If two similar plane numbers make some (number by) multiplying one another then the created (number) will be square.
Let A and B be two similar plane numbers, and let A make C (by) multiplying B.
I say that C is square.
For let A make D (by) multiplying itself.
D is thus square.
Therefore, since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17].
And since A and B are similar plane numbers, one number thus falls (between) A and B in mean proportion [Prop. 8.18].
And if (some) numbers fall between two numbers in continued proportion then, as many (numbers) as fall in (between) them (in continued proportion), so many also (fall) in (between numbers) having the same ratio (as them in continued proportion) [Prop. 8.8].
And hence one number falls (between) D and C in mean proportion.
And D is square.
Thus, C (is) also square [Prop. 8.22].
(Which is) the very thing it was required to show.
Proposition 2
If two numbers make a square (number by) multiplying one another then they are similar plane numbers.
Let A and B be two numbers, and let A make the square (number) C (by) multiplying B.
I say that A and B are similar plane numbers.
For let A make D (by) multiplying itself.
Thus, D is square.
And since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17].
And since D is square, and C (is) also, D and C are thus similar plane numbers.
Thus, one (number) falls (between) D and C in mean proportion [Prop. 8.18].
And as D is to C, so A (is) to B.
Thus, one (number) also falls (between) A and B in mean proportion [Prop. 8.8].
And if one (number) falls (between) two numbers in mean proportion then [the] numbers are similar plane (numbers) [Prop. 8.20].
Thus, A and B are similar plane (numbers).
(Which is) the very thing it was required to show.
Proposition 3
If a cube number makes some (number by) multiplying itself then the created (number) will be cube.
For let the cube number A make B (by) multiplying itself.
I say that B is cube.
For let the side C of A have been taken.
And let C make D by multiplying itself.
So it is clear that C has made A (by) multiplying D.
And since C has made D (by) multiplying itself, C thus measures D according to the units in it [Def. 7.15].
But, in fact, a unit also measures C according to the units in it [Def. 7.20].
Thus, as a unit is to C, so C (is) to D.
Again, since C has made A (by) multiplying D, D thus measures A according to the units in C.
And a unit also measures C according to the units in it.
Thus, as a unit is to C, so D (is) to A.
But, as a unit (is) to C, so C (is) to D.
And thus as a unit (is) to C, so C (is) to D, and D to A.
Thus, two numbers, C and D, have fallen (between) a unit and the number A in continued mean proportion.
Again, since A has made B (by) multiplying itself, A thus measures B according to the units in it.
And a unit also measures A according to the units in it.
Thus, as a unit is to A, so A (is) to B.
And two numbers have fallen (between) a unit and A in mean proportion.
Thus two numbers will also fall (between) A and B in mean proportion [Prop. 8.8].
And if two (numbers) fall (between) two numbers in mean proportion, and the first (number) is cube, then the second will also be cube [Prop. 8.23].
And A is cube.
Thus, B is also cube.
(Which is) the very thing it was required to show.
Proposition 4
If a cube number makes some (number by) multiplying a(nother) cube number then the created (number) will be cube.
For let the cube number A make C (by) multiplying the cube number B.
I say that C is cube.
For let A make D (by) multiplying itself.
Thus, D is cube [Prop. 9.3].
And since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17].
And since A and B are cube, A and B are similar solid (numbers).
Thus, two numbers fall (between) A and B in mean proportion [Prop. 8.19].
Hence, two numbers will also fall (between) D and C in mean proportion [Prop. 8.8].
And D is cube.
Thus, C (is) also cube [Prop. 8.23].
(Which is) the very thing it was required to show.
Proposition 5
If a cube number makes a(nother) cube number (by) multiplying some (number) then the (number) multiplied will also be cube.
For let the cube number A make the cube (number) C (by) multiplying some number B.
I say that B is cube.
For let A make D (by) multiplying itself.
D is thus cube [Prop. 9.3].
And since A has made D (by) multiplying itself, and has made C (by) multiplying B, thus as A is to B, so D (is) to C [Prop. 7.17].
And since D and C are (both) cube, they are similar solid (numbers).
Thus, two numbers fall (between) D and C in mean proportion [Prop. 8.19].
And as D is to C, so A (is) to B.
Thus, two numbers also fall (between) A and B in mean proportion [Prop. 8.8].
And A is cube.
Thus, B is also cube [Prop. 8.23].
(Which is) the very thing it was required to show.
Proposition 6
If a number makes a cube (number by) multiplying itself then it itself will also be cube.
For let the number A make the cube (number) B (by) multiplying itself.
I say that A is also cube.
For let A make C (by) multiplying B.
Therefore, since A has made B (by) multiplying itself, and has made C (by) multiplying B, C is thus cube.
And since A has made B (by) multiplying itself, A thus measures B according to the units in ( A).
And a unit also measures A according to the units in it.
Thus, as a unit is to A, so A (is) to B.
And since A has made C (by) multiplying B, B thus measures C according to the units in A.
And a unit also measures A according to the units in it.
Thus, as a unit is to A, so B (is) to C.
But, as a unit (is) to A, so A (is) to B.
And thus as A (is) to B, (so) B (is) to C.
And since B and C are cube, they are similar solid (numbers).
Thus, there exist two numbers in mean proportion (between) B and C [Prop. 8.19].
And as B is to C, (so) A (is) to B.
Thus, there also exist two numbers in mean proportion (between) A and B [Prop. 8.8].
And B is cube.
Thus, A is also cube [Prop. 8.23].
(Which is) the very thing it was required to show.
Proposition 7
If a composite number makes some (number by) multiplying some (other) number then the created (number) will be solid.
For let the composite number A make C (by) multiplying some number B.
I say that C is solid.
For since A is a composite (number), it will be measured by some number.
Let it be measured by D.
And, as many times as D measures A, so many units let there be in E.
Therefore, since D measures A according to the units in E, E has thus made A (by) multiplying D [Def. 7.15].
And since A has made C (by) multiplying B, and A is the (number created) from (multiplying) D, E, the (number created) from (multiplying) D, E has thus made C (by) multiplying B.
Thus, C is solid, and its sides are D, E, B.
(Which is) the very thing it was required to show.
Proposition 8
If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, then the third from the unit will be square, and (all) those (numbers after that) which leave an interval of one (number), and the fourth (will be) cube, and all those (numbers after that) which leave an interval of two (numbers), and the seventh (will be) both cube and square, and (all) those (numbers after that) which leave an interval of five (numbers).
Let any multitude whatsoever of numbers, A, B, C, D, E, F, be continuously proportional, (starting) from a unit.
I say that the third from the unit, B, is square, and all those (numbers after that) which leave an interval of one (number).
And the fourth (from the unit), C, (is) cube, and all those (numbers after that) which leave an interval of two (numbers).
And the seventh (from the unit), F, (is) both cube and square, and all those (numbers after that) which leave an interval of five (numbers).
For since as the unit is to A, so A (is) to B, the unit thus measures the number A the same number of times as A (measures) B [Def. 7.20].
And the unit measures the number A according to the units in it.
Thus, A also measures B according to the units in A.
A has thus made B (by) multiplying itself [Def. 7.15].
Thus, B is square.
And since B, C, D are continuously proportional, and B is square, D is thus also square [Prop. 8.22].
So, for the same (reasons), F is also square.
So, similarly, we can also show that all those (numbers after that) which leave an interval of one (number) are square.
So I also say that the fourth (number) from the unit, C, is cube, and all those (numbers after that) which leave an interval of two (numbers).
For since as the unit is to A, so B (is) to C, the unit thus measures the number A the same number of times that B (measures) C.
And the unit measures the number A according to the units in A.
And thus B measures C according to the units in A.
A has thus made C (by) multiplying B.
Therefore, since A has made B (by) multiplying itself, and has made C (by) multiplying B, C is thus cube.
And since C, D, E, F are continuously proportional, and C is cube, F is thus also cube [Prop. 8.23].
And it was also shown (to be) square.
Thus, the seventh (number) from the unit is (both) cube and square.
So, similarly, we can show that all those (numbers after that) which leave an interval of five (numbers) are (both) cube and square.
(Which is) the very thing it was required to show.
Proposition 9
If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, and the (number) after the unit is square, then all the remaining (numbers) will also be square.
And if the (number) after the unit is cube, then all the remaining (numbers) will also be cube.
Let any multitude whatsoever of numbers, A, B, C, D, E, F, be continuously proportional, (starting) from a unit.
And let the (number) after the unit, A, be square.
I say that all the remaining (numbers) will also be square.
In fact, it has (already) been shown that the third (number) from the unit, B, is square, and all those (numbers after that) which leave an interval of one (number) [Prop. 9.8].
[So] I say that all the remaining (numbers) are also square.
For since A, B, C are continuously proportional, and A (is) square, C is [thus] also square [Prop. 8.22].
Again, since B, C, D are [also] continuously proportional, and B is square, D is [thus] also square [Prop. 8.22].
So, similarly, we can show that all the remaining (numbers) are also square.
And so let A be cube.
I say that all the remaining (numbers) are also cube.
In fact, it has (already) been shown that the fourth (number) from the unit, C, is cube, and all those (numbers after that) which leave an interval of two (numbers) [Prop. 9.8].
[So] I say that all the remaining (numbers) are also cube.
For since as the unit is to A, so A (is) to B, the unit thus measures A the same number of times as A (measures) B.
And the unit measures A according to the units in it.
Thus, A also measures B according to the units in ( A).
A has thus made B (by) multiplying itself.
And A is cube.
And if a cube number makes some (number by) multiplying itself then the created (number) is cube [Prop. 9.3].
Thus, B is also cube.
And since the four numbers A, B, C, D are continuously proportional, and A is cube, D is thus also cube [Prop. 8.23].
So, for the same (reasons), E is also cube, and, similarly, all the remaining (numbers) are cube.
(Which is) the very thing it was required to show.
Proposition 10
If any multitude whatsoever of numbers is [continuously] proportional, (starting) from a unit, and the (number) after the unit is not square, then no other (number) will be square either, apart from the third from the unit, and all those (numbers after that) which leave an interval of one (number).
And if the (number) after the unit is not cube, then no other (number) will be cube either, apart from the fourth from the unit, and all those (numbers after that) which leave an interval of two (numbers).
Let any multitude whatsoever of numbers, A, B, C, D, E, F, be continuously proportional, (starting) from a unit.
And let the (number) after the unit, A, not be square.
I say that no other (number) will be square either, apart from the third from the unit [and (all) those (numbers after that) which leave an interval of one (number)].
For, if possible, let C be square.
And B is also square [Prop. 9.8].
Thus, B and C have to one another (the) ratio which (some) square number (has) to (some other) square number.
And as B is to C, (so) A (is) to B.
Thus, A and B have to one another (the) ratio which (some) square number has to (some other) square number.
Hence, A and B are similar plane (numbers) [Prop. 8.26].
And B is square.
Thus, A is also square.
The very opposite thing was assumed.
C is thus not square.
So, similarly, we can show that no other (number is) square either, apart from the third from the unit, and (all) those (numbers after that) which leave an interval of one (number).
And so let A not be cube.
I say that no other (number) will be cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers).
For, if possible, let D be cube.
And C is also cube [Prop. 9.8].
For it is the fourth (number) from the unit.
And as C is to D, (so) B (is) to C.
And B thus has to C the ratio which (some) cube (number has) to (some other) cube (number).
And C is cube.
Thus, B is also cube [Prop. 7.13] [Prop. 8.25].
And since as the unit is to A, (so) A (is) to B, and the unit measures A according to the units in it, A thus also measures B according to the units in ( A).
Thus, A has made the cube (number) B (by) multiplying itself.
And if a number makes a cube (number by) multiplying itself then it itself will be cube [Prop. 9.6].
Thus, A (is) also cube.
The very opposite thing was assumed.
Thus, D is not cube.
So, similarly, we can show that no other (number) is cube either, apart from the fourth from the unit, and (all) those (numbers after that) which leave an interval of two (numbers).
(Which is) the very thing it was required to show.
Proposition 11
If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, then a lesser (number) measures a greater according to some existing (number) among the proportional numbers.
Let any multitude whatsoever of numbers, B, C, D, E, be continuously proportional, (starting) from the unit A.
I say that, for B, C, D, E, the least (number), B, measures E according to some (one) of C, D.
For since as the unit A is to B, so D (is) to E, the unit A thus measures the number B the same number of times as D (measures) E.
Thus, alternately, the unit A measures D the same number of times as B (measures) E [Prop. 7.15].
And the unit A measures D according to the units in it.
Thus, B also measures E according to the units in D.
Hence, the lesser (number) B measures the greater E according to some existing number among the proportional numbers (namely, D).
Corollary
And (it is) clear that what(ever relative) place the measuring (number) has from the unit, the (number) according to which it measures has the same (relative) place from the measured (number), in (the direction of the number) before it.
(Which is) the very thing it was required to show.
Proposition 12
If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, then however many prime numbers the last (number) is measured by, the (number) next to the unit will also be measured by the same (prime numbers).
Let any multitude whatsoever of numbers, A, B, C, D, be (continuously) proportional, (starting) from a unit.
I say that however many prime numbers D is measured by, A will also be measured by the same (prime numbers).
For let D be measured by some prime number E.
I say that E measures A.
For (suppose it does) not.
E is prime, and every prime number is prime to every number which it does not measure [Prop. 7.29].
Thus, E and A are prime to one another.
And since E measures D, let it measure it according to F.
Thus, E has made D (by) multiplying F.
Again, since A measures D according to the units in C [Prop. 9.11 corr.], A has thus made D (by) multiplying C.
But, in fact, E has also made D (by) multiplying F.
Thus, the (number created) from (multiplying) A, C is equal to the (number created) from (multiplying) E, F.
Thus, as A is to E, (so) F (is) to C [Prop. 7.19].
And A and E (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, E measures C.
Let it measure it according to G.
Thus, E has made C (by) multiplying G.
But, in fact, via the (proposition) before this, A has also made C (by) multiplying B [Prop. 9.11 corr.].
Thus, the (number created) from (multiplying) A, B is equal to the (number created) from (multiplying) E, G.
Thus, as A is to E, (so) G (is) to B [Prop. 7.19].
And A and E (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, E measures B.
Let it measure it according to H.
Thus, E has made B (by) multiplying H.
But, in fact, A has also made B (by) multiplying itself [Prop. 9.8].
Thus, the (number created) from (multiplying) E, H is equal to the (square) on A.
Thus, as E is to A, (so) A (is) to H [Prop. 7.19].
And A and E are prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, E measures A, as the leading (measuring the) leading.
But, in fact, ( E) also does not measure ( A).
The very thing (is) impossible.
Thus, E and A are not prime to one another.
Thus, (they are) composite (to one another).
And (numbers) composite (to one another) are (both) measured by some [prime] number [Def. 7.14].
And since E is assumed (to be) prime, and a prime (number) is not measured by another number (other) than itself [Def. 7.11], E thus measures (both) A and E.
Hence, E measures A.
And it also measures D.
Thus, E measures (both) A and D.
So, similarly, we can show that however many prime numbers D is measured by, A will also be measured by the same (prime numbers).
(Which is) the very thing it was required to show.
Proposition 13
If any multitude whatsoever of numbers is continuously proportional, (starting) from a unit, and the (number) after the unit is prime, then the greatest (number) will be measured by no [other] (numbers) except (numbers) existing among the proportional numbers.
Let any multitude whatsoever of numbers, A, B, C, D, be continuously proportional, (starting) from a unit.
And let the (number) after the unit, A, be prime.
I say that the greatest of them, D, will be measured by no other (numbers) except A, B, C.
For, if possible, let it be measured by E, and let E not be the same as one of A, B, C.
So it is clear that E is not prime.
For if E is prime, and measures D, then it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12].
The very thing is impossible.
Thus, E is not prime.
Thus, (it is) composite.
And every composite number is measured by some prime number [Prop. 7.31].
Thus, E is measured by some prime number.
So I say that it will be measured by no other prime number than A.
For if E is measured by another (prime number), and E measures D, then this (prime number) will thus also measure D.
Hence, it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12].
The very thing is impossible.
Thus, A measures E.
And since E measures D, let it measure it according to F.
I say that F is not the same as one of A, B, C.
For if F is the same as one of A, B, C, and measures D according to E, then one of A, B, C thus also measures D according to E.
But one of A, B, C (only) measures D according to some (one) of A, B, C [Prop. 9.11].
And thus E is the same as one of A, B, C.
The very opposite thing was assumed.
Thus, F is not the same as one of A, B, C.
Similarly, we can show that F is measured by A, (by) again showing that F is not prime.
For if ( F is prime), and measures D, then it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12].
The very thing is impossible.
Thus, F is not prime.
Thus, (it is) composite.
And every composite number is measured by some prime number [Prop. 7.31].
Thus, F is measured by some prime number.
So I say that it will be measured by no other prime number than A.
For if some other prime (number) measures F, and F measures D, then this (prime number) will thus also measure D.
Hence, it will also measure A, (despite A) being prime (and) not being the same as it [Prop. 9.12].
The very thing is impossible.
Thus, A measures F.
And since E measures D according to F, E has thus made D (by) multiplying F.
But, in fact, A has also made D (by) multiplying C [Prop. 9.11 corr.].
Thus, the (number created) from (multiplying) A, C is equal to the (number created) from (multiplying) E, F.
Thus, proportionally, as A is to E, so F (is) to C [Prop. 7.19].
And A measures E.
Thus, F also measures C.
Let it measure it according to G.
So, similarly, we can show that G is not the same as one of A, B, and that it is measured by A.
And since F measures C according to G, F has thus made C (by) multiplying G.
But, in fact, A has also made C (by) multiplying B [Prop. 9.11 corr.].
Thus, the (number created) from (multiplying) A, B is equal to the (number created) from (multiplying) F, G.
Thus, proportionally, as A (is) to F, so G (is) to B [Prop. 7.19].
And A measures F.
Thus, G also measures B.
Let it measure it according to H.
So, similarly, we can show that H is not the same as A.
And since G measures B according to H, G has thus made B (by) multiplying H.
But, in fact, A has also made B (by) multiplying itself [Prop. 9.8].
Thus, the (number created) from (multiplying) H, G is equal to the square on A.
Thus, as H is to A, (so) A (is) to G [Prop. 7.19].
And A measures G.
Thus, H also measures A, (despite A) being prime (and) not being the same as it.
The very thing (is) absurd.
Thus, the greatest (number) D cannot be measured by another (number) except (one of) A, B, C.
(Which is) the very thing it was required to show.
Proposition 14
If a least number is measured by (some) prime numbers then it will not be measured by any other prime number except (one of) the original measuring (numbers).
For let A be the least number measured by the prime numbers B, C, D.
I say that A will not be measured by any other prime number except (one of) B, C, D.
For, if possible, let it be measured by the prime (number) E.
And let E not be the same as one of B, C, D.
And since E measures A, let it measure it according to F.
Thus, E has made A (by) multiplying F.
And A is measured by the prime numbers B, C, D.
And if two numbers make some (number by) multiplying one another, and some prime number measures the number created from them, then (the prime number) will also measure one of the original (numbers) [Prop. 7.30].
Thus, B, C, D will measure one of E, F.
In fact, they do not measure E.
For E is prime, and not the same as one of B, C, D.
Thus, they (all) measure F, which is less than A.
The very thing (is) impossible.
For A was assumed (to be) the least (number) measured by B, C, D.
Thus, no prime number can measure A except (one of) B, C, D.
(Which is) the very thing it was required to show.
Proposition 15
If three continuously proportional numbers are the least of those (numbers) having the same ratio as them then two (of them) added together in any way are prime to the remaining (one).
Let A, B, C be three continuously proportional numbers (which are) the least of those (numbers) having the same ratio as them.
I say that two of A, B, C added together in any way are prime to the remaining (one), (that is) A and B (prime) to C, B and C to A, and, further, A and C to B.
Let the two least numbers, DE and EF, having the same ratio as A, B, C, have been taken [Prop. 8.2].
So it is clear that DE has made A (by) multiplying itself, and has made B (by) multiplying EF, and, further, EF has made C (by) multiplying itself [Prop. 8.2].
And since DE, EF are the least (of those numbers having the same ratio as them), they are prime to one another [Prop. 7.22].
And if two numbers are prime to one another then the sum (of them) is also prime to each [Prop. 7.28].
Thus, DF is also prime to each of DE, EF.
But, in fact, DE is also prime to EF.
Thus, DF, DE are (both) prime to EF.
And if two numbers are (both) prime to some number then the (number) created from (multiplying) them is also prime to the remaining (number) [Prop. 7.24].
Hence, the (number created) from (multiplying) FD, DE is prime to EF.
Hence, the (number created) from (multiplying) FD, DE is also prime to the (square) on EF [Prop. 7.25].
[For if two numbers are prime to one another then the (number) created from (squaring) one of them is prime to the remaining (number).]
But the (number created) from (multiplying) FD, DE is the (square) on DE plus the (number created) from (multiplying) DE, EF [Prop. 2.3].
Thus, the (square) on DE plus the (number created) from (multiplying) DE, EF is prime to the (square) on EF.
And the (square) on DE is A, and the (number created) from (multiplying) DE, EF (is) B, and the (square) on EF (is) C.
Thus, A, B summed is prime to C.
So, similarly, we can show that B, C (summed) is also prime to A.
So I say that A, C (summed) is also prime to B.
For since DF is prime to each of DE, EF then the (square) on DF is also prime to the (number created) from (multiplying) DE, EF [Prop. 7.25].
But, the (sum of the squares) on DE, EF plus twice the (number created) from (multiplying) DE, EF is equal to the (square) on DF [Prop. 2.4].
And thus the (sum of the squares) on DE, EF plus twice the (rectangle contained) by DE, EF [is] prime to the (rectangle contained) by DE, EF.
By separation, the (sum of the squares) on DE, EF plus once the (rectangle contained) by DE, EF is prime to the (rectangle contained) by DE, EF.
Again, by separation, the (sum of the squares) on DE, EF is prime to the (rectangle contained) by DE, EF.
And the (square) on DE is A, and the (rectangle contained) by DE, EF (is) B, and the (square) on EF (is) C.
Thus, A, C summed is prime to B.
(Which is) the very thing it was required to show.
Proposition 16
If two numbers are prime to one another then as the first is to the second, so the second (will) not (be) to some other (number).
For let the two numbers A and B be prime to one another.
I say that as A is to B, so B is not to some other (number).
For, if possible, let it be that as A (is) to B, (so) B (is) to C.
And A and B (are) prime (to one another).
And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21].
And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, A measures B, as the leading (measuring) the leading.
And ( A) also measures itself.
Thus, A measures A and B, which are prime to one another.
The very thing (is) absurd.
Thus, as A (is) to B, so B cannot be to C.
(Which is) the very thing it was required to show.
Proposition 17
If any multitude whatsoever of numbers is continuously proportional, and the outermost of them are prime to one another, then as the first (is) to the second, so the last will not be to some other (number).
Let A, B, C, D be any multitude whatsoever of continuously proportional numbers.
And let the outermost of them, A and D, be prime to one another.
I say that as A is to B, so D (is) not to some other (number).
For, if possible, let it be that as A (is) to B, so D (is) to E.
Thus, alternately, as A is to D, (so) B (is) to E [Prop. 7.13].
And A and D are prime (to one another).
And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21].
And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, A measures B.
And as A is to B, (so) B (is) to C.
Thus, B also measures C.
And hence A measures C [Def. 7.20].
And since as B is to C, (so) C (is) to D, and B measures C, C thus also measures D [Def. 7.20].
But, A was (found to be) measuring C.
And hence A also measures D.
And ( A) also measures itself.
Thus, A measures A and D, which are prime to one another.
The very thing is impossible.
Thus, as A (is) to B, so D cannot be to some other (number).
(Which is) the very thing it was required to show.
Proposition 18
For two given numbers, to investigate whether it is possible to find a third (number) proportional to them.
Let A and B be the two given numbers.
And let it be required to investigate whether it is possible to find a third (number) proportional to them.
So A and B are either prime to one another, or not.
And if they are prime to one another then it has (already) been show that it is impossible to find a third (number) proportional to them [Prop. 9.16].
And so let A and B not be prime to one another.
And let B make C (by) multiplying itself.
So A either measures, or does not measure, C.
Let it first of all measure ( C) according to D.
Thus, A has made C (by) multiplying D.
But, in fact, B has also made C (by) multiplying itself.
Thus, the (number created) from (multiplying) A, D is equal to the (square) on B.
Thus, as A is to B, (so) B (is) to D [Prop. 7.19].
Thus, a third number has been found proportional to A, B, (namely) D.
And so let A not measure C.
I say that it is impossible to find a third number proportional to A, B.
For, if possible, let it have been found, (and let it be) D.
Thus, the (number created) from (multiplying) A, D is equal to the (square) on B [Prop. 7.19].
And the (square) on B is C.
Thus, the (number created) from (multiplying) A, D is equal to C.
Hence, A has made C (by) multiplying D.
Thus, A measures C according to D.
But ( A) was, in fact, also assumed (to be) not measuring ( C).
The very thing (is) absurd.
Thus, it is not possible to find a third number proportional to A, B when A does not measure C.
(Which is) the very thing it was required to show.
Proposition 19
For three given numbers, to investigate when it is possible to find a fourth (number) proportional to them.
Let A, B, C be the three given numbers.
And let it be required to investigate when it is possible to find a fourth (number) proportional to them.
In fact, ( A, B, C) are either not continuously proportional and the outermost of them are prime to one another, or are continuously proportional and the outermost of them are not prime to one another, or are neither continuously proportional nor are the outermost of them prime to one another, or are continuously proportional and the outermost of them are prime to one another.
In fact, if A, B, C are continuously proportional, and the outermost of them, A and C, are prime to one another, (then) it has (already) been shown that it is impossible to find a fourth number proportional to them [Prop. 9.17].
So let A, B, C not be continuously proportional, (with) the outermost of them again being prime to one another.
I say that, in this case, it is also impossible to find a fourth (number) proportional to them.
For, if possible, let it have been found, (and let it be) D.
Hence, it will be that as A (is) to B, (so) C (is) to D.
And let it be contrived that as B (is) to C, (so) D (is) to E.
And since as A is to B, (so) C (is) to D, and as B (is) to C, (so) D (is) to E, thus, via equality, as A (is) to C, (so) C (is) to E [Prop. 7.14].
And A and C (are) prime (to one another).
And (numbers) prime (to one another are) also the least (numbers having the same ratio as them) [Prop. 7.21].
And the least (numbers) measure those numbers having the same ratio as them (the same number of times), the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, A measures C, (as) the leading (measuring) the leading.
And it also measures itself.
Thus, A measures A and C, which are prime to one another.
The very thing is impossible.
Thus, it is not possible to find a fourth (number) proportional to A, B, C.
And so let A, B, C again be continuously proportional, and let A and C not be prime to one another.
I say that it is possible to find a fourth (number) proportional to them.
For let B make D (by) multiplying C.
Thus, A either measures or does not measure D.
Let it, first of all, measure ( D) according to E.
Thus, A has made D (by) multiplying E.
But, in fact, B has also made D (by) multiplying C.
Thus, the (number created) from (multiplying) A, E is equal to the (number created) from (multiplying) B, C.
Thus, proportionally, as A [is] to B, (so) C (is) to E [Prop. 7.19].
Thus, a fourth (number) proportional to A, B, C has been found, (namely) E.
And so let A not measure D.
I say that it is impossible to find a fourth number proportional to A, B, C.
For, if possible, let it have been found, (and let it be) E.
Thus, the (number created) from (multiplying) A, E is equal to the (number created) from (multiplying) B, C.
But, the (number created) from (multiplying) B, C is D.
And thus the (number created) from (multiplying) A, E is equal to D.
Thus, A has made D (by) multiplying E.
Thus, A measures D according to E.
Hence, A measures D.
But, it also does not measure ( D).
The very thing (is) absurd.
Thus, it is not possible to find a fourth number proportional to A, B, C when A does not measure D.
And so (let) A, B, C (be) neither continuously proportional, nor (let) the outermost of them (be) prime to one another.
And let B make D (by) multiplying C.
So, similarly, it can be show that if A measures D then it is possible to find a fourth (number) proportional to ( A, B, C), and impossible if ( A) does not measure ( D).
(Which is) the very thing it was required to show.
Proposition 20
The (set of all) prime numbers is more numerous than any assigned multitude of prime numbers.
Let A, B, C be the assigned prime numbers.
I say that the (set of all) primes numbers is more numerous than A, B, C.
For let the least number measured by A, B, C have been taken, and let it be DE [Prop. 7.36].
And let the unit DF have been added to DE.
So EF is either prime, or not.
Let it, first of all, be prime.
Thus, the (set of) prime numbers A, B, C, EF, (which is) more numerous than A, B, C, has been found.
And so let EF not be prime.
Thus, it is measured by some prime number [Prop. 7.31].
Let it be measured by the prime (number) G.
I say that G is not the same as any of A, B, C.
For, if possible, let it be (the same).
And A, B, C (all) measure DE.
Thus, G will also measure DE.
And it also measures EF.
(So) G will also measure the remainder, unit DF, (despite) being a number [Prop. 7.28].
The very thing (is) absurd.
Thus, G is not the same as one of A, B, C.
And it was assumed (to be) prime.
Thus, the (set of) prime numbers A, B, C, G, (which is) more numerous than the assigned multitude (of prime numbers), A, B, C, has been found.
(Which is) the very thing it was required to show.
Proposition 21
If any multitude whatsoever of even numbers is added together then the whole is even.
For let any multitude whatsoever of even numbers, AB, BC, CD, DE, lie together.
I say that the whole, AE, is even.
For since everyone of AB, BC, CD, DE is even, it has a half part [Def. 7.6].
And hence the whole AE has a half part.
And an even number is one (which can be) divided in half [Def. 7.6].
Thus, AE is even.
(Which is) the very thing it was required to show.
Proposition 22
If any multitude whatsoever of odd numbers is added together, and the multitude of them is even, then the whole will be even.
For let any even multitude whatsoever of odd numbers, AB, BC, CD, DE, lie together.
I say that the whole, AE, is even.
For since everyone of AB, BC, CD, DE is odd then, a unit being subtracted from each, everyone of the remainders will be (made) even [Def. 7.7].
And hence the sum of them will be even [Prop. 9.21].
And the multitude of the units is even.
Thus, the whole AE is also even [Prop. 9.21].
(Which is) the very thing it was required to show.
Proposition 23
If any multitude whatsoever of odd numbers is added together, and the multitude of them is odd, then the whole will also be odd.
For let any multitude whatsoever of odd numbers, AB, BC, CD, lie together, and let the multitude of them be odd.
I say that the whole, AD, is also odd.
For let the unit DE have been subtracted from CD.
The remainder CE is thus even [Def. 7.7].
And CA is also even [Prop. 9.22].
Thus, the whole AE is also even [Prop. 9.21].
And DE is a unit.
Thus, AD is odd [Def. 7.7].
(Which is) the very thing it was required to show.
Proposition 24
If an even (number) is subtracted from an(other) even number then the remainder will be even.
For let the even (number) BC have been subtracted from the even number AB.
I say that the remainder CA is even.
For since AB is even, it has a half part [Def. 7.6].
So, for the same (reasons), BC also has a half part.
And hence the remainder [ CA has a half part].
[Thus,] AC is even.
(Which is) the very thing it was required to show.
Proposition 25
If an odd (number) is subtracted from an even number then the remainder will be odd.
For let the odd (number) BC have been subtracted from the even number AB.
I say that the remainder CA is odd.
For let the unit CD have been subtracted from BC.
DB is thus even [Def. 7.7].
And AB is also even.
And thus the remainder AD is even [Prop. 9.24].
And CD is a unit.
Thus, CA is odd [Def. 7.7].
(Which is) the very thing it was required to show.
Proposition 26
If an odd (number) is subtracted from an odd number then the remainder will be even.
For let the odd (number) BC have been subtracted from the odd (number) AB.
I say that the remainder CA is even.
For since AB is odd, let the unit BD have been subtracted (from it).
Thus, the remainder AD is even [Def. 7.7].
So, for the same (reasons), CD is also even.
And hence the remainder CA is even [Prop. 9.24].
(Which is) the very thing it was required to show.
Proposition 27
If an even (number) is subtracted from an odd number then the remainder will be odd.
For let the even (number) BC have been subtracted from the odd (number) AB.
I say that the remainder CA is odd.
[For] let the unit AD have been subtracted (from AB).
DB is thus even [Def. 7.7].
And BC is also even.
Thus, the remainder CD is also even [Prop. 9.24].
CA (is) thus odd [Def. 7.7].
(Which is) the very thing it was required to show.
Proposition 28
If an odd number makes some (number by) multiplying an even (number) then the created (number) will be even.
For let the odd number A make C (by) multiplying the even (number) B.
I say that C is even.
For since A has made C (by) multiplying B, C is thus composed out of so many (magnitudes) equal to B, as many as (there) are units in A [Def. 7.15].
And B is even.
Thus, C is composed out of even (numbers).
And if any multitude whatsoever of even numbers is added together then the whole is even [Prop. 9.21].
Thus, C is even.
(Which is) the very thing it was required to show.
Proposition 29
If an odd number makes some (number by) multiplying an odd (number) then the created (number) will be odd.
For let the odd number A make C (by) multiplying the odd (number) B.
I say that C is odd.
For since A has made C (by) multiplying B, C is thus composed out of so many (magnitudes) equal to B, as many as (there) are units in A [Def. 7.15].
And each of A, B is odd.
Thus, C is composed out of odd (numbers), (and) the multitude of them is odd.
Hence C is odd [Prop. 9.23].
(Which is) the very thing it was required to show.
Proposition 30
If an odd number measures an even number then it will also measure (one) half of it.
For let the odd number A measure the even (number) B.
I say that ( A) will also measure (one) half of ( B).
For since A measures B, let it measure it according to C.
I say that C is not odd.
For, if possible, let it be (odd).
And since A measures B according to C, A has thus made B (by) multiplying C.
Thus, B is composed out of odd numbers, (and) the multitude of them is odd.
B is thus odd [Prop. 9.23].
The very thing (is) absurd.
For ( B) was assumed (to be) even.
Thus, C is not odd.
Thus, C is even.
Hence, A measures B an even number of times.
So, on account of this, ( A) will also measure (one) half of ( B).
(Which is) the very thing it was required to show.
Proposition 31
If an odd number is prime to some number then it will also be prime to its double.
For let the odd number A be prime to some number B.
And let C be double B.
I say that A is [also] prime to C.
For if [ A and C ] are not prime (to one another) then some number will measure them.
Let it measure (them), and let it be D.
And A is odd.
Thus, D (is) also odd.
And since D, which is odd, measures C, and C is even, [ D ] will thus also measure half of C [Prop. 9.30].
And B is half of C.
Thus, D measures B.
And it also measures A.
Thus, D measures (both) A and B, (despite) them being prime to one another.
The very thing is impossible.
Thus, A is not unprime to C.
Thus, A and C are prime to one another.
(Which is) the very thing it was required to show.
Proposition 32
Each of the numbers (which is continually) doubled, (starting) from a dyad, is an even-times-even (number) only.
For let any multitude of numbers whatsoever, B, C, D, have been (continually) doubled, (starting) from the dyad A.
I say that B, C, D are even-times-even (numbers) only.
In fact, (it is) clear that each [of B, C, D ] is an even-times-even (number).
For it is doubled from a dyad [Def. 7.8].
I also say that (they are even-times-even numbers) only.
For let a unit be laid down.
Therefore, since any multitude of numbers whatsoever are continuously proportional, starting from a unit, and the (number) A after the unit is prime, the greatest of A, B, C, D, (namely) D, will not be measured by any other (numbers) except A, B, C [Prop. 9.13].
And each of A, B, C is even.
Thus, D is an even-time-even (number) only [Def. 7.8].
So, similarly, we can show that each of B, C is [also] an even-time-even (number) only.
(Which is) the very thing it was required to show.
Proposition 33
If a number has an odd half then it is an even-time-odd (number) only.
For let the number A have an odd half.
I say that A is an even-times-odd (number) only.
In fact, (it is) clear that ( A) is an even-times-odd (number).
For its half, being odd, measures it an even number of times [Def. 7.9].
So I also say that (it is an even-times-odd number) only.
For if A is also an even-times-even (number) then it will be measured by an even (number) according to an even number [Def. 7.8].
Hence, its half will also be measured by an even number, (despite) being odd.
The very thing is absurd.
Thus, A is an even-times-odd (number) only.
(Which is) the very thing it was required to show.
Proposition 34
If a number is neither (one) of the (numbers) doubled from a dyad, nor has an odd half, then it is (both) an even-times-even and an even-times-odd (number).
For let the number A neither be (one) of the (numbers) doubled from a dyad, nor let it have an odd half.
I say that A is (both) an even-times-even and an even-times-odd (number).
In fact, (it is) clear that A is an even-times-even (number) [Def. 7.8].
For it does not have an odd half.
So I say that it is also an even-times-odd (number).
For if we cut A in half, and (then cut) its half in half, and we do this continually, then we will arrive at some odd number which will measure A according to an even number.
For if not, we will arrive at a dyad, and A will be (one) of the (numbers) doubled from a dyad.
The very opposite thing (was) assumed.
Hence, A is an even-times-odd (number) [Def. 7.9].
And it was also shown (to be) an even-times-even (number).
Thus, A is (both) an even-times-even and an even-times-odd (number).
(Which is) the very thing it was required to show.
Proposition 35
If there is any multitude whatsoever of continually proportional numbers, and (numbers) equal to the first are subtracted from (both) the second and the last, then as the excess of the second (number is) to the first, so the excess of the last will be to (the sum of) all those (numbers) before it.
Let A, BC, D, EF be any multitude whatsoever of continuously proportional numbers, beginning from the least A.
And let BG and FH, each equal to A, have been subtracted from BC and EF (respectively).
I say that as GC is to A, so EH is to A, BC, D.
For let FK be made equal to BC, and FL to D.
And since FK is equal to BC, of which FH is equal to BG, the remainder HK is thus equal to the remainder GC.
And since as EF is to D, so D (is) to BC, and BC to A [Prop. 7.13], and D (is) equal to FL, and BC to FK, and A to FH, thus as EF is to FL, so LF (is) to FK, and FK to FH.
By separation, as EL (is) to LF, so LK (is) to FK, and KH to FH [Prop. 7.11] [Prop. 7.13].
And thus as one of the leading (numbers) is to one of the following, so (the sum of) all of the leading (numbers is) to (the sum of) all of the following [Prop. 7.12].
Thus, as KH is to FH, so EL, LK, KH (are) to LF, FK, HF.
And KH (is) equal to CG, and FH to A, and LF, FK, HF to D, BC, A.
Thus, as CG is to A, so EH (is) to D, BC, A.
Thus, as the excess of the second (number) is to the first, so the excess of the last (is) to (the sum of) all those (numbers) before it.
(Which is) the very thing it was required to show.
Proposition 36
If any multitude whatsoever of numbers is set out continuously in a double proportion, (starting) from a unit, until the whole sum added together becomes prime, and the sum multiplied into the last (number) makes some (number), then the (number so) created will be perfect.
For let any multitude of numbers, A, B, C, D, be set out (continuouly) in a double proportion, until the whole sum added together is made prime.
And let E be equal to the sum.
And let E make FG (by) multiplying D.
I say that FG is a perfect (number).
For as many as is the multitude of A, B, C, D, let so many (numbers), E, HK, L, M, have been taken in a double proportion, (starting) from E.
Thus, via equality, as A is to D, so E (is) to M [Prop. 7.14].
Thus, the (number created) from (multiplying) E, D is equal to the (number created) from (multiplying) A, M.
And FG is the (number created) from (multiplying) E, D.
Thus, FG is also the (number created) from (multiplying) A, M [Prop. 7.19].
Thus, A has made FG (by) multiplying M.
Thus, M measures FG according to the units in A.
And A is a dyad.
Thus, FG is double M.
And M, L, HK, E are also continuously double one another.
Thus, E, HK, L, M, FG are continuously proportional in a double proportion.
So let HN and FO, each equal to the first (number) E, have been subtracted from the second (number) HK and the last FG (respectively).
Thus, as the excess of the second number is to the first, so the excess of the last (is) to (the sum of) all those (numbers) before it [Prop. 9.35].
Thus, as NK is to E, so OG (is) to M, L, KH, E.
And NK is equal to E.
And thus OG is equal to M, L, HK, E.
And FO is also equal to E, and E to A, B, C, D, and a unit.
Thus, the whole of FG is equal to E, HK, L, M, and A, B, C, D, and a unit.
And it is measured by them.
I also say that FG will be measured by no other (numbers) except A, B, C, D, E, HK, L, M, and a unit.
For, if possible, let some (number) P measure FG, and let P not be the same as any of A, B, C, D, E, HK, L, M.
And as many times as P measures FG, so many units let there be in Q.
Thus, Q has made FG (by) multiplying P.
But, in fact, E has also made FG (by) multiplying D.
Thus, as E is to Q, so P (is) to D [Prop. 7.19].
And since A, B, C, D are continually proportional, (starting) from a unit, D will thus not be measured by any other numbers except A, B, C [Prop. 9.13].
And P was assumed not (to be) the same as any of A, B, C.
Thus, P does not measure D.
But, as P (is) to D, so E (is) to Q.
Thus, E does not measure Q either [Def. 7.20].
And E is a prime (number).
And every prime number [is] prime to every (number) which it does not measure [Prop. 7.29].
Thus, E and Q are prime to one another.
And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio as them an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
And as E is to Q, (so) P (is) to D.
Thus, E measures P the same number of times as Q (measures) D.
And D is not measured by any other (numbers) except A, B, C.
Thus, Q is the same as one of A, B, C.
Let it be the same as B.
And as many as is the multitude of B, C, D, let so many (of the set out numbers) have been taken, (starting) from E, (namely) E, HK, L.
And E, HK, L are in the same ratio as B, C, D.
Thus, via equality, as B (is) to D, (so) E (is) to L [Prop. 7.14].
Thus, the (number created) from (multiplying) B, L is equal to the (number created) from multiplying D, E [Prop. 7.19].
But, the (number created) from (multiplying) D, E is equal to the (number created) from (multiplying) Q, P.
Thus, the (number created) from (multiplying) Q, P is equal to the (number created) from (multiplying) B, L.
Thus, as Q is to B, (so) L (is) to P [Prop. 7.19].
And Q is the same as B.
Thus, L is also the same as P.
The very thing (is) impossible.
For P was assumed not (to be) the same as any of the (numbers) set out.
Thus, FG cannot be measured by any number except A, B, C, D, E, HK, L, M, and a unit.
And FG was shown (to be) equal to (the sumof) A, B, C, D, E, HK, L, M, and a unit.
Anda perfect number is one which is equal to (the sum of) its own parts [Def. 7.22].
Thus, FG is a perfect (number).
(Which is) the very thing it was required to show.