# $\left(\frac{a}{b}\right)^\frac{1}{n}= \frac{a^\frac{1}{n}}{b^\frac{1}{n}}$

Problem is to prove this for all $$n\in \mathbb{N}$$ $$\left(\frac{a}{b}\right)^\frac{1}{n}= \frac{a^\frac{1}{n}}{b^\frac{1}{n}}$$ My attempt: $$x=\left(\frac{a}{b}\right)^\frac{1}{n}$$ $$x^n=\frac{a}{b}$$ $$b\cdot x^n=a$$ $$b^{(\frac{1}{n})}\cdot x =a^\frac{1}{n}$$ $$x=\frac{a^\frac{1}{n}}{b^\frac{1}{n}}$$ But problem is that I in 4 line use what I need to prove.

• I think you should be very careful to specify which axioms/theorems you're using at each step. – Theo Bendit Jun 25 at 6:27
• Theo is correct, you should specify which axioms you used. – topologicalmagician Jun 25 at 11:42

Fourth line in error. It should be $$b^{(\frac{1}{n})}\cdot x =a^{(\frac{1}{n})}$$ All else OK

• Are you sure it will be ok? Because in previous line $b\cdot x^n=a$ And I use $n$-th root $(bx^n)^\frac{1}{n} = b^\frac{1}{n}\cdot x$ – josf Jun 25 at 6:43
• The same rule should be applied on both sides of the equation. – Narasimham Jun 25 at 7:01

I will give a you a simple proof.

Let $$x = a$$

Let $$y = \frac{1}{b}$$

Let $$z = \frac{1}{n}$$

Therefore your equation becomes $$(xy)^z = x^z * y^z ⇒ xy * xy * xy..... xy, ntimes = (x * x * x..... x, ntimes) * (y * y * y..... y, ntimes)$$ In this equation, LHS and RHS are equal due to commutativity of multiplication.