§9.prop.10ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἀνάλογον ὦσιν, ὁ δὲ μετὰ τὴν μονάδα μὴ ᾖ τετράγωνος, οὐδʼ ἄλλος οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων πάντων.
If any number of numbers starting from a unit be proportional, and the one after the unit be not square, neither will any other be square except the third from the unit and all those which leave out one.
καὶ ἐὰν ὁ μετὰ τὴν μονάδα κύβος μὴ ᾖ, οὐδὲ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων πάντων.
And if the one after the unit be not a cube, neither will any other be a cube except the fourth from the unit and all those which leave out two.
ἔστωσαν ἀπὸ μονάδος ἑξῆς ἀνάλογον ὁσοιδηποτοῦν ἀριθμοὶ οἱ Α, Β, Γ, Δ, Ε, Ζ, ὁ δὲ μετὰ τὴν μονάδα ὁ Α μὴ ἔστω τετράγωνος· λέγω, ὅτι οὐδὲ ἄλλος οὐδεὶς τετράγωνος ἔσται χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος.
Let there be any number of numbers starting from a unit continuously proportional, A, B, Γ, Δ, Ε, Ζ, and let the one after the unit, A, not be square; I say that neither will any other be square except the third from the unit.
εἰ γὰρ δυνατόν, ἔστω ὁ Γ τετράγωνος.
For, if possible, let Γ be square.
ἔστι δὲ καὶ ὁ Β τετράγωνος· οἱ Β, Γ ἄρα πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν.
But B is also square; therefore B, Γ have to one another the ratio which a square number has to a square number.
καί ἐστιν ὡς ὁ Β πρὸς τὸν Γ, ὁ Α πρὸς τὸν Β· οἱ Α, Β ἄρα πρὸς ἀλλήλους λόγον ἔχουσιν, ὃν τετράγωνος ἀριθμὸς πρὸς τετράγωνον ἀριθμόν· ὥστε οἱ Α, Β ὅμοιοι ἐπίπεδοί εἰσιν.
And, as B is to Γ, so is A to B; therefore A, B have to one another the ratio which a square number has to a square number; so that A, B are similar plane numbers.
καί ἐστι τετράγωνος ὁ Β· τετράγωνος ἄρα ἐστὶ καὶ ὁ Α· ὅπερ οὐχ ὑπέκειτο.
And B is square; therefore A is also square; which was contrary to the hypothesis.
οὐκ ἄρα ὁ Γ τετράγωνός ἐστιν.
Therefore Γ is not square.
ὁμοίως δὴ δείξομεν, ὅτι οὐδʼ ἄλλος οὐδεὶς τετράγωνός ἐστι χωρὶς τοῦ τρίτου ἀπὸ τῆς μονάδος καὶ τῶν ἕνα διαλειπόντων.
Similarly indeed we will show that neither is any other square except the third from the unit and those which leave out one.
ἀλλὰ δὴ μὴ ἔστω ὁ Α κύβος.
But indeed let A not be a cube.
λέγω, ὅτι οὐδʼ ἄλλος οὐδεὶς κύβος ἔσται χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων.
I say that neither will any other be a cube except the fourth from the unit and those which leave out two.
εἰ γὰρ δυνατόν, ἔστω ὁ Δ κύβος.
For, if possible, let Δ be a cube.
ἔστι δὲ καὶ ὁ Γ κύβος· τέταρτος γάρ ἐστιν ἀπὸ τῆς μονάδος.
But Γ is also a cube; for it is the fourth from the unit.
καί ἐστιν ὡς ὁ Γ πρὸς τὸν Δ, ὁ Β πρὸς τὸν Γ· καὶ ὁ Β ἄρα πρὸς τὸν Γ λόγον ἔχει, ὃν κύβος πρὸς κύβον.
And, as Γ is to Δ, so is B to Γ; therefore B also has to Γ the ratio which a cube has to a cube.
καί ἐστιν ὁ Γ κύβος· καὶ ὁ Β ἄρα κύβος ἐστίν.
And Γ is a cube; therefore B is also a cube.
καὶ ἐπεί ἐστιν ὡς ἡ μονὰς πρὸς τὸν Α, ὁ Α πρὸς τὸν Β, ἡ δὲ μονὰς τὸν Α μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας, καὶ ὁ Α ἄρα τὸν Β μετρεῖ κατὰ τὰς ἐν αὑτῷ μονάδας· ὁ Α ἄρα ἑαυτὸν πολλαπλασιάσας κύβον τὸν Β πεποίηκεν.
And since, as the unit is to A, so is A to B, and the unit measures A according to the units in it, therefore A also measures B according to the units in itself; therefore A by multiplying itself has made the cube B.
ἐὰν δὲ ἀριθμὸς ἑαυτὸν πολλαπλασιάσας κύβον ποιῇ, καὶ αὐτὸς κύβος ἔσται.
But if a number by multiplying itself make a cube, it itself will also be a cube.
κύβος ἄρα καὶ ὁ Α· ὅπερ οὐχ ὑπόκειται.
Therefore A is also a cube; which is contrary to the hypothesis.
οὐκ ἄρα ὁ Δ κύβος ἐστίν.
Therefore Δ is not a cube.
ὁμοίως δὴ δείξομεν, ὅτι οὐδʼ ἄλλος οὐδεὶς κύβος ἐστὶ χωρὶς τοῦ τετάρτου ἀπὸ τῆς μονάδος καὶ τῶν δύο διαλειπόντων· ὅπερ ἔδει δεῖξαι.
Similarly indeed we will show that neither is any other a cube except the fourth from the unit and those which leave out two; which it was required to prove.
§9.prop.11ἐὰν ἀπὸ μονάδος ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον ὦσιν, ὁ ἐλάττων τὸν μείζονα μετρεῖ κατά τινα τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς.
If any number of numbers starting from a unit be continuously proportional, the less measures the greater according to some one of the numbers existing among the proportional numbers.
ἔστωσαν ἀπὸ μονάδος τῆς Α ὁποσοιοῦν ἀριθμοὶ ἑξῆς ἀνάλογον οἱ Β, Γ, Δ, Ε· λέγω, ὅτι τῶν Β, Γ, Δ, Ε ὁ ἐλάχιστος ὁ Β τὸν Ε μετρεῖ κατά τινα τῶν Γ, Δ.
ἐπεὶ γάρ ἐστιν ὡς ἡ Α μονὰς πρὸς τὸν Β, οὕτως ὁ Δ πρὸς τὸν Ε, ἰσάκις ἄρα ἡ Α μονὰς τὸν Β ἀριθμὸν μετρεῖ καὶ ὁ Δ τὸν Ε·
Let there be any number of numbers starting from the unit A continuously proportional, B, Γ, Δ, Ε; I say that of B, Γ, Δ, Ε the least, B, measures Ε according to some one of Γ, Δ.
ἐναλλὰξ ἄρα ἰσάκις ἡ Α μονὰς τὸν Δ μετρεῖ καὶ ὁ Β τὸν Ε. ἡ δὲ Α μονὰς τὸν Δ μετρεῖ κατὰ τὰς ἐν αὐτῷ μονάδας·
For since, as the unit A is to B, so is Δ to Ε, therefore the unit A measures the number B as many times as Δ measures Ε; therefore, alternately, the unit A measures Δ as many times as B measures Ε.
καὶ ὁ Β ἄρα τὸν Ε μετρεῖ κατὰ τὰς ἐν τῷ Δ μονάδας· ὥστε ὁ ἐλάσσων ὁ Β τὸν μείζονα τὸν Ε μετρεῖ κατά τινα ἀριθμὸν τῶν ὑπαρχόντων ἐν τοῖς ἀνάλογον ἀριθμοῖς.
But the unit A measures Δ according to the units in it; therefore B also measures Ε according to the units in Δ; so that the less B measures the greater Ε according to some number of those existing among the proportional numbers.
Πόρισμα Καὶ φανερόν, ὅτι ἣν ἔχει τάξιν ὁ μετρῶν ἀπὸ μονάδος, τὴν αὐτὴν ἔχει καὶ ὁ καθʼ ὃν μετρεῖ ἀπὸ τοῦ μετρουμένου ἐπὶ τὸ πρὸ αὐτοῦ.
Porism: And it is manifest that, what place the measurer has starting from the unit, the same place also has the number according to which it measures, starting from the measured number and going to the number before it.
¯ὅπερ ἔδει δεῖξαι.
Which it was required to prove.