0
$\begingroup$

If x can be a positive real number (for example a fraction with a numerator and denominator), then why does the following relationship hold true only if and only if a and b are strictly positive real numbers? In other words, why doesn't this relationship also hold true if a and b are complex or negative? See this link for power of product. https://proofwiki.org/wiki/Exponent_Combination_Laws

$\endgroup$
  • $\begingroup$ I think the restriction a, b > 0 ensures that a^x and b^x are defined. If a was allowed to be negative, then a^(1/2) wouldn't be defined, for example. $\endgroup$ – Aegis Apr 18 '16 at 17:34
  • $\begingroup$ So would the relationship still hold true if complex numbers were allowed? $\endgroup$ – W. G. Apr 18 '16 at 17:38
1
$\begingroup$

(I AM COMMENTING IN THE ANSWER SECTION BECAUSE I DON'T HAVE 50 REPUTATION)

1) (2i)^2=-4 and ((2)^2).((i)^2)=4.(-1)=-4

2) (ii)^3=-1 as well as ((i)^3).(((i)^3)=-1

3) By partial deduction we can say (ab)^{x} =a^{x}b^{x} holds for for all a,b and c where a and b belongs to the set of complex number(Z) and x to R+.

4) I am extending his question: Prove or disprove the statement (by any method, most preferably the shortest one): we can say that (ab)^{x} =a^{x}b^{x} holds for for all a,b and c belonging to the set of complex number(Z).

$\endgroup$
  • $\begingroup$ I appreciate Aegis, he says: I think the restriction a, b > 0 ensures that a^x and b^x are defined. If a was allowed to be negative, then a^(1/2) wouldn't be defined, for example., in respond to original question. Aegis, Please, do consider my further extended question. $\endgroup$ – Shivanshu Gupta Apr 18 '16 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.