For two [unequal] numbers, AB and CD, the lesser being continually subtracted, in turn, from the greater, let the remainder never measure the (number) preceding it, until a unit remains. I say that AB and CD are prime to one another---that is to say, that a unit alone measures (both) AB and CD.
For if AB and CD are not prime to one another then some number will measure them. Let (some number) measure them, and let it be E. And let CD measuring BF leave FA less than itself, and let AF measuring DG leave GC less than itself, and let GC measuring FH leave a unit, HA.
In fact, since E measures CD, and CD measures BF, E thus also measures BF. And ( E) also measures the whole of BA. Thus, ( E) will also measure the remainder AF. And AF measures DG. Thus, E also measures DG. And ( E) also measures the whole of DC. Thus, ( E) will also measure the remainder CG. And CG measures FH. Thus, E also measures FH. And ( E) also measures the whole of FA. Thus, ( E) will also measure the remaining unit AH, (despite) being a number. The very thing is impossible. Thus, some number does not measure (both) the numbers AB and CD. Thus, AB and CD are prime to one another. (Which is) the very thing it was required to show.
Let AB and CD be the two given numbers (which are) not prime to one another. So it is required to find the greatest common measure of AB and CD.
In fact, if CD measures AB, CD is thus a common measure of CD and AB, (since CD) also measures itself. And (it is) manifest that (it is) also the greatest (common measure). For nothing greater than CD can measure CD.
But if CD does not measure AB then some number will remain from AB and CD, the lesser being continually subtracted, in turn, from the greater, which will measure the (number) preceding it. For a unit will not be left. But if not, AB and CD will be prime to one another [Prop. 7.1]. The very opposite thing was assumed. Thus, some number will remain which will measure the (number) preceding it. And let CD measuring BE leave EA less than itself, and let EA measuring DF leave FC less than itself, and let CF measure AE. Therefore, since CF measures AE, and AE measures DF, CF will thus also measure DF. And it also measures itself. Thus, it will also measure the whole of CD. And CD measures BE. Thus, CF also measures BE. And it also measures EA. Thus, it will also measure the whole of BA. And it also measures CD. Thus, CF measures (both) AB and CD. Thus, CF is a common measure of AB and CD. So I say that (it is) also the greatest (common measure). For if CF is not the greatest common measure of AB and CD then some number which is greater than CF will measure the numbers AB and CD. Let it (so) measure ( AB and CD), and let it be G. And since G measures CD, and CD measures BE, G thus also measures BE. And it also measures the whole of BA. Thus, it will also measure the remainder AE. And AE measures DF. Thus, G will also measure DF. And it also measures the whole of DC. Thus, it will also measure the remainder CF, the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than CF cannot measure the numbers AB and CD. Thus, CF is the greatest common measure of AB and CD. [(Which is) the very thing it was required to show].
Corollary
So it is manifest, from this, that if a number measures two numbers then it will also measure their greatest common measure. (Which is) the very thing it was required to show.
Let A, B, and C be the three given numbers (which are) not prime to one another. So it is required to find the greatest common measure of A, B, and C.
For let the greatest common measure, D, of the two (numbers) A and B have been taken [Prop. 7.2]. So D either measures, or does not measure, C. First of all, let it measure ( C). And it also measures A and B. Thus, D measures A, B, and C. Thus, D is a common measure of A, B, and C. So I say that (it is) also the greatest (common measure). For if D is not the greatest common measure of A, B, and C then some number greater than D will measure the numbers A, B, and C. Let it (so) measure ( A, B, and C), and let it be E. Therefore, since E measures A, B, and C, it will thus also measure A and B. Thus, it will also measure the greatest common measure of A and B [Prop. 7.2 corr.]. And D is the greatest common measure of A and B. Thus, E measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than D cannot measure the numbers A, B, and C. Thus, D is the greatest common measure of A, B, and C.
So let D not measure C. I say, first of all, that C and D are not prime to one another. For since A, B, C are not prime to one another, some number will measure them. So the (number) measuring A, B, and C will also measure A and B, and it will also measure the greatest common measure, D, of A and B [Prop. 7.2 corr.]. And it also measures C. Thus, some number will measure the numbers D and C. Thus, D and C are not prime to one another. Therefore, let their greatest common measure, E, have been taken [Prop. 7.2]. And since E measures D, and D measures A and B, E thus also measures A and B. And it also measures C. Thus, E measures A, B, and C. Thus, E is a common measure of A, B, and C. So I say that (it is) also the greatest (common measure). For if E is not the greatest common measure of A, B, and C then some number greater than E will measure the numbers A, B, and C. Let it (so) measure ( A, B, and C), and let it be F. And since F measures A, B, and C, it also measures A and B. Thus, it will also measure the greatest common measure of A and B [Prop. 7.2 corr.]. And D is the greatest common measure of A and B. Thus, F measures D. And it also measures C. Thus, F measures D and C. Thus, it will also measure the greatest common measure of D and C [Prop. 7.2 corr.]. And E is the greatest common measure of D and C. Thus, F measures E, the greater (measuring) the lesser. The very thing is impossible. Thus, some number which is greater than E does not measure the numbers A, B, and C. Thus, E is the greatest common measure of A, B, and C. (Which is) the very thing it was required to show.
Let A and BC be two numbers, and let BC be the lesser. I say that BC is either part or parts of A.
For A and BC are either prime to one another, or not. Let A and BC, first of all, be prime to one another. So separating BC into its constituent units, each of the units in BC will be some part of A. Hence, BC is parts of A.
So let A and BC be not prime to one another. So BC either measures, or does not measure, A. Therefore, if BC measures A then BC is part of A. And if not, let the greatest common measure, D, of A and BC have been taken [Prop. 7.2], and let BC have been divided into BE, EF, and FC, equal to D. And since D measures A, D is a part of A. And D is equal to each of BE, EF, and FC. Thus, BE, EF, and FC are also each part of A. Hence, BC is parts of A.
Thus, any number is either part or parts of any (other) number, the lesser of the greater. (Which is) the very thing it was required to show.
For let a number A be part of a [number] BC, and another (number) D (be) the same part of another (number) EF that A (is) of BC. I say that the sum A, D is also the same part of the sum BC, EF that A (is) of BC.
For since which(ever) part A is of BC, D is the same part of EF, thus as many numbers as are in BC equal to A, so many numbers are also in EF equal to D. Let BC have been divided into BG and GC, equal to A, and EF into EH and HF, equal to D. So the multitude of (divisions) BG, GC will be equal to the multitude of (divisions) EH, HF. And since BG is equal to A, and EH to D, thus BG, EH (is) also equal to A, D. So, for the same (reasons), GC, HF (is) also (equal) to A, D. Thus, as many numbers as [are] in BC equal to A, so many are also in BC, EF equal to A, D. Thus, as many times as BC is (divisible) by A, so many times is the sum BC, EF also (divisible) by the sum A, D. Thus, which (ever) part A is of BC, the sum A, D is also the same part of the sum BC, EF. (Which is) the very thing it was required to show.
For let a number AB be parts of a number C, and another (number) DE (be) the same parts of another (number) F that AB (is) of C. I say that the sum AB, DE is also the same parts of the sum C, F that AB (is) of C.
For since which(ever) parts AB is of C, DE (is) also the same parts of F, thus as many parts of C as are in AB, so many parts of F are also in DE. Let AB have been divided into the parts of C, AG and GB, and DE into the parts of F, DH and HE. So the multitude of (divisions) AG, GB will be equal to the multitude of (divisions) DH, HE. And since which(ever) part AG is of C, DH is also the same part of F, thus which(ever) part AG is of C, the sum AG, DH is also the same part of the sum C, F [Prop. 7.5]. And so, for the same (reasons), which(ever) part GB is of C, the sum GB, HE is also the same part of the sum C, F. Thus, which(ever) parts AB is of C, the sum AB, DE is also the same parts of the sum C, F. (Which is) the very thing it was required to show.
For let a number AB be that part of a number CD that a (part) taken away AE (is) of a part taken away CF. I say that the remainder EB is also the same part of the remainder FD that the whole AB (is) of the whole CD.
For which(ever) part AE is of CF, let EB also be the same part of CG. And since which(ever) part AE is of CF, EB is also the same part of CG, thus which(ever) part AE is of CF, AB is also the same part of GF [Prop. 7.5]. And which(ever) part AE is of CF, AB is also assumed (to be) the same part of CD. Thus, also, which(ever) part AB is of GF, ( AB) is also the same part of CD. Thus, GF is equal to CD. Let CF have been subtracted from both. Thus, the remainder GC is equal to the remainder FD. And since which(ever) part AE is of CF, EB [is] also the same part of GC, and GC (is) equal to FD, thus which(ever) part AE is of CF, EB is also the same part of FD. But, which(ever) part AE is of CF, AB is also the same part of CD. Thus, the remainder EB is also the same part of the remainder FD that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show.
For let a number AB be those parts of a number CD that a (part) taken away AE (is) of a (part) taken away CF. I say that the remainder EB is also the same parts of the remainder FD that the whole AB (is) of the whole CD.
For let GH be laid down equal to AB. Thus, which(ever) parts GH is of CD, AE is also the same parts of CF. Let GH have been divided into the parts of CD, GK and KH, and AE into the part of CF, AL and LE. So the multitude of (divisions) GK, KH will be equal to the multitude of (divisions) AL, LE. And since which(ever) part GK is of CD, AL is also the same part of CF, and CD (is) greater than CF, GK (is) thus also greater than AL. Let GM be made equal to AL. Thus, which(ever) part GK is of CD, GM is also the same part of CF. Thus, the remainder MK is also the same part of the remainder FD that the whole GK (is) of the whole CD [Prop. 7.5]. Again, since which(ever) part KH is of CD, EL is also the same part of CF, and CD (is) greater than CF, HK (is) thus also greater than EL. Let KN be made equal to EL. Thus, which(ever) part KH (is) of CD, KN is also the same part of CF. Thus, the remainder NH is also the same part of the remainder FD that the whole KH (is) of the whole CD [Prop. 7.5]. And the remainder MK was also shown to be the same part of the remainder FD that the whole GK (is) of the whole CD. Thus, the sum MK, NH is the same parts of DF that the whole HG (is) of the whole CD. And the sum MK, NH (is) equal to EB, and HG to BA. Thus, the remainder EB is also the same parts of the remainder FD that the whole AB (is) of the whole CD. (Which is) the very thing it was required to show.
For let a number A be part of a number BC, and another (number) D (be) the same part of another EF that A (is) of BC. I say that, also, alternately, which(ever) part, or parts, A is of D, BC is also the same part, or parts, of EF.
For since which(ever) part A is of BC, D is also the same part of EF, thus as many numbers as are in BC equal to A, so many are also in EF equal to D. Let BC have been divided into BG and GC, equal to A, and EF into EH and HF, equal to D. So the multitude of (divisions) BG, GC will be equal to the multitude of (divisions) EH, HF.
And since the numbers BG and GC are equal to one another, and the numbers EH and HF are also equal to one another, and the multitude of (divisions) BG, GC is equal to the multitude of (divisions) EH, HF, thus which(ever) part, or parts, BG is of EH, GC is also the same part, or the same parts, of HF. And hence, which(ever) part, or parts, BG is of EH, the sum BC is also the same part, or the same parts, of the sum EF [Prop. 7.5] [Prop. 7.6]. And BG (is) equal to A, and EH to D. Thus, which(ever) part, or parts, A is of D, BC is also the same part, or the same parts, of EF. (Which is) the very thing it was required to show.
For let a number AB be parts of a number C, and another (number) DE (be) the same parts of another F. I say that, also, alternately, which(ever) parts, or part, AB is of DE, C is also the same parts, or the same part, of F.
For since which(ever) parts AB is of C, DE is also the same parts of F, thus as many parts of C as are in AB, so many parts of F (are) also in DE. Let AB have been divided into the parts of C, AG and GB, and DE into the parts of F, DH and HE. So the multitude of (divisions) AG, GB will be equal to the multitude of (divisions) DH, HE. And since which(ever) part AG is of C, DH is also the same part of F, also, alternately, which(ever) part, or parts, AG is of DH, C is also the same part, or the same parts, of F [Prop. 7.9]. And so, for the same (reasons), which(ever) part, or parts, GB is of HE, C is also the same part, or the same parts, of F [Prop. 7.9]. And so [which(ever) part, or parts, AG is of DH, GB is also the same part, or the same parts, of HE. And thus, which(ever) part, or parts, AG is of DH, AB is also the same part, or the same parts, of DE [Prop. 7.5] [Prop. 7.6]. But, which(ever) part, or parts, AG is of DH, C was also shown (to be) the same part, or the same parts, of F. And, thus] which(ever) parts, or part, AB is of DE, C is also the same parts, or the same part, of F. (Which is) the very thing it was required to show.
Let the whole AB be to the whole CD as the (part) taken away AE (is) to the (part) taken away CF. I say that the remainder EB is to the remainder FD as the whole AB (is) to the whole CD.
(For) since as AB is to CD, so AE (is) to CF, thus which(ever) part, or parts, AB is of CD, AE is also the same part, or the same parts, of CF [Def. 7.20]. Thus, the remainder EB is also the same part, or parts, of the remainder FD that AB (is) of CD [Prop. 7.7] [Prop. 7.8]. Thus, as EB is to FD, so AB (is) to CD [Def. 7.20]. (Which is) the very thing it was required to show.
Let any multitude whatsoever of numbers, A, B, C, D, be proportional, (such that) as A (is) to B, so C (is) to D. I say that as A is to B, so A, C (is) to B, D.
For since as A is to B, so C (is) to D, thus which(ever) part, or parts, A is of B, C is also the same part, or parts, of D [Def. 7.20]. Thus, the sum A, C is also the same part, or the same parts, of the sum B, D that A (is) of B [Prop. 7.5] [Prop. 7.6]. Thus, as A is to B, so A, C (is) to B, D [Def. 7.20]. (Which is) the very thing it was required to show.
Let the four numbers A, B, C, and D be proportional, (such that) as A (is) to B, so C (is) to D. I say that they will also be proportional alternately, (such that) as A (is) to C, so B (is) to D.
For since as A is to B, so C (is) to D, thus which(ever) part, or parts, A is of B, C is also the same part, or the same parts, of D [Def. 7.20]. Thus, alterately, which (ever) part, or parts, A is of C, B is also the same part, or the same parts, of D [Prop. 7.9] [Prop. 7.10]. Thus, as A is to C, so B (is) to D [Def. 7.20]. (Which is) the very thing it was required to show.
Let there be any multitude of numbers whatsoever, A, B, C, and (some) other (numbers), D, E, F, of equal multitude to them, (which are) in the same ratio taken two by two, (such that) as A (is) to B, so D (is) to E, and as B (is) to C, so E (is) to F. I say that also, via equality, as A is to C, so D (is) to F.
For since 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]. Again, since as B is to C, so E (is) to F, thus, alternately, as B is to E, so C (is) to F [Prop. 7.13]. And as B (is) to E, so A (is) to D. Thus, also, as A (is) to D, so C (is) to F. Thus, alternately, as A is to C, so D (is) to F [Prop. 7.13]. (Which is) the very thing it was required to show.
For let a unit A measure some number BC, and let another number D measure some other number EF as many times. I say that, also, alternately, the unit A also measures the number D as many times as BC (measures) EF.
For since the unit A measures the number BC as many times as D (measures) EF, thus as many units as are in BC, so many numbers are also in EF equal to D. Let BC have been divided into its constituent units, BG, GH, and HC, and EF into the (divisions) EK, KL, and LF, equal to D. So the multitude of (units) BG, GH, HC will be equal to the multitude of (divisions) EK, KL, LF. And since the units BG, GH, and HC are equal to one another, and the numbers EK, KL, and LF are also equal to one another, and the multitude of the (units) BG, GH, HC is equal to the multitude of the numbers EK, KL, LF, thus as the unit BG (is) to the number EK, so the unit GH will be to the number KL, and the unit HC to the number LF. And thus, as one of the leading (numbers is) to one of the following, so (the sum of) all of the leading will be to (the sum of) all of the following [Prop. 7.12]. Thus, as the unit BG (is) to the number EK, so BC (is) to EF. And the unit BG (is) equal to the unit A, and the number EK to the number D. Thus, as the unit A is to the number D, so BC (is) to EF. Thus, the unit A measures the number D as many times as BC (measures) EF [Def. 7.20]. (Which is) the very thing it was required to show.
Let A and B be two numbers. And let A make C (by) multiplying B, and let B make D (by) multiplying A. I say that C is equal to D.
For since A has made C (by) multiplying B, B thus measures C according to the units in A [Def. 7.15]. And the unit E also measures the number A according to the units in it. Thus, the unit E measures the number A as many times as B (measures) C. Thus, alternately, the unit E measures the number B as many times as A (measures) C [Prop. 7.15]. Again, since B has made D (by) multiplying A, A thus measures D according to the units in B [Def. 7.15]. And the unit E also measures B according to the units in it. Thus, the unit E measures the number B as many times as A (measures) D. And the unit E was measuring the number B as many times as A (measures) C. Thus, A measures each of C and D an equal number of times. Thus, C is equal to D. (Which is) the very thing it was required to show.
For let the number A make (the numbers) D and E (by) multiplying the two numbers B and C (respectively). I say that as B is to C, so D (is) to E.
For since A has made D (by) multiplying B, B thus measures D according to the units in A [Def. 7.15]. And the unit F also measures the number A according to the units in it. Thus, the unit F measures the number A as many times as B (measures) D. Thus, as the unit F is to the number A, so B (is) to D [Def. 7.20]. And so, for the same (reasons), as the unit F (is) to the number A, so C (is) to E. And thus, as B (is) to D, so C (is) to E. Thus, alternately, as B is to C, so D (is) to E [Prop. 7.13]. (Which is) the very thing it was required to show.
For let the two numbers A and B make (the numbers) D and E (respectively, by) multiplying some number C. I say that as A is to B, so D (is) to E.
For since A has made D (by) multiplying C, C has thus also made D (by) multiplying A [Prop. 7.16]. So, for the same (reasons), C has also made E (by) multiplying B. So the number C has made D and E (by) multiplying the two numbers A and B (respectively). Thus, as A is to B, so D (is) to E [Prop. 7.17]. (Which is) the very thing it was required to show.
Let A, B, C, and D be four proportional numbers, (such that) as A (is) to B, so C (is) to D. And let A make E (by) multiplying D, and let B make F (by) multiplying C. I say that E is equal to F.
For let A make G (by) multiplying C. Therefore, since A has made G (by) multiplying C, and has made E (by) multiplying D, the number A has made G and E by multiplying the two numbers C and D (respectively). Thus, as C is to D, so G (is) to E [Prop. 7.17]. But, as C (is) to D, so A (is) to B. Thus, also, as A (is) to B, so G (is) to E. Again, since A has made G (by) multiplying C, but, in fact, B has also made F (by) multiplying C, the two numbers A and B have made G and F (respectively, by) multiplying some number C. Thus, as A is to B, so G (is) to F [Prop. 7.18]. But, also, as A (is) to B, so G (is) to E. And thus, as G (is) to E, so G (is) to F. Thus, G has the same ratio to each of E and F. Thus, E is equal to F [Prop. 5.9].
So, again, let E be equal to F. I say that as A is to B, so C (is) to D. For, with the same construction, since E is equal to F, thus as G is to E, so G (is) to F [Prop. 5.7]. But, as G (is) to E, so C (is) to D [Prop. 7.17]. And as G (is) to F, so A (is) to B [Prop. 7.18]. And, thus, as A (is) to B, so C (is) to D. (Which is) the very thing it was required to show.
For let CD and EF be the least numbers having the same ratio as A and B (respectively). I say that CD measures A the same number of times as EF (measures) B.
For CD is not parts of A. For, if possible, let it be (parts of A). Thus, EF is also the same parts of B that CD (is) of A [Prop. 7.13] [Def. 7.20]. Thus, as many parts of A as are in CD, so many parts of B are also in EF. Let CD have been divided into the parts of A, CG and GD, and EF into the parts of B, EH and HF. So the multitude of (divisions) CG, GD will be equal to the multitude of (divisions) EH, HF. And since the numbers CG and GD are equal to one another, and the numbers EH and HF are also equal to one another, and the multitude of (divisions) CG, GD is equal to the multitude of (divisions) EH, HF, thus as CG is to EH, so GD (is) to HF. Thus, as one of the leading (numbers is) to one of the following, so will (the sum of) all of the leading (numbers) be to (the sum of) all of the following [Prop. 7.12]. Thus, as CG is to EH, so CD (is) to EF. Thus, CG and EH are in the same ratio as CD and EF, being less than them. The very thing is impossible. For CD and EF were assumed (to be) the least of those (numbers) having the same ratio as them. Thus, CD is not parts of A. Thus, (it is) a part (of A) [Prop. 7.4]. And EF is the same part of B that CD (is) of A [Prop. 7.13] [Def. 7.20]. Thus, CD measures A the same number of times that EF (measures) B. (Which is) the very thing it was required to show.
Let A and B be numbers prime to one another. I say that A and B are the least of those (numbers) having the same ratio as them.
For if not then there will be some numbers less than A and B which are in the same ratio as A and B. Let them be C and D.
Therefore, since the least numbers of those (numbers) having the same ratio measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser---that is to say, the leading (measuring) the leading, and the following the following--- C thus measures A the same number of times that D (measures) B [Prop. 7.20]. So as many times as C measures A, so many units let there be in E. Thus, D also measures B according to the units in E. And since C measures A according to the units in E, E thus also measures A according to the units in C [Prop. 7.16]. So, for the same (reasons), E also measures B according to the units in D [Prop. 7.16]. Thus, E measures A and B, which are prime to one another. The very thing is impossible. Thus, there cannot be any numbers less than A and B which are in the same ratio as A and B. Thus, A and B are the least of those (numbers) having the same ratio as them. (Which is) the very thing it was required to show.
Let A and B be the least numbers of those (numbers) having the same ratio as them. I say that A and B are prime to one another.
For if they are not prime to one another then some number will measure them. Let it (so measure them), and let it be C. And as many times as C measures A, so many units let there be in D. And as many times as C measures B, so many units let there be in E.
Since C measures A according to the units in D, C has thus made A (by) multiplying D [Def. 7.15]. So, for the same (reasons), C has also made B (by) multiplying E. So the number C has made A and B (by) multiplying the two numbers D and E (respectively). Thus, as D is to E, so A (is) to B [Prop. 7.17]. Thus, D and E are in the same ratio as A and B, being less than them. The very thing is impossible. Thus, some number does not measure the numbers A and B. Thus, A and B are prime to one another. (Which is) the very thing it was required to show.
Let A and B be two numbers (which are) prime to one another, and let some number C measure A. I say that C and B are also prime to one another.
For if C and B are not prime to one another then [some] number will measure C and B. Let it (so) measure (them), and let it be D. Since D measures C, and C measures A, D thus also measures A. And ( D) also measures B. Thus, D measures A and B, which are prime to one another. The very thing is impossible. Thus, some number does not measure the numbers C and B. Thus, C and B are prime to one another. (Which is) the very thing it was required to show.
For let A and B be two numbers (which are both) prime to some number C. And let A make D (by) multiplying B. I say that C and D are prime to one another.
For if C and D are not prime to one another then [some] number will measure C and D. Let it (so) measure them, and let it be E. And since C and A are prime to one another, and some number E measures C, A and E are thus prime to one another [Prop. 7.23]. So as many times as E measures D, so many units let there be in F. Thus, F also measures D according to the units in E [Prop. 7.16]. Thus, E has made D (by) multiplying F [Def. 7.15]. But, in fact, A has also made D (by) multiplying B. Thus, the (number created) from (multiplying) E and F is equal to the (number created) from (multiplying) A and B. And if the (rectangle contained) by the (two) outermost is equal to the (rectangle contained) by the middle (two) then the four numbers are proportional [Prop. 6.15]. Thus, as E is to A, so B (is) to F. 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) [Prop. 7.21]. And the least numbers of those (numbers) having the same ratio measure those (numbers) having the same ratio as them an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, E measures B. And it also measures C. Thus, E measures B and C, which are prime to one another. The very thing is impossible. Thus, some number cannot measure the numbers C and D. Thus, C and D are prime to one another. (Which is) the very thing it was required to show.
Let A and B be two numbers (which are) prime to one another. And let A make C (by) multiplying itself. I say that B and C are prime to one another.
For let D be made equal to A. Since A and B are prime to one another, and A (is) equal to D, D and B are thus also prime to one another. Thus, D and A are each prime to B. Thus, the (number) created from (multilying) D and A will also be prime to B [Prop. 7.24]. And C is the number created from (multiplying) D and A. Thus, C and B are prime to one another. (Which is) the very thing it was required to show.
For let two numbers, A and B, both be prime to each of two numbers, C and D. And let A make E (by) multiplying B, and let C make F (by) multiplying D. I say that E and F are prime to one another.
For since A and B are each prime to C, the (number) created from (multiplying) A and B will thus also be prime to C [Prop. 7.24]. And E is the (number) created from (multiplying) A and B. Thus, E and C are prime to one another. So, for the same (reasons), E and D are also prime to one another. Thus, C and D are each prime to E. Thus, the (number) created from (multiplying) C and D will also be prime to E [Prop. 7.24]. And F is the (number) created from (multiplying) C and D. Thus, E and F are prime to one another. (Which is) the very thing it was required to show.
Let A and B be two numbers prime to one another, and let A make C (by) multiplying itself, and let it make D (by) multiplying C. And let B make E (by) multiplying itself, and let it make F by multiplying E. I say that C and E, and D and F, are prime to one another.
For since A and B are prime to one another, and A has made C (by) multiplying itself, C and B are thus prime to one another [Prop. 7.25]. Therefore, since C and B are prime to one another, and B has made E (by) multiplying itself, C and E are thus prime to one another [Prop. 7.25]. Again, since A and B are prime to one another, and B has made E (by) multiplying itself, A and E are thus prime to one another [Prop. 7.25]. Therefore, since the two numbers A and C are both prime to each of the two numbers B and E, the (number) created from (multiplying) A and C is thus prime to the (number created) from (multiplying) B and E [Prop. 7.26]. And D is the (number created) from (multiplying) A and C, and F the (number created) from (multiplying) B and E. Thus, D and F are prime to one another. (Which is) the very thing it was required to show.
For let the two numbers, AB and BC, (which are) prime to one another, be laid down together. I say that their sum AC is also prime to each of AB and BC.
For if CA and AB are not prime to one another then some number will measure CA and AB. Let it (so) measure (them), and let it be D. Therefore, since D measures CA and AB, it will thus also measure the remainder BC. And it also measures BA. Thus, D measures AB and BC, which are prime to one another. The very thing is impossible. Thus, some number cannot measure (both) the numbers CA and AB. Thus, CA and AB are prime to one another. So, for the same (reasons), AC and CB are also prime to one another. Thus, CA is prime to each of AB and BC.
So, again, let CA and AB be prime to one another. I say that AB and BC are also prime to one another.
For if AB and BC are not prime to one another then some number will measure AB and BC. Let it (so) measure (them), and let it be D. And since D measures each of AB and BC, it will thus also measure the whole of CA. And it also measures AB. Thus, D measures CA and AB, which are prime to one another. The very thing is impossible. Thus, some number cannot measure (both) the numbers AB and BC. Thus, AB and BC are prime to one another. (Which is) the very thing it was required to show.
Let A be a prime number, and let it not measure B. I say that B and A are prime to one another. For if B and A are not prime to one another then some number will measure them. Let C measure (them). Since C measures B, and A does not measure B, C is thus not the same as A. And since C measures B and A, it thus also measures A, which is prime, (despite) not being the same as it. The very thing is impossible. Thus, some number cannot measure (both) B and A. Thus, A and B are prime to one another. (Which is) the very thing it was required to show.
For let two numbers A and B make C (by) multiplying one another, and let some prime number D measure C. I say that D measures one of A and B.
For let it not measure A. And since D is prime, A and D are thus prime to one another [Prop. 7.29]. And as many times as D measures C, so many units let there be in E. Therefore, since D measures C according to the units E, D has thus made C (by) multiplying E [Def. 7.15]. But, in fact, A has also made C (by) multiplying B. Thus, the (number created) from (multiplying) D and E is equal to the (number created) from (multiplying) A and B. Thus, as D is to A, so B (is) to E [Prop. 7.19]. And D and A (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser—that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20]. Thus, D measures B. So, similarly, we can also show that if ( D) does not measure B then it will measure A. Thus, D measures one of A and B. (Which is) the very thing it was required to show.
Let A be a composite number. I say that A is measured by some prime number.
For since A is composite, some number will measure it. Let it (so) measure ( A), and let it be B. And if B is prime then that which was prescribed has happened. And if ( B is) composite then some number will measure it. Let it (so) measure ( B), and let it be C. And since C measures B, and B measures A, C thus also measures A. And if C is prime then that which was prescribed has happened. And if ( C is) composite then some number will measure it. So, in this manner of continued investigation, some prime number will be found which will measure (the number preceding it, which will also measure A). And if (such a number) cannot be found then an infinite (series of) numbers, each of which is less than the preceding, will measure the number A. The very thing is impossible for numbers. Thus, some prime number will (eventually) be found which will measure the (number) preceding it, which will also measure A.
Thus, every composite number is measured by some prime number. (Which is) the very thing it was required to show.
Let A be a number. I say that A is either prime or is measured by some prime number.
In fact, if A is prime then that which was prescribed has happened. And if (it is) composite then some prime number will measure it [Prop. 7.31].
Thus, every number is either prime or is measured by some prime number. (Which is) the very thing it was required to show.
Let A, B, and C be any given multitude of numbers. So it is required to find the least of those (numbers) having the same ratio as A, B, and C.
For A, B, and C are either prime to one another, or not. In fact, if A, B, and C are prime to one another then they are the least of those (numbers) having the same ratio as them [Prop. 7.22].
And if not, let the greatest common measure, D, of A, B, and C have be taken [Prop. 7.3]. And as many times as D measures A, B, C, so many units let there be in E, F, G, respectively. And thus E, F, G measure A, B, C, respectively, according to the units in D [Prop. 7.15]. Thus, E, F, G measure A, B, C (respectively) an equal number of times. Thus, E, F, G are in the same ratio as A, B, C (respectively) [Def. 7.20]. So I say that (they are) also the least (of those numbers having the same ratio as A, B, C). For if E, F, G are not the least of those (numbers) having the same ratio as A, B, C (respectively), then there will be [some] numbers less than E, F, G which are in the same ratio as A, B, C (respectively). Let them be H, K, L. Thus, H measures A the same number of times that K, L also measure B, C, respectively. And as many times as H measures A, so many units let there be in M. Thus, K, L measure B, C, respectively, according to the units in M. And since H measures A according to the units in M, M thus also measures A according to the units in H [Prop. 7.15]. So, for the same (reasons), M also measures B, C according to the units in K, L, respectively. Thus, M measures A, B, and C. And since H measures A according to the units in M, H has thus made A (by) multiplying M. So, for the same (reasons), E has also made A (by) multiplying D. Thus, the (number created) from (multiplying) E and D is equal to the (number created) from (multiplying) H and M. Thus, as E (is) to H, so M (is) to D [Prop. 7.19]. And E (is) greater than H. Thus, M (is) also greater than D [Prop. 5.13]. And ( M) measures A, B, and C. The very thing is impossible. For D was assumed (to be) the greatest common measure of A, B, and C. Thus, there cannot be any numbers less than E, F, G which are in the same ratio as A, B, C (respectively). Thus, E, F, G are the least of (those numbers) having the same ratio as A, B, C (respectively). (Which is) the very thing it was required to show.
Let A and B be the two given numbers. So it is required to find the least number which they (both) measure.
For A and B are either prime to one another, or not. Let them, first of all, be prime to one another. And let A make C (by) multiplying B. Thus, B has also made C (by) multiplying A [Prop. 7.16]. Thus, A and B (both) measure C. So I say that ( C) is also the least (number which they both measure). For if not, A and B will (both) measure some (other) number which is less than C. Let them (both) measure D (which is less than C). And as many times as A measures D, so many units let there be in E. And as many times as B measures D, so many units let there be in F. Thus, A has made D (by) multiplying E, and B has made D (by) multiplying F. Thus, the (number created) from (multiplying) A and E is equal to the (number created) from (multiplying) B and F. Thus, as A (is) to B, so F (is) to E [Prop. 7.19]. And A and B are prime (to one another), and prime (numbers) are the least (of those numbers having the same ratio) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20]. Thus, B measures E, as the following (number measuring) the following. And since A has made C and D (by) multiplying B and E (respectively), thus as B is to E, so C (is) to D [Prop. 7.17]. And B measures E. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) measure some number which is less than C. Thus, C is the least (number) which is measured by (both) A and B.
So let A and B be not prime to one another. And let the least numbers, F and E, have been taken having the same ratio as A and B (respectively) [Prop. 7.33]. Thus, the (number created) from (multiplying) A and E is equal to the (number created) from (multiplying) B and F [Prop. 7.19]. And let A make C (by) multiplying E. Thus, B has also made C (by) multiplying F. Thus, A and B (both) measure C. So I say that ( C) is also the least (number which they both measure). For if not, A and B will (both) measure some number which is less than C. Let them (both) measure D (which is less than C). And as many times as A measures D, so many units let there be in G. And as many times as B measures D, so many units let there be in H. Thus, A has made D (by) multiplying G, and B has made D (by) multiplying H. Thus, the (number created) from (multiplying) A and G is equal to the (number created) from (multiplying) B and H. Thus, as A is to B, so H (is) to G [Prop. 7.19]. And as A (is) to B, so F (is) to E. Thus, also, as F (is) to E, so H (is) to G. And F and E are the least (numbers having the same ratio as A and B), and the least (numbers) measure those (numbers) having the same ratio an equal number of times, the greater (measuring) the greater, and the lesser the lesser [Prop. 7.20]. Thus, E measures G. And since A has made C and D (by) multiplying E and G (respectively), thus as E is to G, so C (is) to D [Prop. 7.17]. And E measures G. Thus, C also measures D, the greater (measuring) the lesser. The very thing is impossible. Thus, A and B do not (both) measure some (number) which is less than C. Thus, C (is) the least (number) which is measured by (both) A and B. (Which is) the very thing it was required to show.
For let two numbers, A and B, (both) measure some number CD, and (let) E (be the) least (number measured by both A and B). I say that E also measures CD.
For if E does not measure CD then let E leave CF less than itself (in) measuring DF. And since A and B (both) measure E, and E measures DF, A and B will thus also measure DF. And ( A and B) also measure the whole of CD. Thus, they will also measure the remainder CF, which is less than E. The very thing is impossible. Thus, E cannot not measure CD. Thus, ( E) measures ( CD). (Which is) the very thing it was required to show.
Let A, B, and C be the three given numbers. So it is required to find the least number which they (all) measure.
For let the least (number), D, measured by the two (numbers) A and B have been taken [Prop. 7.34]. So C either measures, or does not measure, D. Let it, first of all, measure ( D). And A and B also measure D. Thus, A, B, and C (all) measure D. So I say that ( D is) also the least (number measured by A, B, and C). For if not, A, B, and C will (all) measure [some] number which is less than D. Let them measure E (which is less than D). Since A, B, and C (all) measure E then A and B thus also measure E. Thus, the least (number) measured by A and B will also measure [ E ] [Prop. 7.35]. And D is the least (number) measured by A and B. Thus, D will measure E, the greater (measuring) the lesser. The very thing is impossible. Thus, A, B, and C cannot (all) measure some number which is less than D. Thus, A, B, and C (all) measure the least (number) D.
So, again, let C not measure D. And let the least number, E, measured by C and D have been taken [Prop. 7.34]. Since A and B measure D, and D measures E, A and B thus also measure E. And C also measures [ E ]. Thus, A, B, and C [also] measure E. So I say that ( E is) also the least (number measured by A, B, and C). For if not, A, B, and C will (all) measure some (number) which is less than E. Let them measure F (which is less than E). Since A, B, and C (all) measure F, A and B thus also measure F. Thus, the least (number) measured by A and B will also measure F [Prop. 7.35]. And D is the least (number) measured by A and B. Thus, D measures F. And C also measures F. Thus, D and C (both) measure F. Hence, the least (number) measured by D and C will also measure F [Prop. 7.35]. And E is the least (number) measured by C and D. Thus, E measures F, the greater (measuring) the lesser. The very thing is impossible. Thus, A, B, and C cannot measure some number which is less than E. Thus, E (is) the least (number) which is measured by A, B, and C. (Which is) the very thing it was required to show.
For let the number A be measured by some number B. I say that A has a part called the same as B.
For as many times as B measures A, so many units let there be in C. Since B measures A according to the units in C, and the unit D also measures C according to the units in it, the unit D thus measures the number C as many times as B (measures) A. Thus, alternately, the unit D measures the number B as many times as C (measures) A [Prop. 7.15]. Thus, which(ever) part the unit D is of the number B, C is also the same part of A. And the unit D is a part of the number B called the same as it (i.e., a B th part). Thus, C is also a part of A called the same as B (i.e., C is the B th part of A). Hence, A has a part C which is called the same as B (i.e., A has a B th part). (Which is) the very thing it was required to show.
For let the number A have any part whatever, B. And let the [number] C be called the same as the part B (i.e., B is the C th part of A). I say that C measures A.
For since B is a part of A called the same as C, and the unit D is also a part of C called the same as it (i.e., D is the C th part of C), thus which(ever) part the unit D is of the number C, B is also the same part of A. Thus, the unit D measures the number C as many times as B (measures) A. Thus, alternately, the unit D measures the number B as many times as C (measures) A [Prop. 7.15]. Thus, C measures A. (Which is) the very thing it was required to show.
Let A, B, and C be the given parts. So it is required to find the least number which will have the parts A, B, and C (i.e., an A th part, a B th part, and a C th part).
For let D, E, and F be numbers having the same names as the parts A, B, and C (respectively). And let the least number, G, measured by D, E, and F, have been taken [Prop. 7.36].
Thus, G has parts called the same as D, E, and F [Prop. 7.37]. And A, B, and C are parts called the same as D, E, and F (respectively). Thus, G has the parts A, B, and C. So I say that ( G) is also the least (number having the parts A, B, and C). For if not, there will be some number less than G which will have the parts A, B, and C. Let it be H. Since H has the parts A, B, and C, H will thus be measured by numbers called the same as the parts A, B, and C [Prop. 7.38]. And D, E, and F are numbers called the same as the parts A, B, and C (respectively). Thus, H is measured by D, E, and F. And ( H) is less than G. The very thing is impossible. Thus, there cannot be some number less than G which will have the parts A, B, and C. (Which is) the very thing it was required to show.