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Unformatted text preview: Rational exponents — an application of the intermediatevalue theorem. It is a consequence of the intermediatevalue theorem. that; given a positive integer n and a real nuaber a > 0, there is exactly one real number b 2 0 such that
bn = e. We denote b by "I3, and cell it the nﬁg root of a. (See Theorem 3.9. p. 145 of Apostol.)
It follows fro. the general theorel about continuity of inverses that the n32 root function, defined by the rule
f(x) = DI? for x 2 0, is continuous. (See Theorem 3.10. p. 147 of Apostol.)
Now (finally!) we can introduce rational exponents. We do
so only when the base is a gositivg real nuaber. Definition. Let r be a rational nulher; let a be e Eositivg real nuaber. We can write r = a/n. where s and n are integers and n is gositive. We then define a” = ("5)! (Here we use the fact that "IE is nonzero, so_ u can be
negative.) He must show that this definition makes sense. A problem
might arise from the fact that the number r can be represented
as a ratio of integers in many different ways. We must show that
the value of ar does not depend on how we represent r. This
is the substance of the following lemma. Lemma 1. Su ose m/n = p/q, where n, n, p, q gig
lateness. and n and q as: Egsitixs 1223 ("I5)“ = (“IE)P. Proof. Let c = "IE and d qIE. Then a = on and a = dq. by definition. Because I/n = p/q, we have nd = np. Using
these facts, we compute \
ap = (en)p = cnp = cmq = (cm)q. and
ap = (dq)P = a“? = (d9)q, so that (c")‘1 = up)“.
(We use here the laws of integral exponents.) We conclude (by
uniqueness of the gig roots) that cl. = dp, or On the basis of Lemma 1, we know that ar is welldefined if r is a rational number and a is positive. In particular, we have the equation al/n = nIE'
by definition. The definition of an/n can then be written in
the forn
an/n = (al/n)l. Consider now the three basic laws of exponents. We already know that these laws hold in the following cases: (1) positive integral exponents; arbitrary bases.
(ii) integral exponents; nonzero bases.
We now consent that these laws also hold in the following case:
(iii) rational exponents; positive bases. The proof is not difficult, but it is tedious. It is given in Theorem 2 following. Later on, we shall extend our definition to arbitrary real
exponents; that is, we shall define sx when x is an arbitrary real nunber (and a is a positive real nunber). Furthermore, we
shall verify that the laws of exponents also holds in this new situation; i.e., in the case: O.
A (iv) real exponents; positive bases.
So you can skip the proof of Theorem 2 if you wish, for we are
going to prove the more general result involving real exponents
later on. Before proving Theorem 2, we make the following remark about
negative bases: If a is negative, one can still define "IE
provided n i_ ggg. For in that case there exists exactly one
real number b such that bn = a. We shall define “I? = b in
this case. It is tempting to use exponent notation in this
situation, defining an/n = (“I3)m if n is odd and a is
negative. However, phi; practice is dangerous! For the laws of exponents do not always hold in these circumstances. For example. if we used this definition, we would have ((8)2)1/6 2, while (8)1/3 = 2. Thus the second law of exponents would not hold in this situation. For this reason, we make the following convention: ﬁg shall use rational exponent notation only when the base 1; positive. Now we verify the laws of exponents for rational exponents and positive bases. Theorem 2. If r and s are rational numbers. and ;§ 3 and b are positive real numbers, then (i) eras = ar+’. (ii) (a’)‘ = a’s.
(iii) a’b’ = (ab)r.
Proof. Let r = n/n and s = p/q, where m, n, p. integers, and where n and q are positive. To prove (i), we note that a a = am/n aP/q amq/nq anP/nq = (an;)q (n9;;)np by definition. = (anE)'q+np by (iii) for integral exponents,
= a("q+“P)/“q by definition,
z ar+s. To prove (ii), we verify first that REF. <"IE)' = by definition. We compute q 6.5 are 6.6 ‘X
by (ii) for integral exponents. By uniqueness of >n££ roots, we
have nlan = cm = (nr~)n’
as desired.
It now follows that
(i) am/n 3 (al/n)n = (an)l/n.
The first eQuation follows from the definition of am/n, and the second from what we just proved. The formula (t) is of course ' ~\
special case of our desired formula (ii). Now we prove (ii) in general: Let c = (at)s : (am/n)p/q. Then
c = («JP/“)5”q by (t) (applied twice) = (<(a')’)1’“)1’q by <:). It follows that Then that To check (iii), (cq)n C let c <ch)n cq qn .. q = (alp)1/nq M an O ab ( a n amp/nq d ((am)p)1/n. and (am)p, by definition, so that up by (ii) for integral exponents. by definition, by definition, by (t). We first note by (iii) for integral exponents; by definition. .7 It follows that cd = nJaE. We then prove (iii) as follows: am/nbm/n II <“J‘E>“<“IF)" = c'd‘“ (cd)n Hm)" (ab)I/n by definition, by (iii) for integral exponents, by (¥). by definition. Thus the three laws hold for rational exponents. a. n ...
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 Spring '09
 BRUBAKER
 Exponents, Apostol

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