Given the ratios of A to B, and of B to C, which compound the ratio of A to C, equal, each to each, to the ratios of D to E, and E to F, which compound the ratio of D to F; to prove that A:C:: D:F. (Dem.) For, first, if the ratio of A to B be equal to that of D to E, and the ratio of B to C equal to that of E to F, by equality (V. 22), A: C:: D: F. And next, if the ratio of A to B be equal to that of E to F, and the ratio of B to C by indirect equality equal to that of D to E, A, B, C, D, E, F. In the same manner may the proposition be demonstrated whatever be the number of ratios. PROPOSITION F. The terms of an analogy are proportional by addition. Given A: B:: C:D; to prove that ::C:C+D. by addition, A: A + B therefore hence, by (Dem.) For B: A:: D: C, by inversion (V. a); A+ BA::C+D: C, by composition (V. 18); inversion, A: A+B = C: C + D (V. ▲). PROPOSITION G. The terms of an analogy are proportional by mixing. Given A: B:: C:D; to prove that D: C - D. A+ B:AB:: C+ (Dem.) For A+B:B=C+D:D by composition (V. 18); and by division, AB: B :: C D:D (V. 17), inversion (V. A). but therefore, by equality (V. 22), A + B : A C-D. If BA, ::C+D:D - C. also by it may be similarly proved that A+B: B-A PROPOSITION H. PROBLEM. To find a common measure of two lines. Given the two lines AB and CD, common measure of them. it is required to find a Find the number of times that CD is contained in AB. If it be contained an exact number of times, then it is a measure of it, and any part of it will also be a common measure. But if it be contained several times in AB, as three times, with a remainder EB; then if EB be a measure of CD, it will also be one of AE, which is a multiple of CD; and therefore it will be also a measure of AB; it would therefore be the common measure required. But if EB be not a measure of CD, let it be contained in it a certain number of times, as 2 times, with the remainder CF; then if CF be a measure of EB, it will also be a measure of DF, which is the multiple of EB; and therefore it will also be a measure of CD, and consequently of AE, and therefore also of then EB2CF; and AB 3CD AB. Let CF be contained 2 times in EB, CD: =2EB + CF = 4CF + CF = 5CF; + EB = 15CF+2CF 17CF. And CF is therefore contained 5 times in CD, and 17 times in AB. In the same manner the common measure of any other two commensurable lines may be found. COR. 1.-If the process for finding a common measure of two lines never terminates, the lines are incommensurable. COR. 2.-Any part of a common measure is also a common measure; and the measure found as above is the greatest common measure. COR. 3.-Any two commensurable lines are to one another, as the numbers denoting the number of times that they respectively contain their common measure. PROPOSITION K. THEOREM. The diagonal and side of a square are incommensurable. and diagonal AC are incommensurable. D the side AB (Dem.) Since the angle at B is a right angle, each of the equal angles at C and A is less than a right angle (I. 32); therefore AB AC. Again, AB + BC AC (I. 20), or 2AB AC; hence AC is greater than once AB, and less than twice AB; and the same may be proved of the diagonal and side of any square. Therefore, when the side of a square is taken once from its diagonal, there is always a remainder less than its side. E B From C as a centre, with the radius CB, describe the arc DB; then CD CB = AB, and AD AB. From D draw DE perpendicular to AC; then ED and EB being tangents (III. 16, Cor.), ED = EB (III. 37, Cor.). But in the triangle ADE, the angle at D is a right angle, and the angle at A is half a right angle ; therefore the angle at E is half a right angle, and therefore AD = DE EB. When the first remainder AD therefore is taken from AB, the remaining part AE, from which AD is still to be taken, is the diagonal of a square, of which AD is the side. But this is the same as the former process; and when it has been performed in regard to AE and AD, the remaining lines to be compared will therefore again be the side and diagonal of a still smaller square; but since, when the side of a square is taken from its diagonal, there is a remainder, therefore in the above process there will always be found to be a remainder; the process therefore will never terminate, or no common measure can ever be found; therefore AC and AB are incommensurable. EXERCISES. 1. If all the terms, or any two homologous terms, or the terms of either of the ratios, of an analogy, be multiplied or divided by the same number, the resulting magnitudes are still proportional. 2. If any number of magnitudes be in continued proportion, the difference between the first and second terms is to the first, as the difference between the first and last to the sum of all the terms except the last. 3. If the same magnitude be added to the terms of a ratio, it will be unchanged, increased, or diminished, according as it is a ratio of equality, minority, or majority. 4. The differences of the successive terms of continued proportionals are also in continued proportion. 5. The first term of an infinite decreasing series of quantities in continued proportion, is a mean proportional between its excess above the second, and the sum of the series. FIFTH воок. SUPPLEMENT. Since the principles of proportion, as delivered in the Fifth Book of Euclid are generally found very difficult, and many teachers prefer to have them established in a more simple way; the following Fifth Book may be substituted for it, in which, it is hoped, the propositions are all proved in as simple a manner as the subject will admit of. In order to effect this, it will be necessary to adopt a different definition of proportion from that given by Euclid-namely, 'If four quantities be proportional, the first divided by the second gives the same quotient as the third divided by the fourth.' But as there is no geometrical method of dividing one line by another, proportion cannot be applied to Geometry by this definition; it will therefore be necessary to prove that if quantities are proportional by the one definition, they are also proportional by the other, and conversely; and then apply the results so obtained by Euclid's definition, which is the tenth of the previous Fifth Book. The definitions of the Fifth Book are also to be used as definitions in this Supplementary Book. ARITHMETICAL PRINCIPLES. " In order to establish the doctrines of Proportion in the manner indicated above, it will be necessary to apply the following arithmetical principles. 12 1st, A fraction is multiplied by a whole number by multiplying its numerator; thus is multiplied by 7, by multiplying its numerator 5 by 7 and retaining the same denominator, thus 5 X 7 35 = ; for the result contains 7 times as many of the 12 same kind of parts as the multiplicand, it is therefore multiplied by 7. 12 35 2d, A fraction is divided by a whole number by multiplying its denominator by that number and retaining the same nume rator; thus 5 is divided by 6 by multiplying its denominator by 6 and retaining the same numerator; thus eighteenth part is only the sixth part of a third part, and hence is the sixth part of the fraction has therefore been divided 18 g' by 6 by multiplying its denominator by 6. 3d, A fraction is multiplied by a fraction by multiplying the numerators together for the numerator of the product, and the denominators together for the denominator of the product; thus, 3 5 to multiply by 3 X 5 15 the numerator of the product, 4 8' and 4 × 8 = 32 the denominator of the product; or, more 3 5 15 = concisely, X 5 1 ; this is evident for is 5 times hence 8 it is the same as to multiply by 5 and then divide by 8, which (by 1 and 2) gives the above result. 3 4 4th, A fraction is divided by a fraction by inverting the divisor, and then multiplying the numerators for the numerator of the quotient and the denominators for the denominator; thus, to 5 3 5 15 invert the by making it then X = 5 4' 7 4 28 3 and to divide the fraction by 4 is to mul3 divide by 7 5' tiply its denominator by 4, which gives (2), but it was only the part of 4 that was the divisor; hence, as it has been divided by 5 times the divisor, the quotient is therefore only the fifth part of the true quotient, therefore the numerator must be multiplied by 3 5 15 5 to obtain the true quotient (1), wherefore X = is the 4 28 true quotient. 5th, A fraction is not altered in value by multiplying or dividing both its numerator and denominator by the same number. 2 12 24 For X if these be each parts of a yard, the first 3 12 36' evidently represents 2 feet; but the second will represent 24 36 24 inches, which is equal to 2 feet, hence its value is not altered. Or, since multiplying the numerator by 12, multiplies (1) the fraction by 12; and multiplying the denominator by 12, divides (2) the fraction by 12; it has therefore been multiplied and divided by the same number, hence its value is not altered. 24 In the same manner, if be divided by 12, both numerator and 36 dénominator, the result is |