§9.prop.33ἐὰν ἀριθμὸς τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις περισσός ἐστι μόνον.
If a number have its half odd, it is even-times odd only.
ἀριθμὸς γὰρ ὁ Α τὸν ἥμισυν ἐχέτω περισσόν· λέγω, ὅτι ὁ Α ἀρτιάκις περισσός ἐστι μόνον.
For let the number A have its half odd; I say that A is even-times odd only.
ὅτι μὲν οὖν ἀρτιάκις περισσός ἐστιν, φανερόν· ὁ γὰρ ἥμισυς αὐτοῦ περισσὸς ὢν μετρεῖ αὐτὸν ἀρτιάκις.
Now that it is even-times odd is manifest; for its half, being odd, measures it an even number of times.
λέγω δή, ὅτι καὶ μόνον.
I say indeed that it is also only so.
εἰ γὰρ ἔσται ὁ Α καὶ ἀρτιάκις ἄρτιος, μετρηθήσεται ὑπὸ ἀρτίου κατὰ ἄρτιον ἀριθμόν· ὥστε καὶ ὁ ἥμισυς αὐτοῦ μετρηθήσεται ὑπὸ ἀρτίου ἀριθμοῦ περισσὸς ὤν· ὅπερ ἐστὶν ἄτοπον.
For if A is also even-times even, it will be measured by an even number according to an even number; so that its half will also be measured by an even number, though it is odd; which is absurd.
ὁ Α ἄρα ἀρτιάκις περισσός ἐστι μόνον· ὅπερ ἔδει δεῖξαι.
Therefore A is even-times odd only; which it was required to prove.
§9.prop.34ἐὰν ἀριθμὸς μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων ᾖ μήτε τὸν ἥμισυν ἔχῃ περισσόν, ἀρτιάκις τε ἄρτιός ἐστι καὶ ἀρτιάκις περισσός.
If a number neither be one of those doubled from a dyad nor have its half odd, it is both even-times even and even-times odd.
ἀριθμὸς γὰρ ὁ Α μήτε τῶν ἀπὸ δυάδος διπλασιαζομένων ἔστω μήτε τὸν ἥμισυν ἐχέτω περισσόν· λέγω, ὅτι ὁ Α ἀρτιάκις τέ ἐστιν ἄρτιος καὶ ἀρτιάκις περισσός.
For let the number A neither be one of those doubled from a dyad nor have its half odd; I say that A is both even-times even and even-times odd.
ὅτι μὲν οὖν ὁ Α ἀρτιάκις ἐστὶν ἄρτιος, φανερόν· τὸν γὰρ ἥμισυν οὐκ ἔχει περισσόν.
Now that A is even-times even is manifest; for it does not have its half odd.
λέγω δή, ὅτι καὶ ἀρτιάκις περισσός ἐστιν.
I say indeed that it is also even-times odd.
ἐὰν γὰρ τὸν Α τέμνωμεν δίχα καὶ τὸν ἥμισυν αὐτοῦ δίχα καὶ τοῦτο ἀεὶ ποιῶμεν, καταντήσομεν εἴς τινα ἀριθμὸν περισσόν, ὃς μετρήσει τὸν Α κατὰ ἄρτιον ἀριθμόν.
For if we bisect A, and bisect its half, and do this continually, we shall come to some odd number which will measure A according to an even number.
εἰ γὰρ οὔ, καταντήσομεν εἰς δυάδα, καὶ ἔσται ὁ Α τῶν ἀπὸ δυάδος διπλασιαζομένων· ὅπερ οὐχ ὑπόκειται.
For, if not, we shall come to a dyad, and A will be one of those doubled from a dyad; which is contrary to the assumption.
ὥστε ὁ Α ἀρτιάκις περισσός ἐστιν.
So that A is even-times odd.
ἐδείχθη δὲ καὶ ἀρτιάκις ἄρτιος.
And it was also proved even-times even.
ὁ Α ἄρα ἀρτιάκις τε ἄρτιός ἐστι καὶ ἀρτιάκις περισσός· ὅπερ ἔδει δεῖξαι.
Therefore A is both even-times even and even-times odd; which it was required to prove.
§9.prop.35ἐὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, ἀφαιρεθῶσι δὲ ἀπό τε τοῦ δευτέρου καὶ τοῦ ἐσχάτου ἴσοι τῷ πρώτῳ, ἔσται ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας.
If there be as many numbers as we please in continued proportion, and there be subtracted from the second and the last numbers equal to the first, then, as the excess of the second is to the first, so will the excess of the last be to all those before itself.
ἔστωσαν ὁποσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, ΒΓ, Δ, ΕΖ ἀρχόμενοι ἀπὸ ἐλαχίστου τοῦ Α, καὶ ἀφῃρήσθω ἀπὸ τοῦ ΒΓ καὶ τοῦ ΕΖ τῷ Α ἴσος ἑκάτερος τῶν ΒΗ, ΖΘ· λέγω, ὅτι ἐστὶν ὡς ὁ ΗΓ πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς τοὺς Α, ΒΓ, Δ.
κείσθω γὰρ τῷ μὲν ΒΓ ἴσος ὁ ΖΚ, τῷ δὲ Δ ἴσος ὁ ΖΛ. καὶ ἐπεὶ ὁ ΖΚ τῷ ΒΓ ἴσος ἐστίν, ὧν ὁ ΖΘ τῷ ΒΗ ἴσος ἐστίν, λοιπὸς ἄρα ὁ ΘΚ λοιπῷ τῷ ΗΓ ἐστιν ἴσος.
Let there be as many numbers as we please in continued proportion, A, BΓ, Δ, EZ, beginning from the least A, and let there be subtracted from BΓ and EZ each of BH, ZΘ equal to A; I say that, as HΓ is to A, so is EΘ to A, BΓ, Δ. For let ZK be set equal to BΓ, and ZΛ equal to Δ. And since ZK is equal to BΓ, of which ZΘ is equal to BH, therefore the remainder ΘK is equal to the remainder HΓ.
καὶ ἐπεί ἐστιν ὡς ὁ ΕΖ πρὸς τὸν Δ, οὕτως ὁ Δ πρὸς τὸν ΒΓ καὶ ὁ ΒΓ πρὸς τὸν Α, ἴσος δὲ ὁ μὲν Δ τῷ ΖΛ, ὁ δὲ ΒΓ τῷ ΖΚ, ὁ δὲ Α τῷ ΖΘ, ἔστιν ἄρα ὡς ὁ ΕΖ πρὸς τὸν ΖΛ, οὕτως ὁ ΛΖ πρὸς τὸν ΖΚ καὶ ὁ ΖΚ πρὸς τὸν ΖΘ. διελόντι, ὡς ὁ ΕΛ πρὸς τὸν ΛΖ, οὕτως ὁ ΛΚ πρὸς τὸν ΖΚ καὶ ὁ ΚΘ πρὸς τὸν ΖΘ. ἔστιν ἄρα καὶ ὡς εἷς τῶν ἡγουμένων πρὸς ἕνα τῶν ἑπομένων, οὕτως ἅπαντες οἱ ἡγούμενοι πρὸς ἅπαντας τοὺς ἑπομένους· ἔστιν ἄρα ὡς ὁ ΚΘ πρὸς τὸν ΖΘ, οὕτως οἱ ΕΛ, ΛΚ, ΚΘ πρὸς τοὺς ΛΖ, ΖΚ, ΘΖ. ἴσος δὲ ὁ μὲν ΚΘ τῷ ΓΗ, ὁ δὲ ΖΘ τῷ Α, οἱ δὲ ΛΖ, ΖΚ, ΘΖ τοῖς Δ, ΒΓ, Α· ἔστιν ἄρα ὡς ὁ ΓΗ πρὸς τὸν Α, οὕτως ὁ ΕΘ πρὸς τοὺς Δ, ΒΓ, Α. ἔστιν ἄρα ὡς ἡ τοῦ δευτέρου ὑπεροχὴ πρὸς τὸν πρῶτον, οὕτως ἡ τοῦ ἐσχάτου ὑπεροχὴ πρὸς τοὺς πρὸ ἑαυτοῦ πάντας· ὅπερ ἔδει δεῖξαι.
And since, as EZ is to Δ, so is Δ to BΓ, and BΓ to A, while Δ is equal to ZΛ, BΓ to ZK, and A to ZΘ, therefore, as EZ is to ZΛ, so is ΛZ to ZK, and ZK to ZΘ. By separation, as EΛ is to ΛZ, so is ΛK to ZK, and KΘ to ZΘ. Therefore it is also the case that, as one of the antecedents is to one of the consequents, so are all the antecedents to all the consequents; therefore, as KΘ is to ZΘ, so are EΛ, ΛK, KΘ to ΛZ, ZK, ΘZ. But KΘ is equal to ΓH, ZΘ to A, and ΛZ, ZK, ΘZ to Δ, BΓ, A; therefore, as ΓH is to A, so is EΘ to Δ, BΓ, A. Therefore, as the excess of the second is to the first, so is the excess of the last to all those before itself; which it was required to prove.