**Archimedes,The Sand-Reckoner (Arenarius), ch.3 (sects. 1-8)**©- translated by Henry Mendell (Cal. State U., L.A.)

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some notes on the translation of numbers (these should be looked at first)

[1] And so these are what I suppose, but I consider it useful for the way of expressing numbers to be stated, so that even those others who haven't come across the book written to Zeuxippus are not at a loss since nothing has yet been said about it in this book. [2] In fact the names of the numbers up to the name of ten-thousand happen to have been provided to us, and beyond the name of ten-thousand we ascertain a number of ten-thousands of units when we say, "even up to ten-thousand myriads." And so let the presently stated numbers up to ten-thousand myriads be called by us first numbers, and let ten-thousand myriads of the first numbers be called a unit of the second numbers, and let units of the second numbers be counted, i.e. from the units decads (10's), hekatontads (100's), chiliads (1000's), myriads (1,0000's) up to ten-thousand myriads. Again let ten-thousand myriads of the second numbers be called a unit of the third numbers, and let units of the third numbers be counted and from the units decads, hekatontads, chiliads, and myriads up to ten-thousand myriads.

[3] In the same manner let ten-thousand myriads of the third numbers also be called a unit of the fourth numbers, and let ten-thousand myriads of the fourth numbers be called a unit of the fifth numbers, and repeatedly proceeding in this way let the numbers have names up to the ten-thousand myriads of the ten-thousand ten-thousandth numbers. And so numbers are adequately ascertained up to so many, but it is also possible to proceed further.

[Keep in mind that the unit is not a number. Hence, the last number of the first numbers is not a number of the second numbers. It measures them. Hence the last number of the first group is 1,0000,0000, while the first number of the second numbers will be 2,0000,0000

The last number in this series will be 1,0000,0000^{1,0000,0000}.]

[4] For let the presently described numbers be called of-the-first-period. Let the last number of the first period be called a unit of the second period of the first numbers. Again, let ten-thousand myriads of the first numbers of the second period be called a unit of the second numbers of the second period. Similarly, let the last of these be called a unit of third numbers of the second period, and repeatedly proceeding in this way the numbers of the second period have names up to ten-thousand myriads of the ten-thousand ten-thousandth numbers. Again let the last number of the second period also be called a unit of the first numbers of the third period, and given that let them repeatedly proceed in this way up to ten-thousand myriads of the ten-thousand-ten-thousandth numbers of the ten-thousand-ten-thousandth period.

Archimedes modestly only seems to produce numbers up
to 1,0000,0000^{1,0000,00001,0000,0000, a fairly large number.}

[5] Given these numbers as expressed in this way, if numbers are set out in continuous proportion from a unit, with a decad next to the unit, then the first eight numbers including the unit will be among what we called the first numbers, while the next eight after them will be among what we called second numbers, and the others in the same way as these will belong to numbers called with the same name by the distance [counted out by] an octad of numbers from the first octad of numbers.

And so in the first octad of numbers, the eighth number is a thousand ten-thousand, while in the second octad, the first, since it is ten-times what preceeds it, will be ten-thousand myriads. But this is the unit of the second numbers. The eighth of the second octad is a thousand myriads of the second numbers. Again the first of the third octad, since it is ten-times what preceeds it, will be ten-thousand myriads of the second numbers. But this is a unit of the third numbers. It is obvious that there will also be many, many octads, as was said.

[6] This too is usefully ascertained. If when numbers are proportional from the unit, some of the numbers from the same proportion multiply one another, the number which arises will be from the same proportion and will be distant from the larger of the numbers which multiplied one another as much as the smaller of the numbers which multiplied one another is distant proportionally from the unit, but it will be distant from the unit by one less than the number of the sum [of the distances] which the numbers which multiplied one another are distant from the unit.

[7] For let there be some numbers in proportion from a unit, A, B, C, D, E, F, G, H, I, J, K, and let A be a unit, and let D be multiplied by H, and let the number that arises be X. Let K be taken from the proportion as being as distant from H as D is from unit A. We must prove that X is equal to K. And so since, given that the numbers are proportional, D from A and K from H have equal distances, D to A has the same ratio as K to H. But D is a multiple of A by D. Therefore, K is also a multiple of H by D. Thus K is equal to X.

[8] And so it is clear that the number which arises is from the proportion and is distant from the larger of the numbers which multiplied one another as much as the smaller is from the unit. But it is obvious that it also is distant from the unit by one less than the number of the sum [of the distances] which D, H are distant from the unit. For A, B, C, D, E, F, G, H are as many as H is distant from the unit, while I, J, K are fewer by one than the distance that D is from the unit. For with H they are as many.

[The continual proportion is (with the terms indexed):

A^{1} : B^{2} = B^{2} : C^{3} = C^{3}
: D^{4} = D^{4} : E^{5} = E^{5} : F^{6}
= F^{6} : G^{7} = G^{7} : H^{8} = H^{8}
: I^{9} = I^{9} : J^{10} = J^{10} : K^{11}

Archimedes treats the distance between, e.g., terms A^{1}^{ and K11
as the number of terms as the terms between them inclusive of } A^{1}^{
and K11, i.e, 11 and not 9.
} D^{4} is 4 after the unit A^{1} and K^{11} is
4 after H^{8}.

H^{8} is 8 after the unit, so that K^{11} is 4 + 8 - 1 = 11
after A^{1}.

The theorem may be proved by counting repeated applications of the more general ex aequali principle: a : b = c : d and b : e = d : f a : e = c : f and by taking alternando so that a*f = e*c, where a is the unit, f = e*c]