table of Contents
1. The power of a positive integer index
Rational Field \ (\ Bbb {Q} \ ) is an ordered domain of the four arithmetic closure (rational numbers do finite addition, subtraction, multiplication, the result of the operations or rational addition). For any \ (r \ in \ Bbb {Q} \) and any \ (n-\ in \ BBB {N} _ + \) , starting from the multiplication may be defined positive integer exponent exponentiation:
\ [R & lt ^ n-= \ underbrace {R & lt \ CDOT R & lt \ cdot r \ cdot \ cdots \ cdot r} _ {\ text {$ n $ items}}. \]
using the commutative and associative law of multiplication, it is easy to prove the following proposition
Proposition 1: for any \ (R & lt, S \ in \ of Bbb {Q} \ setminus \ {0 \} \) , any \ (m, n-\ in \ _ of Bbb {N} + \) , both
\ [( r \ cdot s) ^ n = r ^ n \ cdot s ^ n, \ quad r ^ {mn} = (r ^ m) ^ n, \ quad r ^ {m + n} = r ^ m \ cdot r ^ n. \]
2. The power of positive rational exponent
Rational Field \ (\ Bbb {Q} \ ) is not satisfied on the supremum principles, but on the "square root" operation is not closed, for example,
\ [\ {r \ in \ Bbb {Q} \ | \ r ^ 2 = 2 \ } \]
is the empty set.
Using Dedekind split, rational domain may be \ (\ Bbb {Q} \ ) expanded into real domain \ (\ Bbb {R} \ ) , so \ (\ Bbb {R} \ ) established principle of exact bounds on, and \ ( \ Bbb {Q} \) is \ (\ Bbb {R} \ ) ordered subdomain. four algorithms from rational field \ (\ Bbb {Q} \ ) extended to the field of real numbers \ (\ Bbb {R} \ ) , the corresponding, rational positive integer exponent exponentiation also extends to a positive integer exponent exponentiation real number.
Then, in the real domain \ (\ Bbb {R} \) "square root" operation whether closed on? To answer this question, we first describe the operation about the nature of "evolution."
Proposition 2: for any \ (y> 0 \) and any \ (n-\ in \ of Bbb {N} _ + \) , equation
\ [y = x ^ n \
] there exists a unique positive real root \ (X \) Weigh the positive real root \ (X \) is the \ (Y \) a \ (n-\) th root, referred to as
\ [x = \ quad \ textrm y ^ {frac1n \} { or} \ quad x = \ sqrt [n] {y} . \]
\ (\ red {proof:} \)
Step1 Order.
\ [E = \ {T \ in \ of Bbb {R & lt} \ | T> 0 \ TEXTRM {and} t ^ n <y \}
\.] When \ ( y> 1 \) , the order \ (T_0. 1 = \) ; if \ (0 <y \ leq 1 \) , the order \ (t_0 = \ frac12 y \ ) is easy to prove, in both cases the total. there \ (T_0 \ in E \) , so \ (E \) is a non-empty set of numbers. order \ (L =. 1 + Y>. 1 \) , then \ (L \) is \ (E \) of a upper bound. otherwise, the presence \ (t_1 \ in E \) such that \ (T_l> L =. 1 + Y \) , then
\ [y> t_1 ^ n> L ^ n \ geq L ^ 2 = 1 + 2y + y ^ 2> y, \]
contradiction.
In summary, \ (E \) is a nonempty set number, thus supremum.
Step2. Remember \ (C = \ SUP E \) , then \ (C> 0 \) . Under syndrome \ (Y = n-C ^ \) .
Contradiction, assuming \ (c ^ n <y \ ) or \ (n-C ^> Y \) .
(1) If the \ (C ^ n-<Y \) , then \ (C \ in E \) . And because \ (\ SUP E = C \ in E \) , so \ (C = \ max E \) . for any \ (\ Epsilon> 0 \) , \ (+ \ Epsilon \ C) must not \ (E \) on the other hand, [when \ (\ Epsilon \) very small, \ ((C + \ Epsilon ) ^ n \) and \ (c ^ n \) is very close, it is possible to ensure \ (C ^ n-<(C + \ Epsilon) ^ n-<Y \) , so \ (c + \ epsilon \) in \ (E \ ) to obtain a contradiction]
\[\begin{array}{rcl} &&(c+\epsilon)^n-c^n\\ &=&\epsilon\cdot \left[(c+\epsilon)^{n-1}+ (c+\epsilon)^{n-2}c+\cdots (c+\epsilon)c^{n-2}+c^{n-1}\right]\\ &\leq&\epsilon\cdot \underbrace{\left[(c+\epsilon)^{n-1}+ (c+\epsilon)^{n-1}c+\cdots (c+\epsilon)^{n-1}+(c+\epsilon)^{n-1}\right]}_{\text{$n$项}}\\ &=&n(c+\epsilon)^{n-1}\epsilon. \end{array}\]
当\(0<\epsilon<1\)时, 就有
\[(c+\epsilon)^n-c^n\leq n(c+\epsilon)^{n-1}\epsilon\leq n(c+1)^{n-1}\epsilon.\]
当\(0<\epsilon<1\)并且
\[0<\epsilon <\frac{y-c^n}{ n(c+1)^{n-1}}\]
时, 就有
\[(c+\epsilon)^n-c^n\leq n(c+1)^{n-1}\epsilon<y-c^n,\]
此时\(c^n<(c+\epsilon)^n<y\), 从而\(c+\epsilon \in E\),This \ (c = \ max E \contradiction.
(2) if \ (n-C ^> Y \) , since \ (C = \ SUP E \) , then for any \ (\ Epsilon> 0 \) , \ (the C-\ Epsilon \) is not \ (E \ ) upper bound. on the other hand, [when \ (\ Epsilon \) very small, \ ((the C-\ Epsilon) n-^ \) and \ (c ^ n \) is very close, it is possible to ensure \ ( Y <(the C-\ Epsilon) n-^ <n-C ^ \) , this time again proved \ (c- \ epsilon \) is \ (E \) bound on a give contradictory]
\ [\ begin { array} {rcl} && c ^ n- (c + \ epsilon) ^ n \\ & = & \ epsilon \ cdot \ left [c ^ {n-1} + c ^ {n-2} (c- \ epsilon) + \ cdots c (c + \ epsilon ) ^ {n-2} + (c- \ epsilon) ^ {n-1} \ right] \\ & \ leq & \ epsilon \ cdot \ underbrace {\ left [c ^ {n -1} + c ^ {n- 1} c + \ cdots c ^ {n-1} + c ^ {n-1} \ right]} _ {\ text {$ $ items}} \\ & = & nc n ^ {n-1} \ epsilon.
\ end {array} \] when\ (0 <\ epsilon <c \) and
\ [0 <\ epsilon <\
frac {c ^ ny} {nc ^ {n-1}} \] when there
\ [c- \ epsilon> 0, \ quad y <(c- \ epsilon)
^ n <c ^ n. \] At this time \ (c- \ epsilon \) is \ (E \) an upper bound, or the presence of \ (t \ in E \) such that \ (T> the C-\ Epsilon \) , then
\ [y <(c- \ epsilon
) ^ n <t ^ n <y, \] contradictory, but this in turn \ (c- \ epsilon \) is not \ ( E \) is the upper bound of contradictions.
In summary, \ (C = \ E SUP> 0 \) is the equation
\ [y = x ^ n \
] is a positive real root.
Step 3. Suppose \ (c_1, c_2 \) is the equation \ (y = x ^ n \ ) a positive real root, and \ (c_1 and> c_2 \) , then
\ [y = c_1 ^ n> c_2 ^ n = y, \]
contradiction, so the equation \ [y = x ^ n \ ] positive real root has one and only one.
\ (\ Box \)
推论1. 设\(a>0\), \(m,n,p,q\in \Bbb{N}_+\)并且\(r=\frac{m}{n}=\frac{p}{q}\), 则
\[\left(a^\frac{1}{n}\right)^m=\left(a^m\right)^{\frac1n}=\left(a^p\right)^{\frac1q},\quad \left(a^\frac{1}{n}\right)^\frac{1}{m}=a^\frac{1}{mn}.\]
\ (\ Red {Proof:} \)
Set \ (S = \ left (A ^ m \ right) ^ {\ frac1n} \) , \ (T = \ left (A ^ P \ right) ^ {\ frac1q} \) , according to the conditions and Proposition 2 shows that, \ (s, t> 0 \ ) and
\ [s ^ n = a ^
m, \ quad t ^ q = a ^ p, \] further available
\ [s ^ {nq} = a ^ {mq}, \ quad t ^ {qn} = a
^ {pn}. \] Since \ (\ FRAC {m} {n-} = \ FRAC {P} {Q} \) , then \ (MQ = PN \) , so
\ [ s ^ {nq} = a ^
{mq} = a ^ {np} = t ^ {qn}, \] which represents \ (S \) and \ (T \) is a positive (i.e. \ (A MQ} {^ \) ) of \ (NQ \) th root, proposition 2, can be obtained by the uniqueness
\ (S = T \) , i.e.
\ [\ left (a ^ m \ right) ^ {\ frac1n } = \ left (a ^ p \ right) ^ {\ frac1q}. \]
设\(u=a^{\frac1n}\), 则\(u^n=a\), 进而
\[a^m=(u^n)^m=u^{mn}=(u^m)^n,\]
所以
\[\left(a^\frac{1}{n}\right)^m=u^m=(a^m)^\frac{1}{n}.\]
Set \ (V = A ^ \ FRAC {A} {Mn} \) , then \ (A = U ^ m V ^ {Mn} = \) , so
\ [v = (u ^ m ) ^ \ frac {1 } {mn}. \]
Since \ (\ FRAC {m}} = {Mn \ n-FRAC {} {}. 1 \) , light of the foregoing conclusions, there
\ [a ^ \ frac {1 } {mn} = v = (u ^ m) ^ \ frac {1} {mn} = u ^ \ frac1n = \ left (a ^ \ frac {1} {n} \ right) ^ \ frac {1} {m}. \]
\ (\ Box \)
Inference from 1, we can define
\ [a ^ r = a ^
{\ frac {m} {n}} = \ left (a ^ m \ right) ^ {\ frac1n}, \] it will not rational \ ( R & lt \) different fractional (or about both irreducible) differ.
Corollary 2. set \ (A> 0 \) , \ (R & lt, S \ in \ _ of Bbb {Q} + \) , then
\ [(a ^ r) ^ s = a ^ {rs}, \ quad a ^ {r + s} = a ^ ra ^ s. \]
\ (\ Red {Proof:} \)
设
\[r=\frac{m}{n},\quad s=\frac{p}{q}.\]
根据推论1, 就有
\[\left(a^\frac{m}{n}\right)^\frac{p}{q}=\left[\left(a^\frac{m}{n}\right)^p \right]^\frac{1}{q}=\left[\left(a^\frac{1}{n}\right)^{mp} \right]^\frac{1}{q}=\left[\left(a^{mp}\right)^\frac{1}{n} \right]^\frac{1}{q}=\left(a^{mp}\right)^\frac{1}{nq}=a^\frac{mp}{nq},\]
即
\[(a^r)^s=a^{rs}\]
Set
\ [u = a ^ {\ frac {1} {n}}, \ quad v = a ^ {\ frac {1} {q}}, \ quad w = a ^ \ frac {1} {nq}, \]
the \ (W ^ NQ = A = U ^ n-= V Q ^ \) , so
\ [w ^ {nqmq} = a ^ {mq} = u ^ {nmq}, \ quad w ^ {nq np} = A ^ {NP} = V ^ {QNP}, \]
\ [(W ^ {MQ + NP}) ^ {NQ} = W ^ {nqmq} W ^ {nqnp} = U ^ {NMQ} V ^ { qnp} = (u ^ mv ^
n) ^ {nq}. \] proposition 2 the only conclusions, there
\ [w ^ {mq + np
} = u ^ mv ^ n, \] i.e.
\ [a ^ {r + s} = a ^ \ frac {mq + np} {nq} = w ^ {mq + np} = u ^ mv ^ n = a ^ \ frac {m} {n} a ^ \ frac {p A} = {Q} RA ^ S ^. \]
\ (\ Box \)
3. The power of a positive real number index
to be continued……