XXXIII. 1. 3, 3/3, 9, 9V 3.

2. 75, 375–3, 5025, 28125.3. 3. 12, 243, 144, 288/3.

4. *, &V 5, A., jfV5. 5. 2, 44/7, 791, 29/7.

6. 72, 1447/2, 576, 1152 2. 7. 9V9, 81, 2432/3, 7291/9.

8. *V36, ), 4V6, zi V36. 9. 25/25, 625, 31255, 15625_25. 10. 7+413, 26+15/3, 97+56x3, 362+209/3. 11. 5+26, 11/29/3, 49+206, 1097 2£89V 3. 12. 77 +30/6, 655/2+531/3, 11329+ 4620/6, 98225/2+80187V3.

XXXIV.
{V2}"=(243=2or V2: {V2}}={24}}=2or V2.

(5V 5)=(5.51)=51.5=(595)!=(125)+ or V125,

(5/5)+=(5.51}}=51,5#=(59.5)+=(125), or 125. The square roots of the last four examples can be found as the example in the note, p. 15. The roots are respectively

3+V3, 3/3-2/2, 2+3/-1.

V3+1
72

XXXV.
1. Here 3V 147=3V (49X3)=3x7V3=213

- 3V 75= -37 (25 x 3)=-3X5V 3= -15V3
- 3V f= -3V =-3x}3=-V3

.. 3V 147 – 3V 75 - 31=(21-15-1)/3=573.
2. 4V 3. 3. 18+2012-12/6. 4. 8V2+7V3+5V5+ 2V30.
5. 1+ v3+15. See XXXIX. 8, p. 36.

6. 12.
7. 2 /3. 8. 75V/2. 9. V3+2{/2. 10. V3.
11. Let (5+2)+(V5 - 2)!=2, and put (V5+2)!=a, (5-2)--b,

then w=a+b,
and x3 = (a + b)s=as +63 +3ab(a+b),

but as +63=205, and 3ab=3,

...23=2V5+ 3x, and 28 - 3x=2V5. To find the value of x requires the solution of a cubic equation 12. 17.69. 13. 90.

XXXVI. Since (Va+Vb)(Va-Vb)=a-, and (avx+bvy)(av/x-bvy)=aʻx—bạy; it is obvious that Va+may be made rational by multiplying it by Va-Vb; and va-vb by multiplying it by Va+Vb.

Similarly for the factors a vx+bvy and avx-bvy.

When there are three terms, Na+16+NC, since (Va+Vb) + } {(Va+vb) - Vc=(a+b)-c=(a+b-c) +2 Vab,

and {(a +6–c) + 2avb}. {(a+b-c) – 2Vab}=(a +6—c)2 – 4ab. And two factors will be required in general to render such a denominator a rational quantity

8-572 8-5123+212 24 +12-20 1.

=4 +12. 3- 2723-2V2^3+2V2 -8

83 ) } 8 102

101

3

108

108

(*03375)}= {91.5"}

15

10.10% 2.104 (80)}=(8.10)+=2.108_2x10_20

(-01)= {10} The given expression by these substitutions

hod

19 I 1
1

2 192
{20–1
101

1

{m}'

103"

3 8+

becomes

-+

XXXVII.
1. These reductions offer no difficulty, as
1. V245 +175 7245 - 175_775+573 775-513

+
15-3 75+13 75-13 V5+ V3

50 +12V15 , 50 - 12/15

+ 2

2 100

=50. 2

-100

2

Х

XXXVIII. 1. The expressions 2, 3, 4, 5, 6, are respectively equivalent to (280)8', (320), (415), (512), (610306; and by expanding the powers 280, 320, 415, 575, 610, their numerical values will be known and the given expressions can be arranged in the order of magnitude. 2. 2

13 203

23
1373^13

2V3_2013 V2 216_
x

=16.

=
N2

72 2
(3)*=(+8)=v6.
**=*)-(87) *=(3-376).

12_ V12

3
1 1 12 12
8-1-

811602* 32-32
73={7}or {7}.
37 3

1 1 1
27

9 The extractions and divisions in each example are to be carried out to four places of decimals.

3. 94 is greater or less than 15. According as 4}

5, raising each to the 12th power.
256

125,
but 256 is greater than 125,
... also 4 is greater than V5.

(*)-(.*)-(***)-(32)

(=(43)

=

Similarly 2 is less than 31.

3V2 is greater than 2V/3.

(52)} is less than (53)},
and because (55)1=5!=515,

and 521–51.250...
therefore (58)} is greater than 528

V7 is greater than 2V/3.
1

is less than
2
V3

V3
is greater than

15 4. Here V10+ 17 is greater or less than V19+ V3. According as 17+2V70

22+2V57, by squaring each side. 2070

5+2V57, subtracting 17. 280

25+ 20757 +228, but 280 is obviously less than 25+20V57 + 228,

.. V10+ 17 is less than v 19+V3. (2) V2+V7 is greater than V3+V 5. (3) V 5+V 14 is greater than V 3+3V 2. (4) V 6-V5 is greater than V 8-17. (6) 2+ +53 is less than 3.

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 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.

XXXIX, 1. 2-21

26
3. Since c=6(1 – 62)++ 6(1 - a2)};

.:c-all - 62)=6(1 - a?)},
Co – 2ac(1 – 62)!+a? - a?bo=62 – 62a”,

c? ta’ - 62=2ac(1 -62)},
c* ta+64 + 2aoc? – 202c2 - 2a?bo=4a?c? – 4a2b2c? ;

.. 4a2bc? = 2a2c2 + 26?c+ 2a2b2-a4-64-c.
See Section IV., XIII., Ex. 10, p. 32.
5. Since wla? — 42)++yla? – 22)!=a?,

ac{a? - y2)=a? – y(a? — ), by squaring each side,
then w?a? – ?y? =a* – 2a’y(a? – 22)t+a’y? – xạy“,

and 2y(a? – x2)1=a? – (x? – yo), squaring again;
. . 4y?a? – 4y*x= a* - 2a? (x2 - y2)+(x2 - yo)*,

whence a4 - 2a?(+ y2)+(x2 + y2)=0

.. a? – (x2 + y2) = 0 and 2 + y2 =a'. 6. Since x +ył=al, at + yt-al=0,

and x+y+a+2xtył - 2abxł – 2ałył=0
2+y+a=2(atx! +alyt - alyt)
(x + y + a)? = 4(ax + ay + xy + 2axłyt - 2xałyt - 2yalxł)

+y

= (x+y-a)? +xa(x+y)= 2(x2 + y2 +a?). The truth of this also may be verified by numerical examples, a3, for instance, when (16)+ +(25)1-(81). 8. If (a +66+c+di)text + y + zł,

Then a +8t+c+dt=x+y+z+2(xy)i + 2(x2)++ 2(yz),

and x+y+z=Q, 2(xy)t = 0, 2(xz)t=c, 2(yz)t = dt.
Hence 2(xy).2(x2)+.2(yz)=bichdt, or 8xyz=(bcd)t, and x+y+z=Q,

... 2a(bcd)t=16xyz(x+y+z). But 4xy = b, 4xz=c, 4yz=d, .. bc+bd+cd=16(aoyz + yoxz+zoxy)=16xyz(x+y+z).

Hence, therefore, 2a(bcd)t=bc + bd + cdo 9. Let {r+(2+q*)*}}=a,{r-(p? +q?)}t=b; then i'=(a+b):=a3 + b3 + 3ab(a+b)=2r - 3qx ;

.. 28 +3qx— 2r=0. 10. Find y and z in terms of x, and substitute these values in 2+y-ltd=1, whence z is known, and y, z by substitution. Lastly, substitute these values of X, y, x in ax? +by? + cza.

XL. 4. Since alb - c): - (b+c)e=c;

altch

c aici

bat-c=1.

2

2

s} =

Hence
26 alto 2c a2-ci b altci
and
ii.

cal + c

and
b-
5. In the verification, when x?=ab is found,

(a +x)(6+2)_2ab +(a + b)abb! 2atbi+a+b (ab+b!)?
(a - x)(8 - x)2ab- (a+b)abb! 2a!! - (a + b) -lat-61)a :
(a +2)(6+2) (al + bt)?

{at+bu?! (a - x)(6-2) - (at - 64)? atbi 6. Express the products of the factors in the numerators of the two fractions, as the sums of two squares, and 1+

(a - b)2

=l+

(c-d)? jab - 1)? (cd + 1)?

1 + ax + by 7. Suppose

=+1, {(1+a? +62)(1 + x2 +y)} then(1+ax+by)?=(1+a? +62)(1+2c2 + y2), whence (a -- )* +(6-y)? +(ay - bx)?=0.

8. The quotient is ✓(1 -->)/(1 - y2) – xy.
9. From xy + xz+yz=1, æ=1+yz, and 1+x2– (y+z)? +(1-yz)

;
z+y

(2+y)?
also (++y)(1+3)(1+y),..*1+

=
=?

(1+za)(1+y)
1 +22

{{1+ +y*) }*=z+). Similarly {{1+2°/+ 2*)}*=y(x+2), and z{ (1+2?)(1+y") }'=z(x+y)

.

1+y?

1+z2 10. Since 2? + y2 + 2 + 2xyz=1, 2xyz +22=1 – 2? - y’, and adding x’yto these equals, x?y+ 2xyz +z?=1 – x? - yo +xay', or (1 --?)(1-Y?)=(3+xy)”.

Similarly (1 - y')(1-2)=(x + yz)' and (1 - 2*)(1-z')=(y+xz).

XLI. 1. See Art. 15, p. 16.

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 ai+ałb is greater or less than abi +61, according as ał(a+b) is greater or less than bl(a+b). 7. Since al1 – 62)#+(1-a2)1 <1, then a (1 – 62)<1 – (1 - a?)!,

and a? - a?l? <1 – 2011 – a?)+62 - a'b, or 26(1-a2). </62 -a2)+1, also 462 – 4420? <<02 – az)2 +2102a2)+1, whence (62 +a?ja -2(62 +93) +1>0;

.:.62 +a? - 1>0 and a? +62> 1.
8. Let a>b, b>c, then a? - 2ab +62>0, and a2 + 2ab +62> 4ab;

.. a+b> 2(ab)', similarly b+c> 2(bc)', a+c> 2(ac)t;
.: 2(a+b+c)> 2{(ab)* + (bc)t + (ac)t}, and a +b+c> (ab)* +(bc)*+ (ac)t.
In a similar way may be shewn that a+b+c+d> 4(abcd)..

For a +b> 2(ab)*, c+d> 2(cd)t ;
... a+b+c+d> 2{(ab)i + (cd)t}}> 4{(ab)(cd)t}A> 4(abcd)t.
9. Since x=a+b, and y=c+d,

. . xy=(a+b)(c+d)=(ałct + btdt): +(aldi - bict)? ;

consequently xy> (atch + b1d!)?, and why->abct + bidt. 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

23°y? + 2x2x2 + 2yoz? – ** - y* -z* 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.