Part 6

# Unequal Lines

by Euclid

## Proposition 18

If there are two unequal straight-lines, and a (rect- angle) equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) incom- mensurable [in length], then the square on the greater will be larger than the (square on the) lesser by the (square) on (some straight-line) incommensurable (in length) with the greater.

If the square on the greater is larger than the (square on the) lesser by the (square) on (some straight-line) incommensurable (in) length) with the greater, and a (rectangle) equal to the fourth (part) of the (square) on the lesser, falling short by a square figure, is applied to the greater, then it divides it into (parts which are) incommensurable [in length].

Let A and BC be two unequal straight-lines, of which (let) BC (be) the greater. And let a (rectangle) equal to the fourth [part] of the (square) on the lesser, A, falling short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BDC. And let BD be incommensurable in length with DC. I say that that the square on BC is greater than the (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC).

A B FE D C For, similarly, by the same construction as before, we can show that the square on BC is greater than the (square on) A by the (square) on FD. [Therefore] it must be shown that BC is incommensurable in length with DF. For since BD is incommensurable in length with DC, BC is thus also incommensurable in length with CD [Prop. 10.16]. But, DC is commensurable (in length) with the sum of BF and DC [Prop. 10.6]. And, thus, BC is incommensurable (in length) with the sum of BF and DC [Prop. 10.13]. Hence, BC is also incommen- surable in length with the remainder FD [Prop. 10.16]. And the square on BC is greater than the (square on) A by the (square) on FD. Thus, the square on BC is greater than the (square on) A by the (square) on (some straight-line) incommensurable (in length) with (BC). So, again, let the square on BC be greater than the (square on) A by the (square) on (some straight-line) in- commensurable (in length) with (BC). And let a (rect- angle) equal to the fourth [part] of the (square) on A, falling short by a square figure, have been applied to BC. And let it be the (rectangle contained) by BD and DC. It must be shown that BD is incommensurable in length with DC. For, similarly, by the same construction, we can show that the square on BC is greater than the (square) on A by the (square) on FD. But, the square on BC is greater than the (square) on A by the (square) on (some straight-line) incommensurable (in length) with (BC). Thus, BC is incommensurable in length with FD. Hence, BC is also incommensurable (in length) with the re- maining sum of BF and DC [Prop. 10.16]. But, the sum of BF and DC is commensurable in length with DC [Prop. 10.6]. Thus, BC is also incommensurable in length with DC (Prop. 10.13]. Hence, via separa- tion, BD is also incommensurable in length with DC [Prop. 10.16]. Thus, if there are two… straight-lines, and so on….

• DC, and ẞ = A) then a and a-ase incommensurable when Proposition 19 The rectangle contained by rational straight-lines (which are) commensurable in length is rational. For let the rectangle AC have been enclosed by the rational straight-lines AB and BC (which are) commen- surable in length. I say that AC is rational. D C B For let the square AD have been described on AB. AD is thus rational [Def. 10.4]. And since AB is com- mensurable in length with BC, and AB is equal to BD, BD is thus commensurable in length with BC. And as BD is to BC, so DA (is) to AC [Prop. 6.1]. Thus, DA is commensurable with AC [Prop. 10.11]. And DA (is) rational. Thus, AC is also rational [Def. 10.4]. Thus, the… by rational straight-lines…commensurable, and so on….

## Proposition 20

If a rational (area) is applied to a rational (straight- line) then it produces as breadth a (straight-line which is) rational, and commensurable in length with the (straight- line) to which it is applied. B A For let the rational (area) AC have been applied to the rational (straight-line) AB, producing the (straight-line) BC as breadth. I say that BC is rational, and commen- surable in length with BA. For let the square AD have been described on AB.

AD is thus rational [Def. 10.4]. And AC (is) also ratio- nal. DA is thus commensurable with AC. And as DA is to AC, so DB (is) to BC [Prop. 6.1]. Thus, DB is also commensurable (in length) with BC [Prop. 10.11]. And DB (is) equal to BA. Thus, AB (is) also commen- surable (in length) with BC. And AB is rational. Thus, BC is also rational, and commensurable in length with AB [Def. 10.3]. Thus, if a rational (area) is applied to a rational (straight-line), and so on…. Proposition 21 The rectangle contained by rational straight-lines (which are) commensurable in square only is irrational, and its square-root is irrational-let it be called medial. B A For let the rectangle AC be contained by the rational straight-lines AB and BC (which are) commensurable in square only. I say that AC is irrational, and its square- root is irrational-let it be called medial. For let the square AD have been described on AB. AD is thus rational [Def. 10.4]. And since AB is incom- mensurable in length with BC. For they were assumed to be commensurable in square only. And AB (is) equal to BD. DB is thus also incommensurable in length with BC. And as DB is to BC, so AD (is) to AC [Prop. 6.1]. Thus, DA [is] incommensurable with AC [Prop. 10.11]. And DA (is) rational. Thus, AC is irrational [Def. 10.4]. Hence, its square-root [that is to say, the square-root of the square equal to it] is also irrational [Def. 10.4]-let it be called medial. (Which is) the very thing it was required to show.

### Lemma

If there are two straight-lines then as the first is to the second, so the (square) on the first (is) to the (rectangle contained) by the two straight-lines. Let FE and EG be two straight-lines. I say that as FE is to EG, so the (square) on FE (is) to the (rectangle contained) by FE and EG. For let the square DF have been described on FE. And let GD have been completed. Therefore, since as FE is to EG, so FD (is) to DG [Prop. 6.1], and FD is the (square) on FE, and DG the (rectangle contained) by DE and EG-that is to say, the (rectangle contained) by FE and EG-thus as FE is to EG, so the (square) on FE (is) to the (rectangle contained) by FE and EG. And also, similarly, as the (rectangle contained) by GE and EF is to the (square on) EF-that is to say, as GD (is) to FD-so GE (is) to EF. (Which is) the very thing it was required to show.

## Proposition 22

The square on a medial (straight-line), being ap- plied to a rational (straight-line), produces as breadth a (straight-line which is) rational, and incommensurable in length with the (straight-line) to which it is applied.

A C D E F

Let A be a medial (straight-line), and CB a rational (straight-line), and let the rectangular area BD, equal to the (square) on A, have been applied to BC, producing CD as breadth. I say that CD is rational, and incommen- surable in length with CB. For since A is medial, the square on it is equal to a

(rectangular) area contained by rational (straight-lines which are) commensurable in square only [Prop. 10.21]. Let the square on (A) be equal to GF. And the square on (A) is also equal to BD. Thus, BD is equal to GF. And (BD) is also equiangular with (GF). And for equal and equiangular parallelograms, the sides about the equal an- gles are reciprocally proportional [Prop. 6.14]. Thus, pro- portionally, as BC is to EG, so EF (is) to CD. And, also, as the (square) on BC is to the (square) on EG, so the (square) on EF (is) to the (square) on CD [Prop. 6.22]. And the (square) on CB is commensurable with the (square) on EG. For they are each rational. Thus, the (square) on EF is also commensurable with the (square) on CD [Prop. 10.11]. And the (square) on EF is ratio- nal. Thus, the (square) on CD is also rational [Def. 10.4]. Thus, CD is rational. And since EF is incommensurable in length with EG. For they are commensurable in square only. And as EF (is) to EG, so the (square) on EF (is) to the (rectangle contained) by FE and EG [see previ- ous lemma]. The (square) on EF [is] thus incommen- surable with the (rectangle contained) by FE and EG [Prop. 10.11]. But, the (square) on CD is commensu- rable with the (square) on EF. For they are rational in square. And the (rectangle contained) by DC and CB is commensurable with the (rectangle contained) by FE and EG. For they are (both) equal to the (square) on A. Thus, the (square) on CD is also incommensurable with the (rectangle contained) by DC and CB [Prop. 10.13]. And as the (square) on CD (is) to the (rectangle con- tained) by DC and CB, so DC is to CB [see previous lemma]. Thus, DC is incommensurable in length with CB [Prop. 10.11]. Thus, CD is rational, and incommen- surable in length with CB. (Which is) the very thing it was required to show.

## Proposition 23

A (straight-line) commensurable with a medial (straight- line) is medial. Let A be a medial (straight-line), and let B be com- mensurable with A. I say that B is also a medial (staight- line). Let the rational (straight-line) CD be set out, and let the rectangular area CE, equal to the (square) on A, have been applied to CD, producing ED as width. ED is thus rational, and incommensurable in length with CD [Prop. 10.22]. And let the rectangular area CF, equal to the (square) on B, have been applied to CD, produc- ing DF as width. Therefore, since A is commensurable with B, the (square) on A is also commensurable with the (square) on B. But, EC is equal to the (square) on A, and CF is equal to the (square) on B. Thus, EC is com- mensurable with CF. And as EC is to CF, so ED (is) to DF [Prop. 6.1]. Thus, ED is commensurable in length with DF [Prop. 10.11]. And ED is rational, and incom- mensurable in length with CD. DF is thus also ratio- nal [Def. 10.3], and incommensurable in length with DC [Prop. 10.13]. Thus, CD and DF are rational, and com- mensurable in square only. And the square-root of a (rect- angle contained) by rational (straight-lines which are) commensurable in square only is medial [Prop. 10.21]. Thus, the square-root of the (rectangle contained) by CD and DF is medial. And the square on B is equal to the (rectangle contained) by CD and DF. Thus, B is a me- dial (straight-line). A B E D F Corollary And (it is) clear, from this, that an (area) commensu- rable with a medial area is medial. me, a medial area is expressible as 1

## Proposition 24

A rectangle contained by medial straight-lines (which are) commensurable in length is medial. For let the rectangle AC be contained by the medial straight-lines AB and BC (which are) commensurable in length. I say that AC is medial.

For let the square AD have been described on AB. AD is thus medial [see previous footnote]. And since AB is commensurable in length with BC, and AB (is) equal to BD, DB is thus also commensurable in length with BC. Hence, DA is also commensurable with AC [Props. 6.1, 10.11]. And DA (is) medial. Thus, AC (is) also medial [Prop. 10.23 corr.]. (Which is) the very thing it was required to show.

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