Euclid · Elements §10.prop1.18

Incommensurability of Segments and Square Differences

Passage 173 of 316 · Greek

Summary

Proposition 18 proves that for two unequal straight lines, applying a parallelogram equal to a quarter of the square of the less to the greater, falling short by a square, divides the greater into incommensurable parts if and only if the square on the greater is greater than that on the less by the square on a line incommensurable with itself. The subsequent lemma clarifies the relationship between commensurability in length/square and rational lines.

§10.prop1.18ἐὰν ὦσι δύο εὐθεῖαι ἄνισοι, τῷ δὲ τετάρτῳ μέρει τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, καὶ εἰς ἀσύμμετρα αὐτὴν διαιρῇ, ἡ μείζων τῆς ἐλάσσονος μεῖζον δυνήσεται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.

If there be two unequal straight lines, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and falling short by a square figure, and if it divide it into parts which are incommensurable in length, the greater will be greater in square than the less by the square on a straight line incommensurable with itself.

καὶ ἐὰν ἡ μείζων τῆς ἐλάσσονος μεῖζον δύνηται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος ἴσον παρὰ τὴν μείζονα παραβληθῇ ἐλλεῖπον εἴδει τετραγώνῳ, εἰς ἀσύμμετρα αὐτὴν διαιρεῖ.

And if the greater be greater in square than the less by the square on a straight line incommensurable with itself, and to the greater there be applied a parallelogram equal to the fourth part of the square on the less and falling short by a square figure, it divides it into parts which are incommensurable in length.

῎ἔστωσαν δύο εὐθεῖαι ἄνισοι αἱ Α, ΒΓ, ὧν μείζων ἡ ΒΓ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς ἐλάσσονος τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔΓ, ἀσύμμετρος δὲ ἔστω ἡ ΒΔ τῇ ΔΓ μήκει·

¦10| Let there be two unequal straight lines A, BΓ, of which BΓ is the greater, and to BΓ let there be applied a parallelogram equal to the fourth part of the square on the less A, falling short by a square figure, and let it be the rectangle contained by BΔ, ΔΓ; and let BΔ be incommensurable in length with ΔΓ.

λέγω, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.

I say that the square on BΓ is greater than the square on A by the square on a straight line incommensurable in length with itself.

τῶν γὰρ αὐτῶν κατασκευασθέντων τῷ πρότερον ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. δεικτέον, ὅτι ἀσύμμετρός ἐστιν ἡ ΒΓ τῇ ΔΖ μήκει.

For, with the same construction as before, we shall similarly show that the square on BΓ is greater than the square on A by the square on ZΔ. It must be shown that BΓ is also incommensurable in length with ΔZ.

ἐπεὶ γὰρ ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει, ἀσύμμετρος ἄρα ἐστὶ καὶ ἡ ΒΓ τῇ ΓΔ μήκει.

For since BΔ is incommensurable in length with ΔΓ, therefore BΓ is also incommensurable in length with ΓΔ.

ἀλλὰ ἡ ΔΓ σύμμετρός ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ· καὶ ἡ ΒΓ ἄρα ἀσύμμετρός ἐστι συναμφοτέραις ταῖς ΒΖ, ΔΓ. ὥστε καὶ λοιπῇ τῇ ΖΔ ἀσύμμετρός ἐστιν ἡ ΒΓ μήκει.

But ΓΔ is commensurable in length with the sum of BZ, ΔΓ; therefore BΓ is also incommensurable with the sum of BZ, ΔΓ. So that BΓ is also incommensurable in length with the remainder ZΔ.

καὶ ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ· ἡ ΒΓ ἄρα τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.

And the square on BΓ is greater than the square on A by the square on ZΔ; therefore the square on BΓ is greater than the square on A by the square on a straight line incommensurable with itself.

δυνάσθω δὴ πάλιν ἡ ΒΓ τῆς Α μεῖζον τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ, τῷ δὲ τετάρτῳ τοῦ ἀπὸ τῆς Α ἴσον παρὰ τὴν ΒΓ παραβεβλήσθω ἐλλεῖπον εἴδει τετραγώνῳ, καὶ ἔστω τὸ ὑπὸ τῶν ΒΔ, ΔΓ. δεικτέον, ὅτι ἀσύμμετρός ἐστιν ἡ ΒΔ τῇ ΔΓ μήκει.

Next, let the square on BΓ be greater than the square on A by the square on a straight line incommensurable with itself, and to BΓ let there be applied a parallelogram equal to the fourth part of the square on A and falling short by a square figure, and let it be the rectangle contained by BΔ, ΔΓ. It must be shown that BΔ is incommensurable in length with ΔΓ.

τῶν γὰρ αὐτῶν κατασκευασθέντων ὁμοίως δείξομεν, ὅτι ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ τῆς ΖΔ. ἀλλὰ ἡ ΒΓ τῆς Α μεῖζον δύναται τῷ ἀπὸ ἀσυμμέτρου ἑαυτῇ.

For, with the same construction, we shall similarly show that the square on BΓ is greater than the square on A by the square on ZΔ.

ἀσύμμετρος ἄρα ἐστὶν ἡ ΒΓ τῇ ΖΔ μήκει·

But the square on BΓ is greater than the square on A by the square on a straight line incommensurable with itself; therefore BΓ is incommensurable in length with ZΔ.

ὥστε καὶ λοιπῇ συναμφοτέρῳ τῇ ΒΖ, ΔΓ ἀσύμμετρός ἐστιν ἡ ΒΓ. ἀλλὰ συναμφότερος ἡ ΒΖ, ΔΓ τῇ ΔΓ σύμμετρός ἐστι μήκει· καὶ ἡ ΒΓ ἄρα τῇ ΔΓ ἀσύμμετρός ἐστι μήκει·

So that BΓ is also incommensurable with the remainder, the sum of BZ, ΔΓ. But the sum of BZ, ΔΓ is commensurable in length with ΔΓ; so that BΓ is also incommensurable in length with ΔΓ.

ὥστε καὶ διελόντι ἡ ΒΔ τῇ ΔΓ ἀσύμμετρός ἐστι μήκει.

And therefore, by separation, BΔ is incommensurable in length with ΔΓ.

ἐὰν ἄρα ὦσι δύο εὐθεῖαι, καὶ τὰ ἑξῆς.

Therefore, if there be two straight lines, and the rest.

λῆμμα ἐπεὶ δέδεικται, ὅτι αἱ μήκει σύμμετροι πάντως καὶ δυνάμει, αἱ δὲ δυνάμει οὐ πάντως καὶ μήκει, ἀλλὰ δὴ δύνανται μήκει καὶ σύμμετροι εἶναι καὶ ἀσύμμετροι, φανερόν, ὅτι, ἐὰν τῇ ἐκκειμένῃ ῥητῇ σύμμετρός τις ᾖ μήκει, λέγεται ῥητὴ καὶ σύμμετρος αὐτῇ οὐ μόνον μήκει, ἀλλὰ καὶ δυνάμει, ἐπεὶ αἱ μήκει σύμμετροι πάντως καὶ δυνάμει.

Lemma Since it has been proved that straight lines commensurable in length are always commensurable in square as well, but those commensurable in square are not always commensurable in length, but can be either commensurable or incommensurable in length, it is manifest that, if a certain straight line be commensurable in length with the set-out rational straight line, it is called rational and commensurable with it not only in length but also in square, since straight lines commensurable in length are always commensurable in square as well.

ἐὰν δὲ τῇ ἐκκειμένῃ ῥητῇ σύμμετρός τις ᾖ δυνάμει, εἰ μὲν καὶ μήκει, λέγεται καὶ οὕτως ῥητὴ καὶ σύμμετρος αὐτῇ μήκει καὶ δυνάμει·

And if to the set-out rational straight line there be commensurable in square a certain straight line, if indeed it be so in length also, it is called in this case also rational and commensurable with it in length and in square.

εἰ δὲ τῇ ἐκκειμένῃ πάλιν ῥητῇ σύμμετρός τις οὖσα δυνάμει μήκει αὐτῇ ᾖ ἀσύμμετρος, λέγεται καὶ οὕτως ῥητὴ δυνάμει μόνον σύμμετρος.

But if to the set-out rational straight line a certain straight line being commensurable in square be incommensurable in length with it, it is called in this case also rational, commensurable in square only.

Notes

¦15¦τὸ ὑπὸ τῶν ΒΔΓ — In the manuscript, where we would normally expect the expression `τὸ ὑπὸ τῶν ΒΔ, ΔΓ` (the rectangle contained by BΔ and ΔΓ) listing two genitive nouns, it is written elliptically as `τὸ ὑπὸ τῶν ΒΔΓ`. Semantically, it refers to the rectangle contained by the two segments BΔ and ΔΓ resulting from the division at point Δ.
¦5¦μεῖζον δυνήσεται — The verb `δύναμαι` is used in mathematical texts to mean "to be equal to in square" or "to be greater in square". Here it indicates that the square on the greater line is greater than the square on the less line by the square on a straight line incommensurable with the greater.
¦45¦αἱ μήκει σύμμετροι πάντως καὶ δυνάμει — The phrase functions as the subject clause of the preceding `δέδεικται ὅτι`. It refers to the fundamental theorem of Book X (e.g., Book X, Prop. 10) stating that straight lines commensurable in length are always commensurable in square as well.

Cite this passage

Euclid, Elements §10.prop1.18. Humanitext Reader, https://reader.humanitext.ai/en/text/urn:cts:greekLit:tlg1799.tlg001.humanitext-grc2:10.prop1.18

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