Kitap 10
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Let two numbers B, C be set out which have not to one another the ratio which a square number has to a square number, that is, which are not similar plane numbers; and let it be contrived that, as B is to C, so is the square on A to the square on D —for we have learnt how to do this— [X. 6, Por.] therefore the square on A is commensurable with the square on D. [X. 6]
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And, since B has not to C the ratio which a square number has to a square number, therefore neither has the square on A to the square on D the ratio which a square number has to a square number; therefore A is incommensurable in length with D. [X. 9]
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Let E be taken a mean proportional between A, D; therefore, as A is to D, so is the square on A to the square on E. [V. Def. 9]
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But A is incommensurable in length with D; therefore the square on A is also incommensurable with the square on E; [X. 11] therefore A is incommensurable in square with E.
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Therefore two straight lines D, E have been found incommensurable, D in length only, and E in square and of course in length also, with the assigned straight line A.]
PROPOSITION 11.
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If four magnitudes be proportional, and the first be commensurable with the second, the third will also be commensurable with the fourth; and, if the first be incommensurable with the second, the third will also be incommensurable with the fourth.
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Let A, B, C, D be four magnitudes in proportion, so that, as A is to B, so is C to D, and let A be commensurable with B; I say that C will also be commensurable with D.
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For, since A is commensurable with B, therefore A has to B the ratio which a number has to a number. [X. 5]
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And, as A is to B, so is C to D; therefore C also has to D the ratio which a number has to a number; therefore C is commensurable with D. [X. 6]
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Next, let A be incommensurable with B; I say that C will also be incommensurable with D.
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For, since A is incommensurable with B, therefore A has not to B the ratio which a number has to a number. [X. 7]
