§9.prop.16ἐὰν δύο ἀριθμοὶ πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται ὡς ὁ πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ δεύτερος πρὸς ἄλλον τινά.
If two numbers be prime to each other, it will not be that as the first is to the second, so is the second to some other.
δύο γὰρ ἀριθμοὶ οἱ Α, Β πρῶτοι πρὸς ἀλλήλους ἔστωσαν· λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β πρὸς ἄλλον τινά.
For let two numbers A, B be prime to each other; I say that it is not as A is to B, so is B to some other.
εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Γ. οἱ δὲ Α, Β πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον· μετρεῖ ἄρα ὁ Α τὸν Β ὡς ἡγούμενος ἡγούμενον.
For, if possible, let it be as A is to B, so B to Γ. And A, B are prime, and those which are prime are also least, and the least numbers measure those having the same ratio with them an equal number of times, the antecedent the antecedent and the consequent the consequent; therefore A measures B as antecedent antecedent.
μετρεῖ δὲ καὶ ἑαυτόν· ὁ Α ἄρα τοὺς Α, Β μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἄτοπον.
And it also measures itself; therefore A measures A, B which are prime to each other; which is absurd.
οὐκ ἄρα ἔσται ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Β πρὸς τὸν Γ· ὅπερ ἔδει δεῖξαι.
Therefore it will not be as A is to B, so is B to Γ; which it was required to prove.
§9.prop.17ἐὰν ὦσιν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον, οἱ δὲ ἄκροι αὐτῶν πρῶτοι πρὸς ἀλλήλους ὦσιν, οὐκ ἔσται ὡς ὁ πρῶτος πρὸς τὸν δεύτερον, οὕτως ὁ ἔσχατος πρὸς ἄλλον τινά.
If there be any number of numbers continuously proportional, and their extremes be prime to each other, it will not be as the first is to the second, so is the last to some other.
ἔστωσαν ὁσοιδηποτοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Α, Β, Γ, Δ, οἱ δὲ ἄκροι αὐτῶν οἱ Α, Δ πρῶτοι πρὸς ἀλλήλους ἔστωσαν· λέγω, ὅτι οὐκ ἔστιν ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς ἄλλον τινά.
Let there be any number of numbers continuously proportional A, B, Γ, Δ, and let their extremes A, Δ be prime to each other; I say that it is not as A is to B, so is Δ to some other.
εἰ γὰρ δυνατόν, ἔστω ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε· ἐναλλὰξ ἄρα ἐστὶν ὡς ὁ Α πρὸς τὸν Δ, ὁ Β πρὸς τὸν Ε. οἱ δὲ Α, Δ πρῶτοι, οἱ δὲ πρῶτοι καὶ ἐλάχιστοι, οἱ δὲ ἐλάχιστοι ἀριθμοὶ μετροῦσι τοὺς τὸν αὐτὸν λόγον ἔχοντας ἰσάκις ὅ τε ἡγούμενος τὸν ἡγούμενον καὶ ὁ ἑπόμενος τὸν ἑπόμενον. μετρεῖ ἄρα ὁ Α τὸν Β. καί ἐστιν ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Γ. καὶ ὁ Β ἄρα τὸν Γ μετρεῖ· ὥστε καὶ ὁ Α τὸν Γ μετρεῖ.
For, if possible, let it be as A is to B, so Δ to E; therefore alternately, as A is to Δ, so is B to E. And A, Δ are prime, and those which are prime are also least, and the least numbers measure those having the same ratio with them an equal number of times, the antecedent the antecedent and the consequent the consequent. Therefore A measures B. And it is as A is to B, so is B to Γ. Therefore B also measures Γ; so that A also measures Γ.
καὶ ἐπεί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Γ πρὸς τὸν Δ, μετρεῖ δὲ ὁ Β τὸν Γ, μετρεῖ ἄρα καὶ ὁ Γ τὸν Δ. ἀλλʼ ὁ Α τὸν Γ ἐμέτρει· ὥστε ὁ Α καὶ τὸν Δ μετρεῖ.
And since it is as B is to Γ, so is Γ to Δ, and B measures Γ, therefore Γ also measures Δ. But A was measuring Γ; so that A also measures Δ.
μετρεῖ δὲ καὶ ἑαυτόν.
And it also measures itself.
ὁ Α ἄρα τοὺς α, Δ μετρεῖ πρώτους ὄντας πρὸς ἀλλήλους· ὅπερ ἐστὶν ἀδύνατον.
Therefore A measures A, Δ which are prime to each other; which is impossible.
οὐκ ἄρα ἔσται ὡς ὁ Α πρὸς τὸν Β, οὕτως ὁ Δ πρὸς ἄλλον τινά· ὅπερ ἔδει δεῖξαι.
Therefore it will not be as A is to B, so is Δ to some other; which it was required to prove.
§9.prop.18δύο ἀριθμῶν δοθέντων ἐπισκέψασθαι, εἰ δυνατόν ἐστιν αὐτοῖς τρίτον ἀνάλογον προσευρεῖν.
Two numbers being given, to investigate whether it is possible to find a third proportional to them.
ἔστωσαν οἱ δοθέντες δύο ἀριθμοὶ οἱ Α, Β, καὶ δέον ἔστω ἐπισκέψασθαι, εἰ δυνατόν ἐστιν αὐτοῖς τρίτον ἀνάλογον προσευρεῖν.
Let the two given numbers be A, B, and let it be required to investigate whether it is possible to find a third proportional to them.
οἱ δὴ Α, Β ἤτοι πρῶτοι πρὸς ἀλλήλους εἰσὶν ἢ οὔ.
Now A, B are either prime to each other or not.
καὶ εἰ πρῶτοι πρὸς ἀλλήλους εἰσίν, δέδεικται, ὅτι ἀδύνατόν ἐστιν αὐτοῖς τρίτον ἀνάλογον προσευρεῖν.
And if they are prime to each other, it has been proved that it is impossible to find a third proportional to them.
ἀλλὰ δὴ μὴ ἔστωσαν οἱ Α, Β πρῶτοι πρὸς ἀλλήλους, καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Γ ποιείτω· ὁ Α δὴ τὸν Γ ἤτοι μετρεῖ ἢ οὐ μετρεῖ.
But indeed let A, B not be prime to each other, and let B by multiplying itself make Γ; then A either measures Γ or does not measure it.
μετρείτω πρότερον κατὰ τὸν Δ· ὁ Α ἄρα τὸν Δ πολλαπλασιάσας τὸν Γ πεποίηκεν.
Let it first measure it according to Δ; therefore A by multiplying Δ has made Γ.
ἀλλὰ μὴν καὶ ὁ Β ἑαυτὸν πολλαπλασιάσας τὸν Γ πεποίηκεν· ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Β. ἔστιν ἄρα ὡς ὁ Α πρὸς τὸν Β, ὁ Β πρὸς τὸν Δ· τοῖς Α, Β ἄρα τρίτος ἀριθμὸς ἀνάλογον προσηύρηται ὁ Δ.
ἀλλὰ δὴ μὴ μετρείτω ὁ Α τὸν Γ· λέγω, ὅτι τοῖς Α, Β ἀδύνατόν ἐστι τρίτον ἀνάλογον προσευρεῖν ἀριθμόν.
But indeed B also by multiplying itself has made Γ; therefore the product of A, Δ is equal to the square on B. Therefore, as A is to B, so is B to Δ; therefore a third number proportional to A, B, namely Δ, has been found. But indeed let A not measure Γ; I say that it is impossible to find a third proportional number to A, B.
εἰ γὰρ δυνατόν, προσηυρήσθω ὁ Δ. ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ ἀπὸ τοῦ Β. ὁ δὲ ἀπὸ τοῦ Β ἐστιν ὁ Γ· ὁ ἄρα ἐκ τῶν Α, Δ ἴσος ἐστὶ τῷ Γ. ὥστε ὁ Α τὸν Δ πολλαπλασιάσας τὸν Γ πεποίηκεν· ὁ Α ἄρα τὸν Γ μετρεῖ κατὰ τὸν Δ. ἀλλὰ μὴν ὑπόκειται καὶ μὴ μετρῶν· ὅπερ ἄτοπον.
For, if possible, let Δ have been found; therefore the product of A, Δ is equal to the square on B. But the square on B is Γ; therefore the product of A, Δ is equal to Γ. So that A by multiplying Δ has made Γ; therefore A measures Γ according to Δ. But indeed it is also assumed not to measure it; which is absurd.
οὐκ ἄρα δυνατόν ἐστι τοῖς Α, Β τρίτον ἀνάλογον προσευρεῖν ἀριθμόν, ὅταν ὁ Α τὸν Γ μὴ μετρῇ· ὅπερ ἔδει δεῖξαι.
Therefore it is not possible to find a third proportional number to A, B, when A does not measure Γ; which it was required to prove.