Exponent laws for rational numbers

Question

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?

Answer 1

For the first one you would write

$$\left(a^{\frac{1}{m}}\cdot a^{\frac{1}{n}}\right)^{mn} = a^{\frac{1}{m}\cdot mn}\cdot a^{\frac{1}{n}\cdot mn} = a^{n}\cdot a^{m} = a^{n+m}.$$

Now try to compute

$$\left(a^{\frac{1}{m}+\frac{1}{n}}\right)^{mn}$$

and then use that $z = b^{\frac{1}{mn}}\Leftrightarrow z^{mn} =b$

Answer 2

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.