-a by the rules of multiplication, is +aa or +a2; but if we involve to the 3d power, we have (—a)×(—a)×(—a), which, by the same rules, is -aaa or—a3. It is evident that the 4th power of a is +a*; the 5th power of a is —a3, &c. ; the even powers being positive and the uneven powers negative. We shall thus find that (-a)90=+aoo, and (-a)-a31, &c. &c. 20 15. We see then that the squares of +a and -a are both +a3. If then it be required to extract the square root of +a2, this root may be either +a ora; so that the result is ambiguous, which is expressed by prefixing both signs thus, a; that is, aa. But in extracting the cube root no ambiguity will exist, for the cube of +a and the cube of -a are not the same, the one being +as and the other -a; and, consequently, the cube root of +a3 is+a, and the cube root of —a3 is -ɑ. In the same way the 4th root of +a+ will be either +a or —a, but the 5th root of +ɑ5 will be +a, and the 5th root of -a5 will be -α. Hence the rule, the even root of a positive quantity is either positive or negative; the uneven root of a positive quantity is positive; and the uneven root of a negative quantity is negative. 16. The even root of a negative quantity is impossible. For the square root of a3 is neither a nor -a; since +a squared is a3, and a squared is +a2; and there is no quantity which, multiplied by itself, will produce a3. For the same reason /—a, -,/-1 are impossible. These quantities, or rather expressions, are called imaginary. It is usual to reduce them as follows: in which abc positive quantity. To find the square root of ―a2, we have √/—a3=√a3×(−1)= a-1, (Arts. 11 and 15,) in which last form the imaginary part is the even root of -1; and to this form all imaginary expressions may be reduced. Thus, is a possible quantity, being the even root of a 17. 1. Find the product of -"y"z", x"у3z",—X3у"2". 2. Divide xyz by x11уz. 3. Divide ―axbxx by —ɑnybn≈. 4. Involve -b to the nth power. 5. Involve -b to the (2n)th power. Ans. +ben (for 2n is even whatever be the value of n.) EXAMPLES. Ans. xm+n+rym+n+ezm+nte or (xyz)m+n+r. Ans. (xm―nyn-z—m)P. 6. Involve -b to the (2n+1)th power. 7. Find the 4th root of Ans. +b ifn is even, -b" if n is uneven. 8. Find the 3d root of Ans. -b3n+1 (for whatever be the value of n, 2n is even, and 2n+1 is uneven.) 16x4y8 a3mn am+n7° ax3y3 10. Find the nth root of amn. 2 ———— Ans. a -1. Ans. x2my. 2n+1 Ans. a. Ans. y 2n Ans. my−1. Ans. 3abc/5abc3. CHAPTER II. EXPONENTS IN GENERAL. 18. In the preceding chapter we have applied the rules for the multiplication, division, &c. of powers only to those cases in which the exponents were whole positive numbers. Exponents, however, may be also fractional and negative. 3 19. Let it be required to extract the 3d root of a2. We have seen that this would be a3, but algebraists have agreed to apply to such cases the general rule for the evolution of powers given in Art. 9. To extract the 3d root of a2, according to this rule, we divide the exponent by 3, which gives a3; therefore, ✓a3 and a3 are equivalent expressions, both signifying "the 3d root of a squared;" and in the fractional form, the numerator of the fraction indicates the power, and the denominator the root. In the same 2 3_ n 3 way we find ✅a=a3; Ja=a3; √a°=a3; &a=am, &c. The use of fractional exponents thus renders the radical sign unnecessary, and enables us to express roots and powers by the same general mode of notation. a5 20. Again, let it be required to divide a5 by a7. This is, which a7' fraction reduced by dividing its numerator and denominator by a5, • 1 becomes But by Art. 6, powers of the same quantity are dia2 vided by subtracting their exponents. Hence, to divide a5 by a7, we must subtract 7 from 5; but 5-7-2, so that the quotient 1 will be a with the exponent -2, or a-". Therefore, and a-2 are a2 equivalent expressions; but the latter has the advantage of representing a fraction in the same form with whole numbers. In the same way 1 1 1 we shall find :α-7; α-"; =α-m. 16 EXPONENTS IN GENERAL. We see then that the negative exponent indicates the reciprocal of a power; that is, a-6 is the reciprocal of a®; a is the reciprocal of a7; a-m is the reciprocal of am, &c. 21. We apply the rules given in the preceding chapter also to fractional and negative exponents. Thus, to multiply a3 by a we add the exponents (Art. 5,) and we have ++_a8. axa-a In the same manner we find, m απχα mq-np aman =α ng • 22. Applying the rule for division of powers, (Art. 6,) we have a33 a m m an $ α L mq+np Р mq-np (a−m)n—a—mn. (am)—n—a—mn. (am)"=ɑTMn. 23. Applying the rule for involution of powers, (Art. 8,) we have 3 (a3)3=u&×3=a§. m m mp Xp (an) an =an 24. Applying the rule for evolution of powers, (Art. 9,) we have 3 1 ✓8a=2až 3_2a3. m m 1 p Ba 25. By using fractional exponents involution and evolution are performed by the same rule; that is, by multiplying the exponents. Thus, the above examples will be performed, 3 8a=(8a3) (Art. 19,) =8aaa×† (Art. 8,)=2a* m m m p ✔an=a np. mq √(am)1=a3· mp m ng. mp ng. mp =ang. ÷3 26. We conclude this chapter by exhibiting the regular series of powers of a, decreasing by unity. Beginning at a" and dividing continually by a we have the following series: am, am a¬m+3, a−m+1, a—m. which is a regular series of powers of a from +m tom. Any two terms equally distant from ao are the reciprocals of each other; thus, a1 is the reciprocal of a1, a3 of a-2, &c., and am of a-m. The term ao is found (as the other terms are found) by subtract |