Page 290 - Euclid's Elements of Geometry

ST EW iþ.

ELEMENTS BOOK 10

᾿Επεὶ γὰρ σύμμετρός ἐστιν ἡ Α τῇ Β μήκει, ἡ Α ἄρα πρὸς For since A is commensurable in length with B, A
τὴν Β λόγον ἔχει, ὃν ἀριθμὸς πρὸς ἀριθμόν. ἐχέτω, ὃν ὁ thus has to B the ratio which (some) number (has) to
Γ πρὸς τὸν Δ. ἐπεὶ οὖν ἐστιν ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ (some) number [Prop. 10.5]. Let it have (that) which
Γ πρὸς τὸν Δ, ἀλλὰ τοῦ μὲν τῆς Α πρὸς τὴν Β λόγου δι- C (has) to D. Therefore, since as A is to B, so C (is)
πλασίων ἐστὶν ὁ τοῦ ἀπὸ τῆς Α τετραγώνου πρὸς τὸ ἀπὸ τῆς to D. But the (ratio) of the square on A to the square
Β τετράγωνον· τὰ γὰρ ὅμοια σχήματα ἐν διπλασίονι λόγῳ on B is the square of the ratio of A to B. For similar
ἐστὶ τῶν ὁμολόγων πλευρῶν· τοῦ δὲ τοῦ Γ [ἀριθμοῦ] πρὸς figures are in the squared ratio of (their) corresponding
τὸν Δ [ἀριθμὸν] λόγου διπλασίων ἐστὶν ὁ τοῦ ἀπὸ τοῦ Γ sides [Prop. 6.20 corr.]. And the (ratio) of the square
τετραγώνου πρὸς τὸν ἀπὸ τοῦ Δ τετράγωνον· δύο γὰρ τε- on C to the square on D is the square of the ratio of
τραγώνων ἀριθμῶν εἷς μέσος ἀνάλογόν ἐστιν ἀριθμός, καί the [number] C to the [number] D. For there exits one
ὁ τετράγωνος πρὸς τὸν τετράγωνον [ἀριθμὸν] διπλασίονα number in mean proportion to two square numbers, and
λόγον ἔχει, ἤπερ ἡ πλευρὰ πρὸς τὴν πλευράν· ἔστιν ἄρα (one) square (number) has to the (other) square [num-
καὶ ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β ber] a squared ratio with respect to (that) the side (of the
τετράγωνον, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος [ἀριθμὸς] πρὸς former has) to the side (of the latter) [Prop. 8.11]. And,
τὸν ἀπὸ τοῦ Δ [ἀριθμοῦ] τετράγωνον [ἀριθμόν]. thus, as the square on A is to the square on B, so the
᾿Αλλὰ δὴ ἔστω ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ square [number] on the (number) C (is) to the square
ἀπὸ τῆς Β, οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος πρὸς τὸν ἀπὸ [number] on the [number] D. †
τοῦ Δ [τετράγωνον]· λέγω, ὅτι σύμμετρός ἐστιν ἡ Α τῇ Β And so let the square on A be to the (square) on B as
μήκει. the square (number) on C (is) to the [square] (number)
᾿Επεὶ γάρ ἐστιν ὡς τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ on D. I say that A is commensurable in length with B.
ἀπὸ τῆς Β [τετράγωνον], οὕτως ὁ ἀπὸ τοῦ Γ τετράγωνος For since as the square on A is to the [square] on B, so
πρὸς τὸν ἀπὸ τοῦ Δ [τετράγωνον], ἀλλ᾿ ὁ μὲν τοῦ ἀπὸ τῆς the square (number) on C (is) to the [square] (number)
Α τετραγώνου πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] λόγος δι- on D. But, the ratio of the square on A to the (square)
πλασίων ἐστὶ τοῦ τῆς Α πρὸς τὴν Β λόγου, ὁ δὲ τοῦ ἀπὸ on B is the square of the (ratio) of A to B [Prop. 6.20
τοῦ Γ [ἀριθμοῦ] τετραγώνου [ἀριθμοῦ] πρὸς τὸν ἀπὸ τοῦ Δ corr.]. And the (ratio) of the square [number] on the
[ἀριθμοῦ] τετράγωνον [ἀριθμὸν] λόγος διπλασίων ἐστὶ τοῦ [number] C to the square [number] on the [number] D is
τοῦ Γ [ἀριθμοῦ] πρὸς τὸν Δ [ἀριθμὸν] λόγου, ἔστιν ἄρα the square of the ratio of the [number] C to the [number]
καὶ ὡς ἡ Α πρὸς τὴν Β, οὕτως ὁ Γ [ἀριθμὸς] πρὸς τὸν Δ D [Prop. 8.11]. Thus, as A is to B, so the [number] C
[ἀριθμόν]. ἡ Α ἄρα πρὸς τὴν Β λόγον ἔχει, ὃν ἀριθμὸς ὁ Γ also (is) to the [number] D. A, thus, has to B the ratio
πρὸς ἀριθμὸν τὸν Δ· σύμμετρος ἄρα ἐστὶν ἡ Α τῇ Β μήκει. which the number C has to the number D. Thus, A is
᾿Αλλὰ δὴ ἀσύμμετρος ἔστω ἡ Α τῇ Β μήκει· λέγω, ὅτι commensurable in length with B [Prop. 10.6]. ‡
τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς Β [τετράγωνον] And so let A be incommensurable in length with B. I
λόγον οὐκ ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον say that the square on A does not have to the [square] on
ἀριθμόν. B the ratio which (some) square number (has) to (some)
Εἰ γὰρ ἔχει τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ square number.
τῆς Β [τετράγωνον] λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς For if the square on A has to the [square] on B the ra-
τετράγωνον ἀριθμόν, σύμμετρος ἔσται ἡ Α τῇ Β. οὐκ ἔστι tio which (some) square number (has) to (some) square
δέ· οὐκ ἄρα τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς number then A will be commensurable (in length) with
Β [τετράγωνον] λόγον ἔχει, ὃν τετράγωνος ἀριθμὸς πρὸς B. But it is not. Thus, the square on A does not have
τετράγωνον ἀριθμόν. to the [square] on the B the ratio which (some) square
Πάλιν δὴ τὸ ἀπὸ τῆς Α τετράγωνον πρὸς τὸ ἀπὸ τῆς number (has) to (some) square number.
Β [τετράγωνον] λόγον μὴ ἐχέτω, ὃν τετράγωνος ἀριθμὸς So, again, let the square on A not have to the [square]
πρὸς τετράγωνον ἀριθμόν· λέγω, ὅτι ἀσύμμετρός ἐστιν ἡ Α on B the ratio which (some) square number (has) to
τῇ Β μήκει. (some) square number. I say that A is incommensurable
Εἰ γάρ ἐστι σύμμετρος ἡ Α τῇ Β, ἕξει τὸ ἀπὸ τῆς Α in length with B.
πρὸς τὸ ἀπὸ τῆς Β λόγον, ὃν τετράγωνος ἀριθμὸς πρὸς For if A is commensurable (in length) with B then
τετράγωνον ἀριθμόν. οὐκ ἔχει δέ· οὐκ ἄρα σύμμετρός ἐστιν the (square) on A will have to the (square) on B the ra-
ἡ Α τῇ Β μήκει. tio which (some) square number (has) to (some) square
Τὰ ἄρα ἀπὸ τῶν μήκει συμμέτρων, καὶ τὰ ἑξῆς. number. But it does not have (such a ratio). Thus, A is
not commensurable in length with B.
Thus, (squares) on (straight-lines which are) com-

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