4. Here √10+/7 is greater or less than According as 17+2\/70 2√70 280 22+2 57, by squaring each side. but 280 is obviously less than 25+20√/57+228, ../10+√7 is less than 19+ √3. (2) √2+17 is greater than (3) √5+√14 is greater than √3+3√2. (4) √6-√5 is greater than (5) 2+5 is less than 3. 8-√7. 5. If every two of the three lines be greater than the third, a triangle can be formed with the three lines. 6. It is obvious that if the sides of the equilateral triangle, the square, the pentagon, be given in magnitude, the areas of these regular figures can be determined, and consequently the surfaces of each of the five regular solids. The surface of the tetrahedron is equal to the area of four equal and equilateral triangles. The surface of the hexahedron is the area of six equal squares. The surface of the octahedron is equal to the areas of eight equal and equilateral triangles. The surface of the dodecahedron is equal to the area of twelve equal equilateral and equiangular pentagons. The surface of the icosahedron is equal to the area of twenty equal and equilateral triangles. A sphere can be inscribed within, and another circumscribed about, each of the five regular solids, in the same manner as a circle can be inscribed within and another circumscribed about each of the plane regular figures, the equilateral triangle, the square, the pentagon. Suppose a sphere described about a regular tetrahedron, the surface of the sphere will pass through the four points of the trihedral angles of the tetrahedron. If lines be supposed to be drawn from these points to the centre of the circumscribing sphere, these lines will be each equal to the radius of the sphere. Hence the tetrahedron may be conceived to be divided into four equal triangular pyramids, whose common vertices are at the centre of the sphere, the three edges of each, radii of the sphere, and their four bases the equilateral triangles which make up the surface of the tetrahedron. Since the content of any pyramid is equal to one-third of the content of a prism on the same base and of the same altitude, the contents of the four equal pyramids can be found, and the content, or volume, of the tetrahedron determined. In a similar manner it may be shown that : The volume of the hexahedron is equal to the volumes of six equal pyramids whose bases are equal squares. The volume of the octahedron is equal to the volumes of eight equal pyramids whose bases are equal equilateral triangles. The volume of the dodecahedron is equal to the volumes of twelve equal pyramids whose bases are equal equilateral and equiangular pentagons. The volume of the icosahedron is equal to the volumes of twenty equal pyramids whose bases are equal equilateral triangles. 3. Since c=b(1-b2)1+b(1 − a2 )1 ; .•. c—a(1 − b2)1—b(1 − a2)1, c2 - 2ac(1 − b2)*+a2 − a2b3=b2 – b2a2, c⭑+a+b+2a2c2-2b2c2 - 2a2b3=4a3c2 - 4a2b2c2; See Section IV., XIII., Ex. 10, p. 32. x(a2 — y2)}=a2 − y(aa—x2)1, by squaring each side, 6. Since x+y=a‡, xł+y1—a1=0, - and x+y+a+2x1ył — 2a1xł – 2a1yt=0 =4 1 { (x + y − a) 2 4 a(x+y-a } +a(x+y)+a(a−x−y)+a(x+y− = (x + y − a)2 + xa(x+y)=2(x2+y2+a2). The truth of this also may be verified by numerical examples, as, for instance, when (16)++(25)*=(81)1. 8. If (a+b+c+d})}=x}+y}+z¥‚ Then a+b+c+d3 =x+y+z+2(xy)*+2(xz)*+2(yz)*, and x+y+z=a, 2(xy)*=b1, 2(xz)*=c1, 2(yz)1=d}. Hence 2(xy)1.2(xz)1.2(yz)*=bicid1, or 8xyz=(bcd), and x+y+z=α, But 4xy=b, 4xz=c, 4yz=d, .'. bc+bd+cd=16(x3yz+y2xz+z2xy)=16xyz(x+y+z). 9. Let {r+(r2+q3)t}š=a, {r−(r2+q2)t}}=b; then x3=(a+b)3=a3 +b3+3ab(a+b)=2r−3gx ; •*. x3+3qx-2r=0. 10. Find y and z in terms of x, and substitute these values in a1+y−1+z1=1, whence x is known, and y, z by substitution. Lastly, substitute these values of x, y, x in ax2+by2+cz2. XL. 5. In the verification, when x2=ab is found, (a+x)(b+x) 2ab+(a+b)a§b1 = 2ab+a+b (a+b)3 (a-x)(b-x)2ab − (a+b)ab ̄ ̄2a1b1 − (a+b) { (a + x)(b+x) (a = (ai+bi)2 (a - b) 2 (c-d)2 =1+ =±1, as - bi 6. Express the products of the factors in the numerators of the two fractions, as the sums of two squares, and 1+ then (1+ax+by)2=(1+a2+b2)(1+x2+ y2), whence (a − x)2 + (b − y)2 + (ay — bx)2 = 0. 8. The quotient is √(1-x2)√(1 − y3) — xy. 9. From xy+xz+yz=1, x=: 1+ y2, and 1+x2 - (y + z)2 + (1 — yz)2 (1+x2)(1 +y3) } * =z(x+y). (1+22) (1+x2) } * = y(x+2), and z{ (1+x2)(1+ y2) }' 1+22 10. Since x2+ y2+z2+2xyz=1, 2xyz+z2=1−x2 - ya3, and adding x2y2 to these equals, x2y2+2xyz + z2 = 1 − x2 - y2+x2y2, or (1−x2)(1—y2)=(z+xy)2. Similarly (1 − y2)(1 − z2)=(x+yz)a and (1 − x2)(1 − z2)=(y+xz)2. 1. See Art. 15, p. 16. XLI. 2. See Art. 16, p. 17. 5. The former is the greater or the less, according as x2 is greater or less than ab. 6. Here a1+ab is greater or less than ab+b3, according as a1(a+b) is greater or less than bi(a+b). 7. Since a(1 − b2)*+b(1 − a2)i<1, then a(1 − b2)* <1 – b(1 − a2)*, and a2 — a2b2 <1 −2b(1 − a2)$+b2 – a2b, or 2b(1—a2) <(b2 —a2)+1, also 4b2 - 4a2b2 <(b2 − a2)2 +2(b2-a2)+1, whence (b2 + a2)2 −2(b2 + a2) + 1> 0; ... b2 + a2 - 1> 0 and a2 + b2> 1. 8. Let ab, bc, then a2-2ab+b2> 0, and a2 +2ab+b2> 4ab; .. 2(a+b+c)> 2{(ab)+(bc)+(ac)1?, and a+b+c> (ab)*+ (bc)+(ac)*. ... a+b+c+d>2{(ab)*+(cd)*)}> 4{(ab)*(cd)}}'>4(abcd)1. 9. Since x=a+b, and y=c+d, .. xy=(a+b)(c+d)=(a1c$+b}d1)2+(aìdì — biç1)2 ; consequently xy> (aści+bid1)2, and xłył> aici + bid. 10. If the sum of first and third expressions be taken to be greater than the second, the sum of the squares of the first and third will also be greater than the square of the second. From this inequality may be deduced that 2x2y2+2x2x2+2y2 z2 − x4 − y* -z1 is greater than zero; and as this expression is homogeneous, like expressions may be deduced by assuming the sum of the first and second greater than the third, and the sum of the second and third greater than the first. See Section IV., XIII. Ex. 10, p. 32. (a2+b2)(x2 + y2) <ནཱ? Then (a2+b2)(x2 + y2)> 2ab(x2 — y2) + 2xy(a2 —b2), and (a2 - 2ab+b2)x2+(a2+2ab+b2)y2> 2xy(a3 — b2); x a+b .. (a - b)x− (a+b)y>0; (a−b)x> (a+b)y; and > XLII. У x-b 5. a±b _(a")"±(b")"_"y", which is of the forms Section IV., Art. 13, p. 22. am-b=am - (bm)m=am—x”. Here a"-" and am-am have a common divisor when m and n are odd or even numbers. |