As @J. W. Tanner pointed out in his comment, $a^{m/n}$ is not defined in $\mathbb{R}$ when $a<0$ (and not uniquely defined in $\mathbb{C}$ for all $a$s). That's why softwares usually mess up with things like $(-2)^{2.5} = (-2)^{5/2}$.
Now, you should ask why $a^{m/n}$ is not defined in $\mathbb{R}$ when $a<0$. The motivation of this fact is not trivial, and depends on the importance Mathematicians have attributed to exponentiation identities as $(a^x)^y = a^{xy}$ or $a^{x+y}=a^x\cdot a^y$.
As you should know, the power $a^{1/n}$ (with $a\geq 0$ and $n \in \mathbb{N}$) is defined through the following theorem:
For each $a \geq 0$ and $n \in \mathbb{N}$, there exists a unique $\alpha \geq 0$ s.t. $\alpha^n = a$.
Such an $\alpha$ is called the arithmetic $n$-th root of $a$ and denoted with $\sqrt[n]{a}$ or $a^{1/n}$.
whose proof heavily relies on the completeness of $\mathbb{R}$. Theorem allows you to define the fractional power $a^{m/n}$ with $a\geq 0$ (or $a > 0$ when $m/n < 0$) by letting:
$$a^{m/n} := \sqrt[n]{a^m}\quad \text{(or equivalently } a^{m/n} := (\sqrt[n]{a})^m \text{)}$$
for each $m/n \in \mathbb{Q}$ (it is easy to prove that $\sqrt[n]{a^m} = (\sqrt[n]{a})^m$, hence definition does not depend on the order of application of $m$-th power and $n$-th root).
What happens if constraint $a\geq 0$ is dropped? The theorem cannot remain true for every value of the exponent $n \in \mathbb{N}$: in particular, if $n$ is even (i.e., $n=2,4,6,\ldots$) then $\alpha^n \geq 0$ for all $\alpha \in \mathbb{R}$, therefore equality $\alpha^n = a < 0$ is out of question for even $n$s. On the other hand, the situation for odd $n$s is straightforward:
When $n \in \mathbb{N}$ odd (i.e., $n=1,3,5,\ldots$), for each $a<0$ there exists only one $\alpha < 0$ s.t. $\alpha^n = a$, precisely:
$$\alpha = -\sqrt[n]{-a}\quad \text{(or equivalently } \alpha = -\sqrt[n]{|a|}\text{)}\; . $$
Previous statement allows you to define the arithmetic $n$-th root of $a$ also when $a<0$ and $n \in \mathbb{N}$ is odd by setting:
$$\tag{*} \sqrt[n]{a} := - \sqrt[n]{-a}\; ,$$
but doesn't allow you to define the fractional power $a^{1/n}$, nor $a^{m/n}$ when $a<0$!
In fact, it happens that the definition of rational power with base $a<0$ (by means of $a^{m/n} = \sqrt[n]{a^m}$) is incompatible with usual exponentiation identities, i.e. it causes failure of usual rules like $(a^x)^y = a^{xy}$.
In order to see this, consider $a=-1$ and use (*) to get:
$$(-1)^{1/3} = \sqrt[3]{-1} \stackrel{\text{def.}}{=} - \sqrt[3]{-(-1)} = -\sqrt[3]{1} = -1\; ;$$
if usual exponentiation identities were in force then you would get:
$$-1 = (-1)^{1/3} = (-1)^{2/6} = \left[ (-1)^2 \right]^{1/6} = \left[ 1 \right]^{1/6} = 1$$
which is clearly wrong (for $-1 \neq 1$!), or oddities like:
$$-1 = (-1)^{1/3} = (-1)^{1/6 + 1/6} = (-1)^{1/6} \cdot (-1)^{1/6}$$
whose rightmost side has no meaning at all.
Therefore, there is a problem here: fractional powers with negative base and usual exponentiation identities do not fit together.
Mathematicians think it is way better to choose exponentiation identities to hold over the possibility of giving a definition to the symbol $a^{m/n}$ with $a<0$, because identities are of fundamental importance and almost ubiquitous in every possible kind of computation. ;-)