Let's look again at the length we considered before, i.e. 5 yd., 2 ft., 4 in.
Do or imagine the following:
Take a whiteboard and an indelible marker. Left to right, draw a strip of three rectangles, so the each rectangle abuts the next. Write the largest unit, yards, above the first rectangle, feet above the second, and the smallest unit, inches, above the third. Just below the strip, under the border between the yards rectangle and the feet rectangle, write the number of feet in a yard, i.e. 3. Just under the border between the feet rectangle and the inches rectangle, write the number of inches in a foot, i.e. 12.
What we have just made is a counting board, specialized to help us solve problems using yards, feet, and inches. Counting boards have a long history, going back through medieval times and well into antiquity.
Put away the indelible marker and take out a dry-erase marker. We now get to use the counting board that we have made.
As a first exercise, we are going to convert 5 yd., 2 ft., 4 in. completely into inches, as we did before. Write 5 in the yards rectangle. Write 2 in the feet rectangle. Write 4 in the inches rectangle. What we have just written means 5 yd. + 2 ft. + 4 in.
The first step is to convert the yards into feet. We need to move the 5 yards to the feet rectangle. Move the 5 toward the feet rectangle. As it crosses the border, the 5 is multiplied by the 3 that marks the borderline. It arrives in the feet rectangle as 15. No surprise there---5 yards is the same length as 15 feet. There is already a 2 there. The 15 and the 2 add to give 17. So now we have nothing---or zero---in the yards rectangle, 17 in the feet rectangle, and still 4 in the inches rectangle.
The second step is to convert all these feet to inches. Move the 17 across the border rightwards, multiplying by 12 as we go. This gives 204. Again, no surprise. 17 feet is the same length as 204 inches. Add this to the 4 inches already there, giving 208 inches. In the end, we get nothing in the yard rectangle, nothing in the feet rectangle, and 208 in the inches rectangle.
What we do with the counting board is exactly what happens when we calculate using the wiggle notation `5_{3}2_{12}4`, also known as `5_{3}2_{12}4;`.
Friday, January 22, 2010
Wiggle on the far side.
Recapping, my baby son's weight at his birth yesterday morning was, in ounces,
`\qquad\qquad\qquad 7_{16}12 = 124.`
Late this afternon, he weighed in at
`\qquad\qquad\qquad 7_{16}8 = 120.`
But what if I wanted pounds, rather than ounces? This is easily done, but needs more notation and more thinking.
In both of the preceding wiggle expressions, the last superdigit---the 12 and the 8---appear at face value. We were looking for an answer in ounces, and they represented numbers of ounces. The 7, on the other hand, represented pounds, and so in ounces it was worth 16 times its face value.
With wiggle notation as we have learned it so far, the rightmost superdigit always appears at face value.
Now, we introduce the unit marker. It is a semicolon, placed just to the right of the superdigit that we want to mark as being at face value. Placing a semicolon immediately after the rightmost superdigit gives the same result as leaving it out :
`\qquad\qquad\qquad 7_{16}12; = 7_{16}12 = (7 xx 16) + 12 = 124.`
`\qquad\qquad\qquad 7_{16}8; = 7_{16}8 = (7 xx 16) + 8 = 120.`
To work in pounds instead of ounces, we move the semicolon to the 7, so that
`\qquad\qquad\qquad 7;_{16}12 = 7 + (16 \\ 12) = 7\frac{12}{16} = 7\frac{3}{4}.`
`\qquad\qquad\qquad 7;_{16}8 = 7 + (16 \\ 8) = 7\frac{8}{16} = 7\frac{1}{2}.`
My boy was `7\frac{3}{4}` pounds at birth, but is down to `7\frac{1}{2}` pounds today. (Those are the actual weights I was given. I haven't fudged them for pedagogical advantage.)
There needs to be a better handle on the use of the semicolon, and as shall soon be seen, there is.
`\qquad\qquad\qquad 7_{16}12 = 124.`
Late this afternon, he weighed in at
`\qquad\qquad\qquad 7_{16}8 = 120.`
But what if I wanted pounds, rather than ounces? This is easily done, but needs more notation and more thinking.
In both of the preceding wiggle expressions, the last superdigit---the 12 and the 8---appear at face value. We were looking for an answer in ounces, and they represented numbers of ounces. The 7, on the other hand, represented pounds, and so in ounces it was worth 16 times its face value.
With wiggle notation as we have learned it so far, the rightmost superdigit always appears at face value.
Now, we introduce the unit marker. It is a semicolon, placed just to the right of the superdigit that we want to mark as being at face value. Placing a semicolon immediately after the rightmost superdigit gives the same result as leaving it out :
`\qquad\qquad\qquad 7_{16}12; = 7_{16}12 = (7 xx 16) + 12 = 124.`
`\qquad\qquad\qquad 7_{16}8; = 7_{16}8 = (7 xx 16) + 8 = 120.`
To work in pounds instead of ounces, we move the semicolon to the 7, so that
`\qquad\qquad\qquad 7;_{16}12 = 7 + (16 \\ 12) = 7\frac{12}{16} = 7\frac{3}{4}.`
`\qquad\qquad\qquad 7;_{16}8 = 7 + (16 \\ 8) = 7\frac{8}{16} = 7\frac{1}{2}.`
My boy was `7\frac{3}{4}` pounds at birth, but is down to `7\frac{1}{2}` pounds today. (Those are the actual weights I was given. I haven't fudged them for pedagogical advantage.)
There needs to be a better handle on the use of the semicolon, and as shall soon be seen, there is.
An introduction to wiggle notation.
This afternoon, my day-and-a-half-old boy was weighed again. He is now 7 lb., 8 oz. (Newborns usually lose weight for the first little bit.) So how many ounces is he? Following the setup on the preceding post, we can calculate :
`\qquad\qquad\qquad (7 xx 16) + 8 = 120,`
i. e., today, he weighed in at 120 oz.
We are going to write the left hand side of the calculation in a new way :
`\qquad\qquad\qquad 7_{16}8 = 120.`
This is an example of wiggle notation. The 7 and the 8, written on the line, are called superdigits when we have to be precise, or digits when there is no risk of confusion. The 16, written at the level of a subscript, is called an interbase when we have to be precise, or else a base when there is no risk of confusion. In general, a superdigit can be any quantity, and so can an interbase. The justification for these names must wait.
To see how wiggle notation works, we should use an example with more superdigits and interbases. Consider, then, the quantity 5 yd., 2 ft., 4 in. How many inches is this? Since there are 3 feet in a yard, and 12 inches in a foot, there are 36 inches in a yard. One way to calculate the total number of inches is to convert each unit directly to inches. Using the number convention, we get :
`\qquad\qquad\qquad (5 xx 36) + (2 xx 12) + 4 = 208.`
Another way is to convert the given yards into feet, add to the feet already given, and then convert that into inches and add the given inches:
`\qquad\qquad\qquad (((5 xx 3) + 2) xx 12) + 4 = 208.`
You get 208 inches either way, of course.
The left hand side can be written in wiggle notation, with :
`\qquad\qquad\qquad 5_{3}2_{12}4 = 208.`
Wiggle notation consists of a series of superdigits, separated by interbases written at level of subscripts. Both ends of a wiggle expression are superdigits, written at the regularly. Between them, the numbers wiggle down and up, like a V, or a W, or ...; there is always one fewer of the interbases than the superdigits.
In `5_{3}2_{12}4` the (super)digits are 5, 2, and 4. The (inter)bases are 3 and 12.
Calculation proceeds left to right, multiplying by bases, and adding digits:
`\qquad\qquad\qquad 5 xx 3 = 15,`
`\qquad\qquad\qquad 15+ 2 = 17,`
`\qquad\qquad\qquad 17 xx 12 = 204,`
`\qquad\qquad\qquad 204 + 4 = 208.`
`\qquad\qquad\qquad (7 xx 16) + 8 = 120,`
i. e., today, he weighed in at 120 oz.
We are going to write the left hand side of the calculation in a new way :
`\qquad\qquad\qquad 7_{16}8 = 120.`
This is an example of wiggle notation. The 7 and the 8, written on the line, are called superdigits when we have to be precise, or digits when there is no risk of confusion. The 16, written at the level of a subscript, is called an interbase when we have to be precise, or else a base when there is no risk of confusion. In general, a superdigit can be any quantity, and so can an interbase. The justification for these names must wait.
To see how wiggle notation works, we should use an example with more superdigits and interbases. Consider, then, the quantity 5 yd., 2 ft., 4 in. How many inches is this? Since there are 3 feet in a yard, and 12 inches in a foot, there are 36 inches in a yard. One way to calculate the total number of inches is to convert each unit directly to inches. Using the number convention, we get :
`\qquad\qquad\qquad (5 xx 36) + (2 xx 12) + 4 = 208.`
Another way is to convert the given yards into feet, add to the feet already given, and then convert that into inches and add the given inches:
`\qquad\qquad\qquad (((5 xx 3) + 2) xx 12) + 4 = 208.`
You get 208 inches either way, of course.
The left hand side can be written in wiggle notation, with :
`\qquad\qquad\qquad 5_{3}2_{12}4 = 208.`
Wiggle notation consists of a series of superdigits, separated by interbases written at level of subscripts. Both ends of a wiggle expression are superdigits, written at the regularly. Between them, the numbers wiggle down and up, like a V, or a W, or ...; there is always one fewer of the interbases than the superdigits.
In `5_{3}2_{12}4` the (super)digits are 5, 2, and 4. The (inter)bases are 3 and 12.
Calculation proceeds left to right, multiplying by bases, and adding digits:
`\qquad\qquad\qquad 5 xx 3 = 15,`
`\qquad\qquad\qquad 15+ 2 = 17,`
`\qquad\qquad\qquad 17 xx 12 = 204,`
`\qquad\qquad\qquad 204 + 4 = 208.`
How many pounds? How many ounces? --part 3.
In the preceding post, we found the following equations for the weight of my newborn:
` \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 equations conform to the quantities convention. With the numbers convention, they would be :
` \qquad \qquad \qquad (7 xx 16) + 12 = 124, `
and
` \qquad \qquad \qquad 7 + (16 \\ 12) = 7\frac{3}{4}. `
Here, simply dropping the units in the quantities convention version gives the corresponding version for the numbers convention. In modern technical usage, quantities convention expressions are almost always written so that this is guaranteed to happen: such expressions are to be "consistent" or "homogeneous".
` \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 equations conform to the quantities convention. With the numbers convention, they would be :
` \qquad \qquad \qquad (7 xx 16) + 12 = 124, `
and
` \qquad \qquad \qquad 7 + (16 \\ 12) = 7\frac{3}{4}. `
Here, simply dropping the units in the quantities convention version gives the corresponding version for the numbers convention. In modern technical usage, quantities convention expressions are almost always written so that this is guaranteed to happen: such expressions are to be "consistent" or "homogeneous".
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.
` \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.
How many pounds? How many ounces? --part 1.
My new littlest, a boy, was born yesterday. He weighed 7 lb., 12 oz. If I let his birthweight be $w$, then
amath \qquad\qquad\qquad w = text(7 lb. + 12 \text( oz. endamath
Apposition is the writing of one thing beside another. In my youngest son's birthweight, 7 lb. and 12 oz. are apposed, and this means they should be added. Similarly, in a mixed fraction such as $7\frac34$, the 7 is apposed to the $\frac34$, and that, too, means that they should be added. Notice that this well-established convention conflicts somewhat with the usual algebraic convention that when one thing is put beside another, they should be not added but multiplied, i.e.
$ \qquad\qquad\qquad 7\;a = 7 \times a. $
There is a formal rule for avoiding conflicts between these two different meanings of apposition:
Now that we have dealt with these conventions, let's write my youngest son's birthweight in a a variety of ways:
amath \qquad\qquad\qquad w = 7 \text( lb., 12 \text( oz. = 124 \text( oz. = 7 \frac{3}{4} \text( lb. endamath
The first of these keeps the numbers small, but uses two different units. The second uses just one unit, at the price of a larger number. The third also uses just one unit, at the price of employing a fraction.
amath \qquad\qquad\qquad w = text(7 lb. + 12 \text( oz. endamath
Apposition is the writing of one thing beside another. In my youngest son's birthweight, 7 lb. and 12 oz. are apposed, and this means they should be added. Similarly, in a mixed fraction such as $7\frac34$, the 7 is apposed to the $\frac34$, and that, too, means that they should be added. Notice that this well-established convention conflicts somewhat with the usual algebraic convention that when one thing is put beside another, they should be not added but multiplied, i.e.
$ \qquad\qquad\qquad 7\;a = 7 \times a. $
There is a formal rule for avoiding conflicts between these two different meanings of apposition:
- Quantities with different units are to be added if they are separated by a comma and a space.
- A whole number and a fraction are to be added if the number comes before the fraction and there is no space between them.
- Quantities are to be multiplied, as in algebra, if they are separated simply by a space.
Now that we have dealt with these conventions, let's write my youngest son's birthweight in a a variety of ways:
amath \qquad\qquad\qquad w = 7 \text( lb., 12 \text( oz. = 124 \text( oz. = 7 \frac{3}{4} \text( lb. endamath
The first of these keeps the numbers small, but uses two different units. The second uses just one unit, at the price of a larger number. The third also uses just one unit, at the price of employing a fraction.
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