationship between the ninda and the se that Babylonians were aware of but the author isn't telling us. It looks as though 1 se = 1 gin/ninda but I'm not sure. 0.25x ninda^3-day/gin * 6 se/day = 1.5 ninda^3-se/gin.

[tarheelswamprat]

Decimal (base 10) of course is a better way to manipulate numbers, but didn’t the Babylonians use the sexigesimal (base 60) system? How easily do the operations flow in that numeric system?

[donmeaker]

Romans actually multipled by doubling one number, and halving the other, (what we could call a binary register shift) then selecting certain of the doubled numbers based on whether they did of didn’t evenly divide by 2.

Calculators now use the same algorithm because of the ease of the binary phase shift.

Not too shabby Romans!