TITLE: Exponent laws for rational numbers
QUESTION [0 upvotes]: this seems to be easy but I dont manage to prove it properly.
You are only allowed to use the following laws:
$$\left(z^m\right)^n=z^{m\cdot n}\quad \text{for $n,m\in \mathbb N$}$$
and
$$z=a^{\frac 1n}\quad\Leftrightarrow\quad z^n=a\quad\text{for }n\in \mathbb N. $$
Now prove that for all  $a,b\in \mathbb R$ and for all $m,n\in \mathbb N$ we have
$(a)\quad a^{\frac 1m}\cdot a^{\frac 1n}=a^{\frac 1m+\frac 1n}$
$(b)\quad \left(a^{\frac 1m}\right)^{\frac 1n}=a^{\frac 1m\cdot \frac 1n}$
$(c)\quad a^{\frac 1n}\cdot b^{\frac 1n}=(a\cdot b)^\frac 1n$
$(d)\quad (a^m)^{\frac 1n}=\left(a^{\frac 1n}\right)^m$
Do you know how to do these?

REPLY [1 votes]: b) $z=\left(a^{1/m}\right)^{1/n}\iff z^n=a^{1/m}\iff (z^n)^m=a\iff z^{nm}=a\iff z=a^{1/nm}$ and
$$\frac1{nm}=\frac1n\frac1m.$$
d) proceed as above up to $z^{nm}=a$, exchange $n,m$ and backtrack.
a), c) cannot be proven with only the two given rules.
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