How is this $\frac{m}{n}$ equals to $m \times \frac{1}{n}$? Any logical proof for this? Which draws this conclusion:
A fractional exponent like $\frac{m}{n}$ means do the $m^{\text{th}}$ power, then take the $n^{\text{th}}$ root or take the $n^{\text{th}}$ root and then do the $m^{\text{th}}$ power.
I assume we're taking $m, n \in \mathbb{Z}$, $n \neq 0$.
First, note that $$m\times \frac{1}{n} = \frac{m}{1}\times\frac{1}{n} = \frac{m\times 1}{1\times n} = \frac{m}{n}.$$
Now $x^{\frac{m}{n}} = x^{m\times\frac{1}{n}}$ as $\frac{m}{n} = m\times\frac{1}{n}$. Now, using the rule $x^{a\times b} = (x^a)^b$ we see that
$$x^{\frac{m}{n}} = x^{m\times\frac{1}{n}} = (x^m)^{\frac{1}{n}}.$$
Now we use the rule that $x^{\frac{1}{n}} = \sqrt[n]{x}$ and see that $x^{\frac{m}{n}} = (x^m)^{\frac{1}{n}} = \sqrt[n]{x^m}$.
To see that we also have $x^{\frac{m}{n}} = (\sqrt[n]{x})^m$, note that multiplication is commutative (the order doesn't matter), so $m\times\frac{1}{n} = \frac{1}{n}\times m$. Therefore $$x^{\frac{m}{n}} = x^{m\times\frac{1}{n}} = x^{\frac{1}{n}\times m} = (x^{\frac{1}{n}})^m = (\sqrt[n]{x})^m.$$
It is the same; you just change the order of the operations but it does not affect the result. Using logarithms would show it in a simple way.
If you are making a number bigger by multiplying by $m$, then making it smaller by dividing it by $n$, it will be the same as making it smaller by $n$ then making it bigger by $m$. I guess it's just how numbers work.
