Sometimes, numbers and other quantities are used to count segments : "chapter one, chapter two, chapter three"; "first mile, second mile, third mile"; "first year, second year, third year".
Sometimes, they are used to mark waypoints: "start (milestone 0), milestone 1, milestone 2, milestone 3"; "0 years old, first birthday, second birthday, third birthday".
Sometimes, board games are played with pieces placed in the spaces: chess, draughts (checkers), snakes and ladders, noughts and crosses (tic-tac-toe).
Sometimes, board games are played with pieces placed at the intersections: nine men's morris, go.
In all of our multiplication tables so far, numbers have occupied cells in an array.
Let us now fashion a multiplication table where the numbers are now thought of as labels for points.
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These four points are the corners of a square. The idea now is to think of every point in the square as having a number, so that all the numbers form a continuous stripy multiplication table. The line joining through the two `10` points consists of points that all have the value ten. This is a stripe of the continuous table.
We can mark some other points, to help see how things are working :
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c[2]=[10.25,-11]; text([10.25,-11],"17.8",above);
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This is clearly still a multiplication table of sorts. To better emphasize continuity, we can drop all of our marker points except those on the perimeter of the square, and draw marker lines :
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These marker lines are just a few of the continuous infinity of parallel lines that form the stripes of this table.
Let the foregoing table be a single quilt square. We can make copies of this square, some exact, and some multiplied by various whole powers of ten, and put them together like a quilt, to form a continuous, infinite, stripy multiplication table.
To see how this is done, change the scale a bit, and consider the following powers-of-ten stripy multiplication table :
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Put a copy of the preceding "single quilt square" table in the middle, and fit appropriate other quilt squares around it in all directions, selecting the appropriate power of ten multiplier for the quilt square so that edges match.
The result is an infinite, continuous, stripy multiplication table.
Take a horizontal or vertical line through this infinite table. Each point on such a line has a number associated with it. The numbers run from arbitrarily small positive numbers at one end of the line, to arbitrarily large positive numbers at the other. A line so marked is called a logarithmic scale. On a logarithmic scale, the integer powers of any given positive number are evenly spaced.
The single quilt square multiplication table devised above, if given a quarter turn anticlockwise, is apparently the original log-log plot. For details, see the article on Lalanne's Universal Calculator in Ron Doerfler's beautiful 2010 "Graphical Computing" calendar.
I ran into Doerfler's calendar, and was introduced to Lalanne, three or four days after coming up with and using stripy continuous multiplication tables in lectures to give my liberal arts major classes a feel for logarithmic units. I am familiar with log-log plots, but hadn't approached this in that light.
As my father used to say, great minds think alike, but greater minds think first.
Nor is this the first time Doerfler and I have crossed paths mentally. A piece of Doerfler's writing, on calculating inverse hyperbolic tangents, proved useful in a physics paper I got published in the American Journal of Physics a few years ago. That, too, was related to logarithms.
Wednesday, March 3, 2010
Tuesday, March 2, 2010
Stripy multiplication tables
Consider the following (somewhat contrived) stripy multiplication table:
`\qquad\qquad\qquad[(1,100, 10000),(100,10000,1000000),(10000,1000000,100000000)]`
It is easy to interleave new rows and new columns, to get another stripy multiplication table :
`\qquad[(1,10,100,1000,10000),(10,100,1000,10000,100000),(100,1000,10000,100000,1000000),(1000,10000,100000,1000000,10000000),(10000,100000,1000000,10000000,100000000)]`
Now look at this very simple table, which is a subtable of the last one :
`\qquad\qquad\qquad[(1,10),(10,100)]`.
Can we interleave a new row and a new column into this table, preserving the existing values, and producing a new stripy multiplication table? In other words, can we fill in the missing values in this,
`\qquad\qquad\qquad[(1,"*",10),("*","*","*"),(10,"*",100)]`,
to make the result a stripy proportion table? Stripiness is constancy along the diagonals :
`\qquad\qquad\qquad[(1,u,10),(u,10,v),(10,v,100)]`.
The proportionality condition requires e.g. that
`\qquad\qquad\qquad1/u = u/10,`
i.e. that `u^2 = 10,` i.e. that `u = \pm\sqrt10`. Let us choose `u = \sqrt{10}.` Then we have
`\qquad\qquad\qquad[(1,\sqrt10,10),(\sqrt10,10,v),(10,v,100)]`.
Looking at the subtable in the top right hand corner,
`\qquad\qquad\qquad[(\sqrt10,10),(10,v)]`,
we can solve for `v` :
`\qquad\qquad\qquad v = \frac{10 xx 10}{\sqrt10},`
`\qquad\qquad\qquad\qquad = \frac{100}{\sqrt10},`
`\qquad\qquad\qquad\qquad = \frac{100 xx \sqrt10}{\sqrt10 xx \sqrt10},`
`\qquad\qquad\qquad\qquad = \frac{100 xx \sqrt10}{10},`
`\qquad\qquad\qquad\qquad = 10\sqrt10.`
So now we have a complete stripy multiplication table :
`\qquad\qquad\qquad[(1,\sqrt10,10),(\sqrt10,10,10\sqrt10),(10,10\sqrt10,100)]`.
This can be more revealingly presented by writing all the entries as powers of `10`:
`\qquad\qquad\qquad[(10^0,10^{1/2},10^1),(10^{1/2},10^1,10^{1 1/2}),(10^1,10^{1 1/2},10^2)]`,
or if we write the exponents fully in decimal,
`\qquad\qquad\qquad[(10^0,10^{0.5},10^1),(10^{0.5},10^1,10^{1.5}),(10^1,10^{1.5},10^2)]`.
This illustrates a general truth. A stripy proportion table is made by raising a single number (i.e. a base) to powers (exponents) coming from a stripy glide table. Here, for instance, we can pick `10` as the base, and
`\qquad\qquad\qquad[(0,0.5,1),(0.5,1,1.5),(1,1.5,2)]`
as the table of exponents.
The choice is not unique, however. For instance, we could get the same result by using `\sqrt10` or `10^{0.5}` for the base, and
`\qquad\qquad\qquad[(0,1,2),(1,2,3),(2,3,4)]`
for the table of exponents. Either way, we the result is the multiplication table
`\qquad\qquad\qquad[(10^0,10^{0.5},10^1),(10^{0.5},10^1,10^{1.5}),(10^1,10^{1.5},10^2)]`.
Let us rewrite this in decimals, rounding to three significant figures :
`\qquad\qquad\qquad[(1,3.16,10),(3.16,10,31.6),(10,31.6,100)]`.
When working with stripy multiplication tables, lets us adopt the understanding that rounded values are shorthand for the exact values.
We can fill in new rows between these. Stripiness constrains what is possible :
`\qquad\qquad\qquad[(1,a,3.16,b,10),(a,3.16,b,10,c),(3.16,b,10,c,31.6),(b,10,c,31.6,d),(10,c,31.6,d,100)]`.
Consider, now, just the first two columns :
`\qquad\qquad\qquad[(1,a),(a,3.16),(3.16,b),(b,10),(10,c)]`.
By stripiness, the second column is essentially the first column, displaced upward one place. Applying the proportionality condition to the first two rows, we have :
`\qquad\qquad\qquad1/a = a/3.16`.
Then `a^2 = 3.16` and `a = \pm1.78`, where we are using `3.16` as shorthand for `\sqrt{10}` and `1.78` as shorthand for `\sqrt\sqrt10`. Selecting the positive sign, `a = 1.78`.
Now, considering the second and third rows, we have
`\qquad\qquad\qquad1.78/3.16 = 3.16/b`.
This determines `b` fully, including its sign. Alternatively, we can look at the third and fourth rows, taking
`\qquad\qquad\qquad3.16/b = b/10`,
and this determines that the magnitude of `b` is `\sqrt{3.16 xx 10} = 5.62`, i.e. that it is the geometric mean of `3.16` and `10`.
Looking back, we see that with the choice of positive sign, `a` is also the geometric mean of the `1` and `3.16`, the quantities above and below it.
`\qquad\qquad\qquad[(1,1.78),(1.78,3.16),(3.16,5.62),(5.62,10),(10,c)]`.
The last value can be conveniently found from applying the proportionality condition to the first and last rows, i.e. to the four corners of the preceding table,
`\qquad\qquad\qquad1/10 = 1.78/c`,
so that `c = 10 xx 1.78 = 17.8`, and the table is
`\qquad\qquad\qquad[(1,1.78),(1.78,3.16),(3.16,5.62),(5.62,10),(10,17.8)]`.
By stripiness, we can generate most of the third column,
`\qquad\qquad\qquad[(1,1.78,3.16),(1.78,3.16,5.62),(3.16,5.62,10),(5.62,10,17.8),(10,17.8,"d")]`.
The proportionality condition on the four corners of this table is
`\qquad\qquad\qquad1/10 = 3.16/d`,
whence `d = 10 xx 3.16 = 31.6,` and the table becomes
`\qquad\qquad\qquad[(1,1.78,3.16),(1.78,3.16,5.62),(3.16,5.62,10),(5.62,10,17.8),(10,17.8,31.6)]`.
We can find the remaining columns in the same way, getting, to three significant digits,
`\qquad\qquad\qquad[(1,1.78,3.16,5.62,10),(1.78,3.16,5.62,10,17.8),(3.16,5.62,10,17.8,31.6),(5.62,10,17.8,31.6,56.2),(10,17.8,31.6,56.2,100)]`.
As before, we can express this as
`\qquad\qquad\qquad[(10^0,10^0.25,10^0.5,10^0.75,10^1),(10^0.25,10^0.5,10^0.75,10^1,10^1.25),(10^0.5,10^0.75,10^1,10^1.25,10^1.5),(10^0.75,10^1,10^1.25,10^1.5,10^1.75),(10^1,10^1.25,10^1.5,10^1.75,10^2)]`.
i.e, as `10` to the power of exponents from the stripy glide table
`\qquad\qquad\qquad[(0,0.25,0.5,0.75,1),(0.25,0.5,0.75,1,1.25),(0.5,0.75,1,1.25,1.5),(0.75,1,1.25,1.5,1.75),(1,1.25,1.5,1.75,2)]`.
Using stripiness and proportionality, the proportion table
`\qquad\qquad\qquad[(1,1.78,3.16,5.62,10),(1.78,3.16,5.62,10,17.8),(3.16,5.62,10,17.8,31.6),(5.62,10,17.8,31.6,56.2),(10,17.8,31.6,56.2,100)]`
can be extended infinitely in all directions. Here's a start :
`\qquad[(0.1,0.178,0.316,0.562,1,1.78,3.16,5.62,10),(0.178,0.316,0.562,1,1.78,3.16,5.62,10,17.8),(0.316,0.562,1,1.78,3.16,5.62,10,17.8,31.6),(0.562,1,1.78,3.16,5.62,10,17.8,31.6,56.2),(1,1.78,3.16,5.62,10,17.8,31.6,56.2,100),(1.78,3.16,5.62,10,17.8,31.6,56.2,100,178.),(3.16,5.62,10,17.8,31.6,56.2,100,178.,316.),(5.62,10,17.8,31.6,56.2,100,178.,316.,562.),(10,17.8,31.6,56.2,100,178.,316.,562.,1000)]`.
This infinite multiplication table can be built from very little.
First, we note that every row and every column consists of the same infinite-in-both-directions sequence of values,
`\qquad\ldots,0.316,0.562,1,1.78,3.16,5.62,10,17.8,31.6,\ldots`
Second, we note that this is what is classically called a geometric progression, a sequence of numbers with a common ratio. Each number is `10^0.25 \approx 1.78` times the one that precedes it. Since `1` is in this sequence, the sequence consists of all numbers `10^{n/4}`, with `n` an integer.
Third, we note that, up to round factors of ten, there are only four distinct numbers here, even in the infinite version of the table. To three significant digits, they are : `1, 1.78, 3.16, 5.62`. Every number in the table consists of one of these digit strings, differing only where the decimal point is placed.
To have this infinite table, to three digit accuracy, at one's mental service, all one need do is remember this sequence of numbers and what they mean :
`\qquad\qquad\qquad[(\ulx,\ul{10^x}),(0,1),(1//4,1.78),(1//2,3.16),(3//4,5.62),(1,10)]`.
To find, for instance, `56.2 xx 3.16`, one reasons
`\qquad\qquad\qquad 56.2 xx 3.16 = 10 xx 5.62 xx 3.16,`
`\qquad\qquad\qquad\qquad = 10^1 xx 10^\frac{3}{4} xx 10^\frac{1}{2},`
`\qquad\qquad\qquad\qquad = 10^(1 + 3/4 + 1/2),`
`\qquad\qquad\qquad\qquad = 10^(2 + 1/4),`
`\qquad\qquad\qquad\qquad = 10^2 xx 10^{1/4},`
`\qquad\qquad\qquad\qquad = 100 xx 1.78,`
`\qquad\qquad\qquad\qquad = 178.`
`\qquad\qquad\qquad[(1,100, 10000),(100,10000,1000000),(10000,1000000,100000000)]`
It is easy to interleave new rows and new columns, to get another stripy multiplication table :
`\qquad[(1,10,100,1000,10000),(10,100,1000,10000,100000),(100,1000,10000,100000,1000000),(1000,10000,100000,1000000,10000000),(10000,100000,1000000,10000000,100000000)]`
Now look at this very simple table, which is a subtable of the last one :
`\qquad\qquad\qquad[(1,10),(10,100)]`.
Can we interleave a new row and a new column into this table, preserving the existing values, and producing a new stripy multiplication table? In other words, can we fill in the missing values in this,
`\qquad\qquad\qquad[(1,"*",10),("*","*","*"),(10,"*",100)]`,
to make the result a stripy proportion table? Stripiness is constancy along the diagonals :
`\qquad\qquad\qquad[(1,u,10),(u,10,v),(10,v,100)]`.
The proportionality condition requires e.g. that
`\qquad\qquad\qquad1/u = u/10,`
i.e. that `u^2 = 10,` i.e. that `u = \pm\sqrt10`. Let us choose `u = \sqrt{10}.` Then we have
`\qquad\qquad\qquad[(1,\sqrt10,10),(\sqrt10,10,v),(10,v,100)]`.
Looking at the subtable in the top right hand corner,
`\qquad\qquad\qquad[(\sqrt10,10),(10,v)]`,
we can solve for `v` :
`\qquad\qquad\qquad v = \frac{10 xx 10}{\sqrt10},`
`\qquad\qquad\qquad\qquad = \frac{100}{\sqrt10},`
`\qquad\qquad\qquad\qquad = \frac{100 xx \sqrt10}{\sqrt10 xx \sqrt10},`
`\qquad\qquad\qquad\qquad = \frac{100 xx \sqrt10}{10},`
`\qquad\qquad\qquad\qquad = 10\sqrt10.`
So now we have a complete stripy multiplication table :
`\qquad\qquad\qquad[(1,\sqrt10,10),(\sqrt10,10,10\sqrt10),(10,10\sqrt10,100)]`.
This can be more revealingly presented by writing all the entries as powers of `10`:
`\qquad\qquad\qquad[(10^0,10^{1/2},10^1),(10^{1/2},10^1,10^{1 1/2}),(10^1,10^{1 1/2},10^2)]`,
or if we write the exponents fully in decimal,
`\qquad\qquad\qquad[(10^0,10^{0.5},10^1),(10^{0.5},10^1,10^{1.5}),(10^1,10^{1.5},10^2)]`.
This illustrates a general truth. A stripy proportion table is made by raising a single number (i.e. a base) to powers (exponents) coming from a stripy glide table. Here, for instance, we can pick `10` as the base, and
`\qquad\qquad\qquad[(0,0.5,1),(0.5,1,1.5),(1,1.5,2)]`
as the table of exponents.
The choice is not unique, however. For instance, we could get the same result by using `\sqrt10` or `10^{0.5}` for the base, and
`\qquad\qquad\qquad[(0,1,2),(1,2,3),(2,3,4)]`
for the table of exponents. Either way, we the result is the multiplication table
`\qquad\qquad\qquad[(10^0,10^{0.5},10^1),(10^{0.5},10^1,10^{1.5}),(10^1,10^{1.5},10^2)]`.
Let us rewrite this in decimals, rounding to three significant figures :
`\qquad\qquad\qquad[(1,3.16,10),(3.16,10,31.6),(10,31.6,100)]`.
When working with stripy multiplication tables, lets us adopt the understanding that rounded values are shorthand for the exact values.
We can fill in new rows between these. Stripiness constrains what is possible :
`\qquad\qquad\qquad[(1,a,3.16,b,10),(a,3.16,b,10,c),(3.16,b,10,c,31.6),(b,10,c,31.6,d),(10,c,31.6,d,100)]`.
Consider, now, just the first two columns :
`\qquad\qquad\qquad[(1,a),(a,3.16),(3.16,b),(b,10),(10,c)]`.
By stripiness, the second column is essentially the first column, displaced upward one place. Applying the proportionality condition to the first two rows, we have :
`\qquad\qquad\qquad1/a = a/3.16`.
Then `a^2 = 3.16` and `a = \pm1.78`, where we are using `3.16` as shorthand for `\sqrt{10}` and `1.78` as shorthand for `\sqrt\sqrt10`. Selecting the positive sign, `a = 1.78`.
Now, considering the second and third rows, we have
`\qquad\qquad\qquad1.78/3.16 = 3.16/b`.
This determines `b` fully, including its sign. Alternatively, we can look at the third and fourth rows, taking
`\qquad\qquad\qquad3.16/b = b/10`,
and this determines that the magnitude of `b` is `\sqrt{3.16 xx 10} = 5.62`, i.e. that it is the geometric mean of `3.16` and `10`.
Looking back, we see that with the choice of positive sign, `a` is also the geometric mean of the `1` and `3.16`, the quantities above and below it.
`\qquad\qquad\qquad[(1,1.78),(1.78,3.16),(3.16,5.62),(5.62,10),(10,c)]`.
The last value can be conveniently found from applying the proportionality condition to the first and last rows, i.e. to the four corners of the preceding table,
`\qquad\qquad\qquad1/10 = 1.78/c`,
so that `c = 10 xx 1.78 = 17.8`, and the table is
`\qquad\qquad\qquad[(1,1.78),(1.78,3.16),(3.16,5.62),(5.62,10),(10,17.8)]`.
By stripiness, we can generate most of the third column,
`\qquad\qquad\qquad[(1,1.78,3.16),(1.78,3.16,5.62),(3.16,5.62,10),(5.62,10,17.8),(10,17.8,"d")]`.
The proportionality condition on the four corners of this table is
`\qquad\qquad\qquad1/10 = 3.16/d`,
whence `d = 10 xx 3.16 = 31.6,` and the table becomes
`\qquad\qquad\qquad[(1,1.78,3.16),(1.78,3.16,5.62),(3.16,5.62,10),(5.62,10,17.8),(10,17.8,31.6)]`.
We can find the remaining columns in the same way, getting, to three significant digits,
`\qquad\qquad\qquad[(1,1.78,3.16,5.62,10),(1.78,3.16,5.62,10,17.8),(3.16,5.62,10,17.8,31.6),(5.62,10,17.8,31.6,56.2),(10,17.8,31.6,56.2,100)]`.
As before, we can express this as
`\qquad\qquad\qquad[(10^0,10^0.25,10^0.5,10^0.75,10^1),(10^0.25,10^0.5,10^0.75,10^1,10^1.25),(10^0.5,10^0.75,10^1,10^1.25,10^1.5),(10^0.75,10^1,10^1.25,10^1.5,10^1.75),(10^1,10^1.25,10^1.5,10^1.75,10^2)]`.
i.e, as `10` to the power of exponents from the stripy glide table
`\qquad\qquad\qquad[(0,0.25,0.5,0.75,1),(0.25,0.5,0.75,1,1.25),(0.5,0.75,1,1.25,1.5),(0.75,1,1.25,1.5,1.75),(1,1.25,1.5,1.75,2)]`.
Using stripiness and proportionality, the proportion table
`\qquad\qquad\qquad[(1,1.78,3.16,5.62,10),(1.78,3.16,5.62,10,17.8),(3.16,5.62,10,17.8,31.6),(5.62,10,17.8,31.6,56.2),(10,17.8,31.6,56.2,100)]`
can be extended infinitely in all directions. Here's a start :
`\qquad[(0.1,0.178,0.316,0.562,1,1.78,3.16,5.62,10),(0.178,0.316,0.562,1,1.78,3.16,5.62,10,17.8),(0.316,0.562,1,1.78,3.16,5.62,10,17.8,31.6),(0.562,1,1.78,3.16,5.62,10,17.8,31.6,56.2),(1,1.78,3.16,5.62,10,17.8,31.6,56.2,100),(1.78,3.16,5.62,10,17.8,31.6,56.2,100,178.),(3.16,5.62,10,17.8,31.6,56.2,100,178.,316.),(5.62,10,17.8,31.6,56.2,100,178.,316.,562.),(10,17.8,31.6,56.2,100,178.,316.,562.,1000)]`.
This infinite multiplication table can be built from very little.
First, we note that every row and every column consists of the same infinite-in-both-directions sequence of values,
`\qquad\ldots,0.316,0.562,1,1.78,3.16,5.62,10,17.8,31.6,\ldots`
Second, we note that this is what is classically called a geometric progression, a sequence of numbers with a common ratio. Each number is `10^0.25 \approx 1.78` times the one that precedes it. Since `1` is in this sequence, the sequence consists of all numbers `10^{n/4}`, with `n` an integer.
Third, we note that, up to round factors of ten, there are only four distinct numbers here, even in the infinite version of the table. To three significant digits, they are : `1, 1.78, 3.16, 5.62`. Every number in the table consists of one of these digit strings, differing only where the decimal point is placed.
To have this infinite table, to three digit accuracy, at one's mental service, all one need do is remember this sequence of numbers and what they mean :
`\qquad\qquad\qquad[(\ulx,\ul{10^x}),(0,1),(1//4,1.78),(1//2,3.16),(3//4,5.62),(1,10)]`.
To find, for instance, `56.2 xx 3.16`, one reasons
`\qquad\qquad\qquad 56.2 xx 3.16 = 10 xx 5.62 xx 3.16,`
`\qquad\qquad\qquad\qquad = 10^1 xx 10^\frac{3}{4} xx 10^\frac{1}{2},`
`\qquad\qquad\qquad\qquad = 10^(1 + 3/4 + 1/2),`
`\qquad\qquad\qquad\qquad = 10^(2 + 1/4),`
`\qquad\qquad\qquad\qquad = 10^2 xx 10^{1/4},`
`\qquad\qquad\qquad\qquad = 100 xx 1.78,`
`\qquad\qquad\qquad\qquad = 178.`
Why is the standard multiplication table harder to learn than the usual addition tables?
A standard times table, in proportion table form, looks like this :
`\qquad\qquad\qquad[(1,2,3,4,5,6,7,8,9,10),(2,4,6,8,10,12,14,16,18,20),(3,6,9,12,15,18,21,24,27,30),(4,8,12,16,20,24,28,32,36,40),(5,10,15,20,25,30,35,40,45,50),(6,12,18,24,30,36,42,48,54,60),(7,14,21,28,35,42,49,56,63,70),(8,16,24,32,40,48,56,64,72,80),(9,18,27,36,45,54,63,72,81,90),(10,20,30,40,50,60,70,80,90,100)]`
A standard addition table (not a proportion table of course, but rather a glide table,) looks like this :
`\qquad\qquad\qquad[(0,1,2,3,4,5,6,7,8,9),(1,2,3,4,5,6,7,8,9,10),(2,3,4,5,6,7,8,9,10,11),(3,4,5,6,7,8,9,10,11,12),(4,5,6,7,8,9,10,11,12,13),(5,6,7,8,9,10,11,12,13,14),(6,7,8,9,10,11,12,13,14,15),(7,8,9,10,11,12,13,14,15,16),(8,9,10,11,12,13,14,15,16,17),(9,10,11,12,13,14,15,16,17,18)]`
A glide table is to addition and subtraction what a proportion table is to multiplication and division.
In a glide table, all the rows, and equivalently all the columns, differ by fixed amounts.
The way to reconstruct a value at one corner of a box (`2 xx 2` subtable) also differs, but in an obvious way : to find the value at one corner, one adds the values at the neighboring corners and subtracts the value at the opposite corner.
So now we have presented two tables, one for multiplication, and the other for addition. It is reasonably common for people to not remember all multiplications in the times table, but rather rarer for them not to know all that is in the addition table.
There are a number of reasons for this, of course. The numbers in the multiplication table are mostly larger, for instance.
Two of the reasons, however, for the relative ease of learning the addition table can help us construct multiplication tables that are easy, for what they do:
`\qquad\qquad\qquad[(1,2,4,8,16,32),(2,4,8,16,32,64),(4,8,16,32,64,128),(8,16,32,64,128,256),(16,32,64,128,256,512),(32,64,128,256,512,1024)]`
In the next posting, we shall begin constructing some useful stripy multiplication tables.
`\qquad\qquad\qquad[(1,2,3,4,5,6,7,8,9,10),(2,4,6,8,10,12,14,16,18,20),(3,6,9,12,15,18,21,24,27,30),(4,8,12,16,20,24,28,32,36,40),(5,10,15,20,25,30,35,40,45,50),(6,12,18,24,30,36,42,48,54,60),(7,14,21,28,35,42,49,56,63,70),(8,16,24,32,40,48,56,64,72,80),(9,18,27,36,45,54,63,72,81,90),(10,20,30,40,50,60,70,80,90,100)]`
A standard addition table (not a proportion table of course, but rather a glide table,) looks like this :
`\qquad\qquad\qquad[(0,1,2,3,4,5,6,7,8,9),(1,2,3,4,5,6,7,8,9,10),(2,3,4,5,6,7,8,9,10,11),(3,4,5,6,7,8,9,10,11,12),(4,5,6,7,8,9,10,11,12,13),(5,6,7,8,9,10,11,12,13,14),(6,7,8,9,10,11,12,13,14,15),(7,8,9,10,11,12,13,14,15,16),(8,9,10,11,12,13,14,15,16,17),(9,10,11,12,13,14,15,16,17,18)]`
A glide table is to addition and subtraction what a proportion table is to multiplication and division.
In a glide table, all the rows, and equivalently all the columns, differ by fixed amounts.
The way to reconstruct a value at one corner of a box (`2 xx 2` subtable) also differs, but in an obvious way : to find the value at one corner, one adds the values at the neighboring corners and subtracts the value at the opposite corner.
So now we have presented two tables, one for multiplication, and the other for addition. It is reasonably common for people to not remember all multiplications in the times table, but rather rarer for them not to know all that is in the addition table.
There are a number of reasons for this, of course. The numbers in the multiplication table are mostly larger, for instance.
Two of the reasons, however, for the relative ease of learning the addition table can help us construct multiplication tables that are easy, for what they do:
- the standard addition table, even though it is the same size, has fewer distinct values than the standard, multiplication table, and
- the standard addition table has a simpler structure---it is stripy.
`\qquad\qquad\qquad[(1,2,4,8,16,32),(2,4,8,16,32,64),(4,8,16,32,64,128),(8,16,32,64,128,256),(16,32,64,128,256,512),(32,64,128,256,512,1024)]`
In the next posting, we shall begin constructing some useful stripy multiplication tables.
Thursday, February 25, 2010
Long division taken apart, continued
Now that we have taken long division apart, in the last posting, and made a multiplication table for the divisor `21`, in the posting before that, we are ready to look at the standard long division algorithm, with Jakow Trachtenberg's trick.
We are dividing `1543` by `21`, using the multiplication table
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105),(6,126),(7,147),(8,168),(9,189)]`.
Start by writing
`\qquad\qquad\qquad\qquad\quad"****."`
`\qquad\qquad\qquad 21 )\bar1543.`
From the table, see that no whole `21`s go into `1` or into `15`, but that `7` whole `21`s, but not `8`, go into `154` :
`\qquad\qquad\qquad\qquad\qquad\quad7"*."`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad147 .`
Subtract `1470` from `1543` leaving `73`.
`\qquad\qquad\qquad\qquad\qquad\quad7"*."`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad\ul1470.`
`\qquad\qquad\qquad\qquad\qquad\quad73.`
From the table, again, `3` whole `21`s, i.e. `63`, go into `73`, leaving `10`.
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad\ul1470`
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad\qquad\qquad\quad\ul63.`
`\qquad\qquad\qquad\qquad\qquad\quad10.`
This is the stopping point for divmod division, so that `1543 :- 21 = 73 " remainder " 10.`
For decimal division, one continues `21`s into `100` go `4` times, i.e. `84` leaving `16`.
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad\ul1470`
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad\qquad\qquad\quad\ul63.`
`\qquad\qquad\qquad\qquad\qquad\quad10.0`
`\qquad\qquad\qquad\qquad\qquad\qquad\ul8.4`
`\qquad\qquad\qquad\qquad\qquad\qquad1.6`
This continues until one has reached the desired degree of accuracy.
We are dividing `1543` by `21`, using the multiplication table
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105),(6,126),(7,147),(8,168),(9,189)]`.
Start by writing
`\qquad\qquad\qquad\qquad\quad"****."`
`\qquad\qquad\qquad 21 )\bar1543.`
From the table, see that no whole `21`s go into `1` or into `15`, but that `7` whole `21`s, but not `8`, go into `154` :
`\qquad\qquad\qquad\qquad\qquad\quad7"*."`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad147 .`
Subtract `1470` from `1543` leaving `73`.
`\qquad\qquad\qquad\qquad\qquad\quad7"*."`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad\ul1470.`
`\qquad\qquad\qquad\qquad\qquad\quad73.`
From the table, again, `3` whole `21`s, i.e. `63`, go into `73`, leaving `10`.
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad\ul1470`
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad\qquad\qquad\quad\ul63.`
`\qquad\qquad\qquad\qquad\qquad\quad10.`
This is the stopping point for divmod division, so that `1543 :- 21 = 73 " remainder " 10.`
For decimal division, one continues `21`s into `100` go `4` times, i.e. `84` leaving `16`.
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad 21 )\bar1543.`
`\qquad\qquad\qquad\qquad\quad\ul1470`
`\qquad\qquad\qquad\qquad\qquad\quad73.`
`\qquad\qquad\qquad\qquad\qquad\quad\ul63.`
`\qquad\qquad\qquad\qquad\qquad\quad10.0`
`\qquad\qquad\qquad\qquad\qquad\qquad\ul8.4`
`\qquad\qquad\qquad\qquad\qquad\qquad1.6`
This continues until one has reached the desired degree of accuracy.
Long division taken apart
When we divide, we are trying to find out how many times the divisor (e.g. 21) goes into the dividend (e.g. 1543).
We could just repeatedly subtract, keeping a tally of the number of times we have subtracted, stopping when further subtraction would give a negative result. At least, that is how we could do divmod division.
Keeping such a tally would be tedious, however. It is better to subtract large but convenient multiples of the divisor, than smaller multiples, keeping tally of these separately as we go.
In our usual decimal system of numeration, powers of ten are particularly convenient multipliers.
In 1543, the highest nonzero column is the thousands column. We can therefore try subtracting thousands of `21`s. We could not subtract `21` thousand even once from `1` thousand and anything without the result being negative. So we need `0` thousands of `21`s, and we still have `1` thousand and something.
Next, we try hundreds of `21`s. Again, we cannot subtract even `21` hundreds even once from `15` hundred and something, without the result being negative. So we need `0` hundreds of `21`s, and we still have `15` hundred and something.
Next, we try tens of `21`s. We can subtract `21` tens from `154` tens and something. Indeed, we can subtract `7` times, a total of `147` tens, without getting negative result. So now we have that `154` tens and `3` is the same as `7` lots of `21` tens, and then `7` tens and `3`.
Next we try whole `21`s. We can subtract `21` from `73` just `3` times without going negative. This subtracts a total of `63` from `73`, leaving `10`.
So now we have that `1543` is the same as `7` lots of `21` tens, and `3` lots of `21` ones, and a further `10`.
This can instead be understood as `70` lots of `21`, and `3` lots of `21`, and `10` more. This in turn is `73 xx 21` and `10`. i.e.
`\qquad\qquad\qquad 1543 = (73 xx 21) + 10.`
If we are doing divmod division, we can stop :
`\qquad\qquad\qquad 1543 -: 21 = 73" remainder "10.`
But if, instead of a remainder, we want the quotient to have a fractional part expressed as a decimal, then we keep going.
We have `10` that still needs to be divided by `21`. So now we try subtracting tenths of `21`. `10` is `100` tenths. We can subtract `4` lots of tenths of `21`, i.e. `84` tenths, from `100` tenths, leaving `16` tenths.
Next, we try subtracting hundredths of `21` from `160` hundredths. `7` hundredths of `21` is `147` hundredths, which when subtracted from `160` hundredths leaves `13` hundredths.
Stopping at this point, we find that
`\qquad\qquad\qquad 1543 = (73.47 xx 21) + 0.13`,
which we can write
`\qquad\qquad\qquad 1543 -: 21 = 73.47" remainder "0.13`.
The preceding discussion has been rather lengthy--such work is usually set out in a much more compressed format. Unfortunately, long division seems to have been taught procedurally without adequate preparation, so that the elided computational form entrains elided thinking. Even among the few in these calculator-infested days who can still actually do long division, a substantial fraction can give no convincing account of why the standard long division algorithm works.
We could just repeatedly subtract, keeping a tally of the number of times we have subtracted, stopping when further subtraction would give a negative result. At least, that is how we could do divmod division.
Keeping such a tally would be tedious, however. It is better to subtract large but convenient multiples of the divisor, than smaller multiples, keeping tally of these separately as we go.
In our usual decimal system of numeration, powers of ten are particularly convenient multipliers.
In 1543, the highest nonzero column is the thousands column. We can therefore try subtracting thousands of `21`s. We could not subtract `21` thousand even once from `1` thousand and anything without the result being negative. So we need `0` thousands of `21`s, and we still have `1` thousand and something.
Next, we try hundreds of `21`s. Again, we cannot subtract even `21` hundreds even once from `15` hundred and something, without the result being negative. So we need `0` hundreds of `21`s, and we still have `15` hundred and something.
Next, we try tens of `21`s. We can subtract `21` tens from `154` tens and something. Indeed, we can subtract `7` times, a total of `147` tens, without getting negative result. So now we have that `154` tens and `3` is the same as `7` lots of `21` tens, and then `7` tens and `3`.
Next we try whole `21`s. We can subtract `21` from `73` just `3` times without going negative. This subtracts a total of `63` from `73`, leaving `10`.
So now we have that `1543` is the same as `7` lots of `21` tens, and `3` lots of `21` ones, and a further `10`.
This can instead be understood as `70` lots of `21`, and `3` lots of `21`, and `10` more. This in turn is `73 xx 21` and `10`. i.e.
`\qquad\qquad\qquad 1543 = (73 xx 21) + 10.`
If we are doing divmod division, we can stop :
`\qquad\qquad\qquad 1543 -: 21 = 73" remainder "10.`
But if, instead of a remainder, we want the quotient to have a fractional part expressed as a decimal, then we keep going.
We have `10` that still needs to be divided by `21`. So now we try subtracting tenths of `21`. `10` is `100` tenths. We can subtract `4` lots of tenths of `21`, i.e. `84` tenths, from `100` tenths, leaving `16` tenths.
Next, we try subtracting hundredths of `21` from `160` hundredths. `7` hundredths of `21` is `147` hundredths, which when subtracted from `160` hundredths leaves `13` hundredths.
Stopping at this point, we find that
`\qquad\qquad\qquad 1543 = (73.47 xx 21) + 0.13`,
which we can write
`\qquad\qquad\qquad 1543 -: 21 = 73.47" remainder "0.13`.
The preceding discussion has been rather lengthy--such work is usually set out in a much more compressed format. Unfortunately, long division seems to have been taught procedurally without adequate preparation, so that the elided computational form entrains elided thinking. Even among the few in these calculator-infested days who can still actually do long division, a substantial fraction can give no convincing account of why the standard long division algorithm works.
Building ad hoc multiplication tables
Proportion tables are sometimes useful for doing long division.
For many people, the most difficult part of that standard long division algorithm is estimating which multiple of the divisor to subtract. Jakow Trachtenberg taught a simple trick for avoiding this difficulty.
In the next posting, we are going to divide 1543 by 21, using the standard long division algorithm and Trachtenberg's trick.
The first step is to invest some time making a table of multiples of `21`, up to the `9 xx 21`. The `0` and `1` rows are obvious :
`\qquad\qquad\qquad[(0, 0),(1, 21)]`.
The `2` row can be found by doubling the second row :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42)]`.
The `3` row can be found by adding the `1` row and the `2` row :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63)]`.
The `4` row can be found either by adding the `1` row to the `3` row, or else by doubling the `2` row. One picks whichever is more convenient :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84)]`.
The `5` row can be found either by adding the `2` row and the `3` row, or else by adding the `1` row and the `5` row. Again, one picks whichever is more convenient :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105)]`.
One continues in this way, constructing the next row opportunistically, until at last one has :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105),(6,126),(7,147),(8,168),(9,189)]`.
This can be checked by casting out nines, if one knows how to do that. (If not, it needs to be the subject of yet another post.)
Now we are ready divide anything by `21`.
`\qquad\qquad\qquad[(0, 0,(0)),(1, 21,(3)),(2,42,(6)),(3,63,(0)),(4,84,(3)),(5,105,(6)),(6,126,(0)),(7,147,(3)),(8,168,(6)),(9,189,(0))]`.
This checks out, so we can rely on the table we have made.
For many people, the most difficult part of that standard long division algorithm is estimating which multiple of the divisor to subtract. Jakow Trachtenberg taught a simple trick for avoiding this difficulty.
In the next posting, we are going to divide 1543 by 21, using the standard long division algorithm and Trachtenberg's trick.
The first step is to invest some time making a table of multiples of `21`, up to the `9 xx 21`. The `0` and `1` rows are obvious :
`\qquad\qquad\qquad[(0, 0),(1, 21)]`.
The `2` row can be found by doubling the second row :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42)]`.
The `3` row can be found by adding the `1` row and the `2` row :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63)]`.
The `4` row can be found either by adding the `1` row to the `3` row, or else by doubling the `2` row. One picks whichever is more convenient :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84)]`.
The `5` row can be found either by adding the `2` row and the `3` row, or else by adding the `1` row and the `5` row. Again, one picks whichever is more convenient :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105)]`.
One continues in this way, constructing the next row opportunistically, until at last one has :
`\qquad\qquad\qquad[(0, 0),(1, 21),(2,42),(3,63),(4,84),(5,105),(6,126),(7,147),(8,168),(9,189)]`.
This can be checked by casting out nines, if one knows how to do that. (If not, it needs to be the subject of yet another post.)
Now we are ready divide anything by `21`.
`\qquad\qquad\qquad[(0, 0,(0)),(1, 21,(3)),(2,42,(6)),(3,63,(0)),(4,84,(3)),(5,105,(6)),(6,126,(0)),(7,147,(3)),(8,168,(6)),(9,189,(0))]`.
This checks out, so we can rely on the table we have made.
Wednesday, February 24, 2010
Egyptian division prolonged---fractions
Last time, we used Egyptian division to find a whole number quotient and a remainder for the division `1543 -: 21`, in the process formed the table
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73)]`.
To calculate the fractional part in decimal form, we begin by subtracting the last row from the first, and append the result at the bottom:
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21)]`.
In practice, of course, don't actually subtract, but just write the remainder we have already found, and that same remainder divided by the divisor in fractional form in the next column.
Now we take the divisor row, `[21, 1]`, halve it to get `[10.5, 0.5]`, and append at the bottom :
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(10.5, 0.5)]`.
We now continue, halving the bottom row and appending, halving the bottom row and appending, until the second number in the bottom row is comparable to the accuracy we require. Suppose we want the fractional part of the quotient to be accurate to two decimal places, then :
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(10.5, 0.5), (5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.328125,0.015625), (0.1640625, 0.0078125)]`.
We now make a sum as close as we can to the `10` in the target row `[10, 10/21]`, by adding together some of the rows below it. `10.5` is larger than `10`, so we ignore that row. `5.25` is smaller, so that row get used. `10 - 5.25 = 4.75` left to account for. This is larger than `2.625`, so we use that row too, leaving `4.75 - 2.625 = 2.125` to account for. Continuing, `2.125 - 1.3125 = 0.8125`; `0.8125 - 0.65625 = 0.15625`; skip `0.328125`. For the last entry, we perform the subtraction, even though the result is negative, `0.15625 - 0.1640625 = -0.0078125`. The reason is that it leaves an error `-0.0078125` that is smaller in magnitude than the error we would have without the subtraction, i.e. `0.1640625`. This gives us more accuracy than we expected.
Deleting (or striking out) the rows to be ignored, we get :
`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.1640625, 0.0078125)]`.
Adding the contributing rows, we get
`5.25+2.625+1.3125+0.65625+0.1640625 = 10.0078125`,
(or, easier, `10 - -0.0078125 = 10.0078125`) and
`0.25+0.125+0.0625+0.03125+0.0078125 = 0.4765625`,
so the table finally becomes
`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.1640625, 0.0078125),(10.0078125, 0.4765625)]`.
Since `10.0078125` is within a tenth of a percent of `10`, we must have that `0.4765625` is within a tenth of a percent of `10/21`, i.e. `10/21 \approx 0.477`, this compares well with the accurate answer of `0.476190...` i.e. we have almost three decimal places of accuracy.
So `1543/71 \approx 73.477.`
Although it makes relatively few demands in terms of times tables, and this part requires only that we know how to halve decimal quantities, one has to put in considerable labor for each decimal place.
The method is much better adapted to binary fractions than to decimals. These are sometimes useful, e.g. when one is working to sixteenths or thirty-seconds of an inch, say. In binary fractions, the final table looks like this,
`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5 1/4, 1/4), (2 5/8, 1/8), (1 5/16, 1/16), (21/32,1/32), (21/64, 1/64),(21/128, 1/128),(10 1/128, 61/128)]`,
so that `1543/71 \approx 73 61/128.`
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73)]`.
To calculate the fractional part in decimal form, we begin by subtracting the last row from the first, and append the result at the bottom:
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21)]`.
In practice, of course, don't actually subtract, but just write the remainder we have already found, and that same remainder divided by the divisor in fractional form in the next column.
Now we take the divisor row, `[21, 1]`, halve it to get `[10.5, 0.5]`, and append at the bottom :
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(10.5, 0.5)]`.
We now continue, halving the bottom row and appending, halving the bottom row and appending, until the second number in the bottom row is comparable to the accuracy we require. Suppose we want the fractional part of the quotient to be accurate to two decimal places, then :
`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(10.5, 0.5), (5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.328125,0.015625), (0.1640625, 0.0078125)]`.
We now make a sum as close as we can to the `10` in the target row `[10, 10/21]`, by adding together some of the rows below it. `10.5` is larger than `10`, so we ignore that row. `5.25` is smaller, so that row get used. `10 - 5.25 = 4.75` left to account for. This is larger than `2.625`, so we use that row too, leaving `4.75 - 2.625 = 2.125` to account for. Continuing, `2.125 - 1.3125 = 0.8125`; `0.8125 - 0.65625 = 0.15625`; skip `0.328125`. For the last entry, we perform the subtraction, even though the result is negative, `0.15625 - 0.1640625 = -0.0078125`. The reason is that it leaves an error `-0.0078125` that is smaller in magnitude than the error we would have without the subtraction, i.e. `0.1640625`. This gives us more accuracy than we expected.
Deleting (or striking out) the rows to be ignored, we get :
`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.1640625, 0.0078125)]`.
Adding the contributing rows, we get
`5.25+2.625+1.3125+0.65625+0.1640625 = 10.0078125`,
(or, easier, `10 - -0.0078125 = 10.0078125`) and
`0.25+0.125+0.0625+0.03125+0.0078125 = 0.4765625`,
so the table finally becomes
`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.1640625, 0.0078125),(10.0078125, 0.4765625)]`.
Since `10.0078125` is within a tenth of a percent of `10`, we must have that `0.4765625` is within a tenth of a percent of `10/21`, i.e. `10/21 \approx 0.477`, this compares well with the accurate answer of `0.476190...` i.e. we have almost three decimal places of accuracy.
So `1543/71 \approx 73.477.`
Although it makes relatively few demands in terms of times tables, and this part requires only that we know how to halve decimal quantities, one has to put in considerable labor for each decimal place.
The method is much better adapted to binary fractions than to decimals. These are sometimes useful, e.g. when one is working to sixteenths or thirty-seconds of an inch, say. In binary fractions, the final table looks like this,
`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5 1/4, 1/4), (2 5/8, 1/8), (1 5/16, 1/16), (21/32,1/32), (21/64, 1/64),(21/128, 1/128),(10 1/128, 61/128)]`,
so that `1543/71 \approx 73 61/128.`
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