Then p, p√1-k2 are the straight lines required. According as p is of the form a or A, the straight lines are (1) or (2) √A, NA-RA. PROPOSITION 30. To find two rational straight lines commensurable only and such that the square on the greater is gre the square on the less by the square on a straight l mensurable in length with the greater. Let there be set out a rational straight line AB, and two square numbers CE, ED such that their sum CD is not Then, in a similar manner to the preceding, we that BA, AF are rational straight lines commer square only. And since, as DC is to CE, so is the square the square on AF, therefore, convertendo, as CD is to DE, so is the AB to the square on BF. [v. 19, Por., But CD has not to DE the ratio which a squa has to a square number; X. 30, 31] PROPOSITIONS 29-31 69 therefore neither has the square on AB to the square on BF the ratio which a square number has to a square number; therefore AB is incommensurable in length with BF. [x. 9] And the square on AB is greater than the square on AF by the square on FB incommensurable with AB. Therefore AB, AF are rational straight lines commensurable in square only, and the square on AB is greater than the square on AF by the square on FB incommensurable in length with AB. In this case we take m2, n2 such that m2 + n2 is not square. whence or m2 + n2 : m2 = p2 : x3, Q. E. D. n m The proof is after the manner of the proof of the preceding proposition and need not be repeated. According as p is of the form a or A, the straight lines take the that is, a, Va B, or (2) JA, NA B and PROPOSITION 31. To find two medial straight lines commensurable in square only, containing a rational rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater. Let there be set out two rational straight lines A, B commensurable in square only and such that the square on A, being the greater, is greater than the square on B the less by the square on a straight line commensurable in length with A. [x. 29] And let the square on C be equal to the rectangle A, B. Now the rectangle A, B is medial; therefore the square on C is also medial; therefore C is also medial. [x. 21] A [x. 21] 70 BOOK X [x. 31 Let the rectangle C, D be equal to the square on B. therefore the rectangle C, D is also rational. And since, as A is to B, so is the rectangle A, B to the square on B, while the square on C is equal to the rectangle A, B, and the rectangle C, D is equal to the square on B, therefore, as A is to B, so is the square on C to the rectangle C, D. But, as the square on C is to the rectangle C, D, so is C to D; therefore also, as A is to B, so is C to D. But A is commensurable with B in square only; therefore C is also commensurable with D in square only. [x. 11] And C is medial; therefore D is also medial. [x. 23, addition] And since, as A is to B, so is C to D, and the square on A is greater than the square on B by the square on a straight line commensurable with A, therefore also the square on C is greater than the square on D by the square on a straight line commensurable with C. [x. 14] Therefore two medial straight lines C, D, commensurable in square only and containing a rational rectangle, have been found, and the square on C is greater than the square on D by the square on a straight line commensurable in length with C. Similarly also it can be proved that the square on C exceeds the square on D by the square on a straight line incommensurable with C, when the square on A is greater than the square on B by the square on a straight line incommensurable with A. [x. 30] I. Take the rational straight lines commensurable in square only found. in x. 29, i.e. p, p√ I-k2. Take the mean proportional p (1 − k2)1 and x such that Then p (1 − k2)1, x, or p (1 − k2)3, p ( 1 − k) are straight lines satisfying the given conditions. ՆԱՆԱ наз to prove a swiat Touaout way by meal lemma after x. 21 to the effect that a : b = ab : b2.) From (2) it follows [x. 11] that x ~ p(1 − k2); whence, since p( medial, x or p (1 – k2)3 is medial also. (y) From (2), since p, p√1-k2 satisfy the remaining conditio problem, p(1 − k2), p(1 - k2) do so According as p is of the form a or also [x. 14]. A, the straight lines take the a2 - b2 Ja Ja2 – b2' A-k2A A (A-k2A) II. To find medial straight lines commensurable in square only ing a rational rectangle, and such that the square on one exceeds th on the other by the square on a straight line incommensurable with th we simply begin with the rational straight lines having the corre property [x. 30], viz. p, and we arrive at the straight lin ρ VI + k2 To find two medial straight lines commensurable in only, containing a medial rectangle, and such that the on the greater is greater than the square on the less square on a straight line commensurable with the great A D B E Therefore the square on D is medial; therefore D is also medial. Let the rectangle D, E be equal to the rectangle Then since, as the rectangle A, B is to the rectan so is A to C, while the square on D is equal to the rectangle A, and the rectangle D, E is equal to the rectangle B, therefore, as A is to C, so is the square on D to the D, E. But, as the square on D is to the rectangle D, I to E; therefore also, as A is to C, so is D to E. But A is commensurable with C in square only therefore D is also commensurable with E in square of But D is medial; therefore E is also medial. [x. And, since, as A is to C, so is D to E, while the square on A is greater than the square the square on a straight line commensurable with A therefore also the square on D will be greater than on E by the square on a straight line commensurabl I say next that the rectangle D, E is also media For, since the rectangle B, C is equal to the rectar while the rectangle B, C is medial, therefore the rectangle D, E is also medial. Therefore two medial straight lines D, E, comm in square only, and containing a medial rectangle, 1 found such that the square on the greater is greater |