Return to Vignettes of Ancient Mathematics

Go to abbreviations for forms
Go to rules for manipulations of forms
Go to prob. 1, go to prob. 2, go to prob. 3,
go to prob. 4, go to prob. 5, go to prob. 6.

The text used is the edition of Tannery (1893), but I have also consulted the translation of ver Eecke (1959) and the paraphrase of Heath (1910).

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Notes on translation. I resist translating ὕπαρξις as ‘positive’ and λεῖψις as ‘negative’, as that might imply a conception of positive and negative numfbers. ὕπαρξις will be ‘existence’ and λεῖψις ‘deficiency’. The term ὑπόστασις is also problematic, as it refers to the values in the particular solution. What is the exact sense of this word that had been a technical term for general existence since the early Stoa? The contrast would seem to be the numbers expressed with the indeterminate number (the variable) with the actual numbers. So ‘actualities’ is not bad. There is only one indeterminate variable, but, as you will see, it has a difference symbol for each power, or form. The number is then expressed as a sequence of forms, each followed by a number of how many there are (the coefficients), in descending order, except that the existents are grouped first followed by the deficiencies. For example, in modern notation, 3x5 - 4x4 - 5x3+7unit would be grouped 3x5 + 7units - 4x4 - 5x3.

The system of fractions in Diophantus is ultimately derived from Demotic Egyptian (script from 7th cent. BCE on). Here fractions were commonly presented as a sequence of parts or nths. For example, 2/5 (1/3 + 1/15) would be 3rd 15th, which we can represent as 3' 15'. Occasionally fractions would be represented by the denominator and numerator. Greeks, beginning about 300 BCE adopted both representations. Following the convention, Diophantus expresses fractions the reverse of what we do, the part (denominator) is on top, the whole (numerator) is on the bottom. I write 1/5 as 5´ and this denominator as a superscript, 8/5 will be written, 8. One-half is special, written 𐅵.

Book I

Seeing that you are zealous to learn the discovery of problems in numbers, my most esteemed Dionysus, I have attempted [to organize the science/procedure], starting from the foundations from which matters are constituted, to establish the nature and power in numbers. And so the matter seems perhaps more difficult since it is not yet known (for the souls of those starting out lack hope for success); nevertheless, it will become easily grasped by you due to your eagerness and my presentation/demonstration. For a swift passion for learning supplements teaching.

But also in addition to these it stands as obvious to you, perceiving that all the numbers as composed from some plêthos of units have their existence to infinity, and so, in fact, their happening to be among them: some being squares, which are from some number multiplied times itself—and this number is called ‘side of the square’; others cubes, which are from squares being multiplied times their sides; others power-powers, which are from squares by multiplied times themselves; others power-cubes, which are from squares being multiplied by cubes from the same sides as they; others cube-cubes, which are from cubes being multiplied times themselves, and from the adding of these or excess or multiplication or ratio that’s to one another or even each to the particular sides, it happens that most arithmetical problems are woven. They are solved when you walk the indicated road.

And so, each of the these numbers was judged fit to be an element of the arithmetic science/inquiry by acquiring a briefer derived-name. And so the square is called ‘power’ and its symbol is Δ with a sign on it, Υ, ΔΥ power. (This is an abbreviation for ‘dynamis’ or δύναμις, Greek for ‘power’, effectively ‘square’) And next is ‘cube’ and its symbol is Κ with a sign on it, Υ, ΚΥ cube. (This is an abbreviation for ‘kybos’ or κῦβος, Greek for ‘cube’)And next is power-power, from a square being multiplied times itself, and its symbol is two delta’s, with a sign on it, Υ, ΔΥΔ power-power. And next is the power-cube, from a square being multiplied times the cube from the same side as it, and ts symbol is ΔΚ, with a sign on it, Υ, ΔΚΥ power-cube. And next is the cube-cube, which is from a cube being multiplied times itself, and its symbol is two kappa’s, with a sign on it, Υ, ΚΥΚ cube-cube.

But what gets none of these properties, while having an undetermined plêthos of units in it, is arithmos. But there is another symbol that’s immutable (not inflected), the unit, and its symbol is M, with a sign on it, Ο, units. (This is an abbreviation for ‘monas’ or μονάς, Greek for ‘unit’) But just as the homonymous parts of arithmoi are called by comparison with the arithmoi, of three the third, of four the fourth, so too of the arithmoi now named the homonymous parts will be called by comparison with the arithmoi. of arithmos, the arithmost
and of power, the powerth,
and of cube, the cubeth
and of power-power, the power-powerth
and of power-cube, the power-cubeth
and of cube-cube, the cube-cubeth.

And each of them has a line fraction on the symbol of the homonymous arithmos that separates the form.

And so, having displayed for you the namesakes of the arithmoi, I will step over to their multiplications. But they will be very evident to you since they nearly made plain through their naming.

An arithmos multiplied times an arithmos makes a power;
and times a power a cube;
and times a cube, a power-power;
and times a power-power, a power-cube;
and times a power-cube, a cube-cube.
And a power times a power, a power-power;
times a cube, a power-cube;
times a power-power, cube-cube.
And a cube times a cube, a cube-cube.
Every arithmos multiplied times the homonymous part of it makes a unit.

Given that the unit is unalterable and always stands, the form multiplied times it will be the form itself.

But the homonymous parts multiplied times themselves will make parts homonymous with the the arithmoi.
For example, the arithmost;
times the arithmost makes a powerth;
and times a powerth, the cubeth;
and times the cubeth, the power-powerth;
and times the power-powerth, the power-cubeth;
and times the power-cube, the cube-cubeth;
and this follows according to the homonym (lit: homonymously).

And arithmost
times a power, arithmost;
and times a cube, power;
and times a power-power, cube;
and times a power-cube, power-power;
and times a cube-cube, power-cube.
And a powerth
times an arithmos, arithmost;
and times a cube, arithmos;
and times a power-power, power;
and times a power-cube, cube;
and times a cube-cube, power-power.

And a cubeth
times an arithmos, powerth;
and times a power, arithmost;
and times a power-power, arithmos;
and times a power-cube, power;
and times a cube-cube, cube.

And a power-powerth
times an arithmos, cubeth
and times a power, powerth;
and times a cube, arithmost;
and times a power-cubeth, arithmos;
and times a cube-cube, power.

And a power-cubeth
and times an arithmos, power-powerth,
and times a power, cubeth
and times a cubeth, powerth
and times a power-power, arithmost,
and times a cube-cubeth, arithmos.

And the cube-cubeth
times an arithmos, power-cubeth;
and times a power, power-powerth;
and times a cube, cubeth;
and times a power-power, powerth;
and times a power-cube, arithmost.

A deficiency multiplied times a deficiency makes an existence. And a deficiency times an existence makes a deficiency, and the symbol of deficiency is Ψ, defective and inclining downwards [i.e. upside down], 𐅢.

And when these multiplications have been made clear to you, the divisions of the preceding forms are obvious. And so, it does well, in engaging in the subject matter, to get practice in addition and subtraction and multiplications with respect to the forms, and how you might add existent and deficient forms, not of the same plêthos, with the other forms, whether themselves also existent or deficient, and how you might subtract some, whether existent or both similarly existent and deficient, from existent forms and other deficient forms.

After these, if from some problem certain forms become equal to the same forms, but not of the same plêthos, it will be required to subtract similars from similars from both parts, until one form becomes equal to one form. But if somehow in whichever part or in both, certain forms exist in deficiencies, it will be required to add the deficient forms in both parts, until the forms of both parts become existent, and again to subtract similars from similars until one form is left for each of the parts.

Let this be arranged artfully in actualities of the propositions, if possible, until one form is left equal to one form. Later we will also show you how when two forms are equal leaving out one, such a problem is solved.

But now let us proceed on the road to the propositions, while holding the matter connected to the forms themselves as very great. But given that they are most large in number and most large in weight, for this reason, too, when the are slowly made certain by those who have undertaken them, although they are also difficult in themselves to remember, I have judged it right to divide the things admitted among them, and most of all to divide the things at the beginning that are elementary from the simpler to the more obscure, as is fitting For, in this way, they will become easily traveled to those setting out, and their course will be recollected, with the subject matter of them coming about in thirteen books.

1. To divide the prescribed arithmos into two arithmoi in an excess that’s given.

In fact, let the given arithmos be 100, and the excess units 40. To find the arithmoi. Let the smaller be assigned, arithmos 1. Therefore, the larger will be arithmos 1 units 40. Therefore, together they become arithmos 2 units 40. But units 100 are given. Therefore, units 100 are equal to arithmos 2 units 40. And from similars are similars. I subtract from 100, units 40 [and from the 2 arithmoi and of 40 units similarly 40 units]. Remainders arithmos 2 equal units 60. Therefore each arithmos becomes units 30. For the actualities. The smaller will be units 30 and the larger units 70, and the demonstration is obvious.

2. To divide the prescribed arithmos into two arithmoi in a ratio that’s given.

In fact, let it be prescribed to divide 60 into two arithmoi in a ratio 3-times. Let the smaller be assigned arithmos 1. Therefore, the larger will be arithmos 3, and the larger is triple the smaller. It is required the two be equal to units 60. But the two added are arithmos 4. Therefore, arithmos 4 are equal to units 60. Therefore, arithmos is units 15. Therefore, the smaller will be units 15 and the larger units 45.

3. To divide the prescribed arithmos into two arithmoi in a ratio and an excess that’s given.

In fact, let it be prescribed to divide 80 into two arithmoi so that the larger is 3-times the smaller and furthermore exceeds by units 4. Let the smaller be assigned arithmos 1.. Therefore, the larger will be arithmos 3 and units 4. Remaining, I want the two to be equal to units 80. But the two added are arithmos 2 and units 4. Therefore, arithmos 4 and units 4 are equal to units 80.

And I subtract similars from similars. Therefore, remainders units 76 are equal to units 4. And arithmos becomes units 19.

For the actualities. Therefore the smaller arithmos will be units 19, but the larger units 61. [upon there being added 4 units, which I subtracted from 80 units. For I subtracted into order to find of how many units each arithmos will be but later I add to the 4 units to the larger arithmos after know of how many each [is].

4. To find two arithmoi in a given ratio so that the excess of them is also given.

In fact, let it be prescribed that the larger is 5-times the smaller, and the excess of them make units 20. Let the smaller be assigned arithmos 1. Therefore, the larger will be arithmos 5. For the rest, I want arithmos 5 to exceed arithmos 1, units 20. But the excess of them is arithmos 4. These equal units 20. But the smaller arithmos will be units 5, the larger 25. And the larger remains being 5-times the smaller, and the excess becomes units 20.

5. To divide the prescribed arithmos into two arithmoi so that given parts, not the same, of each of arithmoi resulting from the division, in being added make the given arithmos.

In fact, it is required that the given arithmos be given so that if the given parts, not the same, be taken out of the initial prescribed arithmos it be in the place between the two arithmoi that come about.

Note: if m´and n´ are the given parts, and a the initial arithmos, and b the given sum, b needs to be between a m´ and a·n´, in order to avoid, as we would say, negative numbers and 0.

In fact, let it be prescribed that 100 divide into two arithmoi so that a 3rd of the 1st and the 5th of the 2nd times the same, being added, make units 30. I assigned the 5th of the 2nd, arithmos 1. Therefore, it will be arithmos 5. Therefore, the 3rd of the 1st will be units 30 𐅢 arithmos 1. Therefore, it will be units 90 𐅢 arithmos 3. For the rest, I want the two added to make units 100. But the two added make arithmos 2 and units 90. These are equal tounits 100.

And similars from similars. Therefore, remainders units 10 equal arithmos 2. [Therefore, arithmos will be units 5]

For the actualities. I assigned the 5th of the 2nd, arithmos 1. It will be units 5. Therefore, it (the 1st)] is units 25. And the 3rd of the 1st, units 30 𐅢 arithmos 1, will be units 25, therefore it [the 3rd] will be units 75. And the 3rd of the 1st and the 5th of the 2nd remain units 30, [which being added in common make the prescribed arithmos]

6. To divide the prescribed arithmos into two arithmoi so that the given part of the first exceeds the given part of the other by a given arithmos.

In fact, it is required to the given arithmos be smaller than the arithmos that comes about if the given part in which is the excess is taken from the initial prescribed arithmos.

Let the prescribed, in fact, to divide 100 into two arithmoi so that the 4th of 1st exceeds the 6th of 2nd by units20.

I assigned the 6th of 2nd, arithmos 1 . Therefore, it will be arithmos 6. Therefore, the 4th of 1st will be arithmos 1 and units 20. Therefore, it will be arithmos 4 and units 80. Remaining, I want the two added to make units 100. But the two added make arithmos 10 and units 80. These are equal to units 100.

From similars are similars taken away. Remainder arithmos 10 are equal to units 20, and arithmos becomes units 2.

For the actualities. I assigned the 6th of 2nd, arithmos 1. It will be units 2. Therefore, it will be units 12. But the 4th of 1st, arithmos 1 and units 20. It will be units 22. Therefore, it will be units 88. And there remains the 4th of 1st exceeding the 6th of 2nd by units 20, [which being added in common make the prescribed arithmos.]

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