Prove $\forall n\in \mathbb{N} ,\,(ab)^{1/n} = a^{1/n} \times b^{1/n}$

Question

Prove $\forall n\in \mathbb{N} ,\,(ab)^{1/n} = a^{1/n} \times b^{1/n}$

I tried putting

\begin{align}(ab)^{1/n} &= c\\ a^{1/n}&=d\\ b^{1/n}&= e\end{align}

and then \begin{align}ab &= c^n\\a &= d^n\\b &= e^n\end{align}

That leads to $$d^n \times e^n = c^n$$ and now I've got stuck here

Answer 1

You need only to show that

• $ \left( a^{\frac{1}{n}}\times b^{\frac{1}{n}} \right)^n = ab$

Using rules for exponents you get $$\left( a^{\frac{1}{n}}\times b^{\frac{1}{n}} \right)^n = \left( a^{\frac{1}{n}} \right)\times \left(b^{\frac{1}{n}} \right)^n = ab$$

Edit after comment:

Using associativity you get $$\left( c \cdot d \right)^n = \underbrace{(c \cdot d) \cdot \ldots \cdot ( c \cdot d)}_{n} = \underbrace{(c \cdot \ldots \cdot c)}_{n} \cdot \underbrace{(d \cdot \ldots \cdot d)}_{n} = c^n \cdot d^n$$