Therefore, by division, since b+b'+b" is positive, which must be true since both factors are positive or negative according as a is greater or less than b. 2. Prove that (x2 + y2+ z2) (x'2 + y'2 + z'2)>(xx' + yy' + zz')2. It can easily be shown that (x2 + y2+ z2) (x2 + y'2 + z'2) - (xx' + yy' + zz')2= The second member of this identity can never be negative because it is the sum of three squares and can only be zero when each of the parentheses is zero, i. e., when (1) yz'y'z = 0, (2) zx' z'x 0, (3) xy' x'y 0. Divide equation (1) by y'z', equation (2) by z'x', and equation (3) by x'y' and obtain 12 Hence (x2+ y2 + z2) (x22 + y22 + z12) > (xx' + yy' + zz')2, 1st. Let us suppose that x does not lie between 2 and 5, is not equal to either of these values. and Then (x − 2) (x — 5) is positive, and we may-multiply by this factor without reversing the signs of the inequality (8329, 332, Ex. 2). according as (3x-1)(x-5)+(2x-3) (x-2)> or <5(x-2) (x-5), according as 5x23x + 11 > according as according as or <5x35x+50, Under the present supposition, a can not have the value 31, but it follows from the preceding inequalities that if x 5, then F> 5, and if x2, then F 5. 2d. Suppose 2 < x < 5. In this case (x 2) (x-5) is negative and the sign of inequality is reversed (2330) and we must reverse all the signs in the preceding inequalities after multiplying by (x − 2) (x −5). It follows therefore that if 2 < x < 31, then F> 5, and if 34 < x < 5, then F < 5. 335. Powers and Roots. The following principles are deduced by means of the lemmas (i). If both members of an inequality are positive and are raised to the same integral power, the resulting inequality subsists in the same sense; that is, if ab, then a" > b", in which a and b are positive, and n is a positive integer. For, if a"b", then by definition, 324, or an (1) (a—b) (a-1+an-2b+a"-312 + ··· +abn−2+bn-1)>0. [102, i] Since ab then ab is positive by definition, and since a and b are positive by hypothesis, the second parenthesis in inequality (1) is positive, Q. E. D. E. g., 75, and 49 > 25. A similar proof holds for the two following principles: (ii). If both members of an inequality are negative and are raised to the same positive odd power, the resulting inequality subsists in the same sense; that is, if E. g., a>—b, then (— a)2n+1>(—b)2n+1 when n=1, 2, 3, 4 -5-6, and (-5)>(-6)3, or -125-216. (iii). If both members of an inequality are negative and are raised to the same positive even power, the resulting inequality will be of the opposite species; that is, if E. g., a b, then (-a)2n < (— b)2n. -35, and (3) (-5), or 81 < 625. (iv). If the same principal root of both members of an inequality is taken, the resulting inequality subsists in the same sense; that is, E. g., 169 49, and 137; -6427, and - 4 <- 3. 16949, (v). If the same negative even root of both members of an inequality is taken, the resulting inequality subsists in contrary sense; that is, if E. g., 2n a>b, then "Va < −2" Vb. 16949, then 13 <-2. The proof is similar to that of (iv). EXERCISE LVIII Prove the following inequalities, supposing that all the letters represent positive quantities: 3. 6x+7x-3> 6 x2 + 17 x 13 if x < 1. 4. (x+2)(x+3) > (x − 4) (x-5) if x > 2. Find the limits of x in the following: 5. (6x+1)+15 > (2x-3) (18x+5). 6. (5x-1) 20 (3x+4)2 + (4 x − 3)2. 7. (x+2) (x-3) (x + 4) < (x − 1) (x + 2) (x + 4). 8. For what value of y is +b positive, and a > b? a y а +2, if a and b are 11. Find the limits of x when 6x+8 < 9 x 16x 25 < 12x + 5. 12. When will a3b + ab3 > 2a2b2? Prove that 13. a6b> b (2a + 5 b) if a> b. 13 and 14. (mn) · (m1 + n1) > (m3 + n3)2 if m>n. 2 > ab. 15. ab2+al>2 ab3 if a>b. 16. (a + b) * >> 2 17. ab (a+b) + bc (b + c) + ca (c + a) > 6 abc. 27. ax+by+cz <1 if a2+b2+ c2 = 1 and x2 + y2 + z2 = 1. 28. bc (b+c) + ca (c + a) + ab (a + b) < 2 (a3 + b3 + c3) · 29. x3+ y3+23 >3 xyz if x + y + z > 0. 30. If x=a+b2 and y=c+d2, xy> ac+bd, or xy> ad+be. 31. If ab,am_bmmam-1(a - b), m being a positive integer. 32. - (a+b-c) (a+c—b) (b+c-a) <abc. 33. x1y + y1z + z1x > xy1 + yz1 + zx1. 34. (y-z) (z — x) + ( z − x) (x − y) + (x − y) (y — z) ≥ 0. 35. yzzx + xy = x2 + y2 + z2. CHAPTER V IRRATIONAL NUMBERS AND LIMITS 336. The System of Rational Numbers Insufficient.-We have seen in Books I and II that a system of rational numbers is sufficient for the use of the four fundamental operations and supplies the means for expressing the solution of all problems which can be solved by these operations. However, the system of rational numbers does not fully meet the needs of Algebra. A great central problem in Algebra is the equation. A number system which is algebraically complete should supply the means for expressing the solution of all possible equations. The system of rational numbers enables one to express the solution of equations of the first degree in one, two, or more unknown quantities (Book II); but it does not even contain symbols for the roots of such elementary equations of higher degrees as To solve equation (1) extract the square root of both members, then x= ±√2= ± 1.4142 [8291] The value of the symbol V2 can only be found approximately by the device of extracting the square root. According to the number of decimal places, to which this operation may be carried, the V2 is expressed approximately by a rational number; if to three places, then by To solve equation (2), extract the square root of both members, then The V-1 can not be expressed in terms of rational or irrational numbers, and belongs to another system of numbers called imaginary or complex numbers (2104, Ex. 3). But how is the system of rational numbers to be enlarged into a system of algebraic numbers which will give us the means to express the roots of all possible equations, and at the same time be sufficiently simple? |