38 BOOK X [x. 14 Therefore, as the squares on E, B are to the square on B, so are the squares on D, F to the square on D ; therefore, separando, as the square on E is to the square on B, so is the square on F to the square on D; 30 therefore also, as E is to B, so is F to D; therefore, inversely, as B is to E, so is D to F. But, as A is to B, so also is C to D; [v. 17] [VI. 22] therefore, ex aequali, as A is to E, so is C to F. [v. 22] Therefore, if A is commensurable with E, C is also com 35 mensurable with F, and, if A is incommensurable with E, C is also incommensurable with F. Therefore etc. [x. 11] 3, 5, 8, 10. Euclid speaks of the square on the first (third) being greater than the square on the second (fourth) by the square on a straight line commensurable (incommensurable) "with itself (avry),” and similarly in all like phrases throughout the Book. For clearness' sake I substitute "the first,' 66 the third," or whatever it may be, for "itself" in these cases. Suppose a, b, c, d to be straight lines such that Euclid has to use a somewhat roundabout method owing to the absence of a convertendo proposition in his Book v. (which omission Simson supplied by his Prop. E). If a ~ √a2 – b2, we may put √a-b2 = ka, where k is of the form m/n and m, n are integers. And if a b2 ka, it follows in this case that √c2 - d2 = kc. x. 15] PROPOSITIONS 14, 15 PROPOSITION 15. 39 If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable. For let the two commensurable magnitudes AB, BC be added together; B For, since AB, BC are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC, it will also measure the whole AC. But it measures AB, BC also; therefore D measures AB, BC, AC; therefore AC is commensurable with each of the magnitudes AB, BC. I Next, let AC be commensurable with AB; say that AB, BC are also commensurable. [x. Def. 1] For, since AC, AB are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures CA, AB, it will also measure the remainder BC. But it measures AB also; therefore D will measure AB, BC; therefore AB, BC are commensurable. Therefore etc. [x. Def. 1] (1) If a, b be any two commensurable magnitudes, they are of the form mc, nc, where c is a common measure of a, b and m, n some integers. с a+b= (m + n) c ; It follows that therefore (a + b), being measured by ‹, is commensurable with both a and b. (2) If a + b is commensurable with either a or b, say a, we may put a + b = mc, a = nc, where c is a common measure of (a + b), a, and m, n are integers. Subtracting, we have whence ba. b= (m-n) c, 40 BOOK X PROPOSITION 16. [x. 16 If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable. For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC. For, if CA, AB are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D. therefore it will also measure the remainder BC. But it measures AB also; therefore D measures AB, BC. Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable : which is impossible. Therefore no magnitude will measure CA, AB ; therefore CA, AB are incommensurable. Α B [x. Def. 1] Similarly we can prove that AC, CB are also incommensurable. Therefore AC is incommensurable with each of the magnitudes AB, BC. Next, let AC be incommensurable with one of the magnitudes AB, BC. I First, let it be incommensurable with AB; say that AB, BC are also incommensurable. For, if they are commensurable, some magnitude will measure them. Let it measure them, and let it be D. Since then D measures AB, BC, therefore it will also measure the whole AC. But it measures AB also; therefore D measures CA, AB. LEMMA. If to any straight line there be applied a paralle deficient by a square figure, the applied parallelogram to the rectangle contained by the segments of the strai resulting from the application. For let there be applied to the straight line parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB. This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectang CB. Therefore etc. If a be the given straight line, and x the side of the square by applied rectangle is to be deficient, the rectangle is equal to ax of course equal to x (ax). The rectangle may be written a x + y = a. Given the area x (ax), or xy (where x+y=a), two applications will give rectangles equal to this area, the sides of t being x or a-x (x or y) respectively; but the second mode of e shows that the rectangles do not differ in form but only in position. PROPOSITION 17. If there be two unequal straight lines, and to the there be applied a parallelogram equal to the fourth the square on the less and deficient by a square figure, it divide it into parts which are commensurable in leng 5 the square on the greater will be greater than the sq the less by the square on a straight line commensurab the greater. And, if the square on the greater be greater than the on the less by the square on a straight line commensural 15 the greater, and let there be applied to BC a parallelogram equal to the fourth part of the square on the less, A, that is, equal to the square on the half of A, and deficient 20 by a square figure. Let this be the rectangle BD, DC, [cf. Lemma] 25 and let BD be commensurable in length with DC; I say that the square on BC is greater than the squa by the square on a straight line commensurable with For let BC be bisected at the point E, and let EF be made equal to DE. Therefore the remainder DC is equal to BF. And, since the straight line BC has been cut in parts at E, and into unequal parts at D, 30 therefore the rectangle contained by BD, DC, toget the square on ED, is equal to the square on EC; And the same is true of their quadruples; therefore four times the rectangle BD, DC, toget four times the square on DE, is equal to four times t 35 on EC. 40 45 But the square on A is equal to four times the BD, DC; and the square on DF is equal to four times the s DE, for DF is double of DE. And the square on BC is equal to four times t on EC, for again BC is double of CE. Therefore the squares on A, DF are equal to t on BC, so that the square on BC is greater than the square the square on DF. It is to be proved that BC is also commensurable Since BD is commensurable in length with DC therefore BC is also commensurable in length with |