defining the value of real numbers raised to rational exponents if $b^{1/n}$ where $b$ is a real number and $n$ even is defined as the positive real solution to the equation $x^n=b,$ how did they suddenly decide to define the value of $b^{m/n}$ where $m$ and $n$ are integers (rational exponent). did they calculate $(b^{1/n})^m$ and then used the multiplicative property of exponents assuming its still applies even when $m\ne n.$
Thanks
 A: When $m,n$ are positive integers we have $x^m\cdot x^n=x^{(m+n)}$ and $(x^m)^n=x^{(mn)}$ for every real $x.$
It was found interesting and useful to extend this property to all  integers $m,n.$ To be consistent, this requires $x^0=1$ when $x\ne 0,$ and $x^{(-m)}\cdot x^m=1$ when $x\ne 0.$
It was found to be interesting and useful to further extend this property to all rational $m,n.$  To be consistent, this requires $(x^{(1/m)})^m=x$  when $x>0$ and $m\in \Bbb Z^+,$ and $(x^{(1/n)})^m=x^{(m/n)}$ when $x>0.$  
It was  found interesting and useful to further extend this property to all $m,n \in \Bbb R^+$ by defining $x^y=\lim_{k\to \infty}x^{(q_k)}$ for any sequence $(q_k)_{k\in \Bbb N}$ of rationals converging to $y,$ when $x>0.$
Some restrictions are necessary. For example, there is no real number $x^{(1/2n)}$ when $x<0$ and $n \in \Bbb Z^+.$ And $0^y$ is not defined when $y< 0.$
It is sometimes convenient to define $0^0=1,$ because $x^x$ converges to $1$ when $x$ converges to $0$ through positive values. This definition is often implicitly assumed : When an author writes "Consider a polynomial $p(x)=\sum_{j=1}^nA_jx^j$ " you may very well ask what this means when $x=0$ because the term $A_0x^0$ is $A_0\cdot 0^0$ when $x=0.$ The author is defining $0^0=1$.
