- Diophantus,
*Arithmetica*I Introduction and Props. 1-6: (Draft)© - trans. by Henry Mendell, Cal. State U., L.A.

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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, 3x^{5} - 4x^{4} - 5x^{3}+7unit would be grouped 3x^{5} + 7units - 4x^{4} - 5x^{3}.

Diophantus, as is not uncommon, 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^{5´}.

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, Δ^{Υ} power. And next is ‘cube’ and its symbol is Κ with a sign on it, Υ, Κ^{Υ} 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, Δ^{Υ}Δ 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, ΔΚ^{Υ} 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, Κ^{Υ}Κ cube-cube.

But what gets none of these properties, while having an undetermined plêthos of units in it, is . But there is another symbol that’s immutable (not inflected), the unit, and its symbol is M, with a sign on it, Ο, .
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 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.

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

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

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 4. Let the smaller be assigned 1.. Therefore, the larger will be 3 and 4. Remaining, I want the two to be equal to 80. But the two added are 2 and 4. Therefore, 4 and 4 are equal to 80.

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

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

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

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 30. I assigned the 5th of the 2nd, 1. Therefore, it will be 5. Therefore, the 3rd of the 1st will be 30 𐅢 1. Therefore, it will be 90 𐅢 3. For the rest, I want the two added to make 100. But the two added make 2 and 90. These are equal to 100.

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

For the actualities. I assigned the 5th of the 2nd, 1. It will be 5. Therefore, it (the 1st)] is 25. And the 3rd of the 1st, 30 𐅢 1, will be 25, therefore it [the 3rd] will be 75. And the 3rd of the 1st and the 5th of the 2nd remain 30, [which being added in common make the prescribed 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 20.

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

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

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