How many pounds? How many ounces? --part 2.

It will be helpful to us to make these conversion calculations explicit :

` \qquad \qquad \qquad (7 " lb."  xx   \frac{16 " oz."}{"lb."})  +  12 " oz." =  112 " oz."  +  12 " oz." = 124 " oz.,"  `

and 

` \qquad \qquad \qquad 7 " lb." +  (12 " oz."  xx   \frac{"lb."}{16 " oz."}) =  7 " lb."  +  \frac{12}{16} " lb." = 7\frac{3}{4} " lb."  `

Let us alter the form of the later conversion calculation a little, so that it uses the same conversion ratio as in earlier conversion calculation.   We have to rewrite the multiplication as a division :

` \qquad \qquad \qquad 7 " lb. " +  (12 " oz." -:  \frac{16 " oz."}{"lb."}) =  7 " lb." + \frac{12}{16} " oz." = 7\frac{3}{4} " lb."  `

Second, we can rewrite the division using the "into" or "reverse division" operator "\".  The into operator divides one quantity by another, but does it in the opposite order from the usual division operator.  So, for instance

`\qquad\qquad\qquad 3\\6 = 2 = 6//3 = 6-:3 = \frac{6}{3}.`

As an aid to memory, it is convenient to think of the 3 in 3\6 as being 'on the bottom', just as it is in `\frac{6}{3}` and 6/3.  Perversely, the reverse division operator is rarely seen or used, although it follows the same order as pencil and paper calculations, at least in the English speaking world :

`\qquad\qquad\qquad 3)\bar{"  "6" "}`

 Using the reverse division operator, then, the conversion calculation becomes :

` \qquad \qquad \qquad 7 " lb." + ( \frac{16 " oz."}{"lb."} "  "\\"  " 12 " oz.") =  7 " lb." + \frac{12}{16} " oz." = 7\frac{3}{4} " lb."  `

At last, then, we have have the two expressions, converting my newborn son's weight of 7 lb., 12 oz. into all ounces or all pounds respectively : 

` \qquad \qquad \qquad (7 " lb."  xx  \frac{16 " oz."}{"lb."})+  12 " oz." = 124 " oz.,"  `

and

` \qquad \qquad \qquad 7 " lb." + ( \frac{16 " oz."}{"lb."} "  "\\"  " 12 " oz.") = 7\frac{3}{4} " lb."  `

These two expressions have a similarity of form that we shall make much use of subsequently.