~~指數運算律疑問~~

Actually, a^(mn)=(a^m)^n is true for m,n being rational numbers and a>0.

According to Bolzano's Theorem:
If a function f(x) is defined and is continuous over a closed bounded interval [a,b] and the values of the endpoints a and b are of different signs, then at least one point c exists in the open interval (a, b) where the value of the function is 0 and the graph cuts the X-axis.

Then, we have the deduction: x^n=a (a>0) has one real positive root x=a^(1/n), where n is a positive integer. However, this does not hold when a<0 and also when a<0, the equation x^n=a may not have real root as the root may appear as a complex number, as a result, it is not practical for we to use the rule a^(mn)=(a^m)^n when a<0 and m/n is a frational number.

Hence, to avoid the presence of complex numbers, we have to define that a^(mn)=(a^m)^n is for a>0 when m/n is a fractional number.

2010-09-06 22:47:28 補充：
Actually, my reference is the first word file at http://hk.search.yahoo.com/search?p=1-1+%E6%8C%87%E6%95%B8.doc&fr2=sb-top&fr=FP-tab-web-t&rd=r1. You can take a look~

早為你計
1 ^ N時
你已把N當作大於一ge數
所以你將1/2作入N會唔同左
1^(1/2)
=±√1
=±1

2010-09-06 22:10:30 補充：
因為
唔係早