# How to transform fraction powers between numerator and denominator

I apologize in advance if this question is trivial to most of you but I'd like to verify if my understanding is correct. I want to verify the following rules:

1. If we have $\dfrac{1}{\sqrt{x}}$, then it can be rewritten as $x\cdot x^2$ (because when the denominator $x^{1/2}$ is moved to numerator, we flip the fraction).

2. $\dfrac{x}{x^2}$ can be rewritten as $x\cdot x^{-2}$ (because moving the denominator to numerator, we change the sign).

Are these mathematically correct ?

Thank you

• (1) No. (2) Yes. – arctic tern Oct 6 '16 at 21:59
• Note that if this were true then (1) would imply $\frac {x} {x^2}= x ·x^{1/2} = x·x^{-2}$ which is a contradiction – Ron Oct 6 '16 at 22:13

## 2 Answers

Here is a rule that says you flip fractions when moving from bottom to top:

$$\frac{A/B}{\color{Blue}{C/D}} = \frac{A}{B}\times \color{Blue}{\frac{D}{C}}.$$

Notice that the fraction $C/D$ does not appear in any exponent. Your "rule"

$$\frac{A}{B^{\color{Red}{C/D}}}=A\times B^{\color{Red}{D/C}}$$

is made-up and wrong. It is not correct. What is correct is that moving powers across the fraction bar end up changing the sign of the exponent:

$$\frac{A}{ B^C}=A\times B^{-C}.$$

Assuming we're working in $\mathbb R$ (2) can be proven directly from the field axioms, using the existence of the multiplicative inverse for a non zero element: let $x\ne 0$ and $n\in \mathbb R$ $\rightarrow$ for $x^n$ there exist $y\in \mathbb R$ such that $y·x^n=1$ , we note $y=(x^n)^{-1}$ and by convention this is often written as $x^{-n}$ or $\frac {1}{x^n}$. Now to prove (2) we use the multiplicative inverse axiom and the multiplicative identity axiom: $$\frac{x}{x^2}=\frac{x}{x^2}·1=\frac{x}{x^2}\frac{x^{-2}}{x^{-2}}=\frac{x·x^{-2}}{x^{2}·x^{-2}}=\frac{x·x^{-2}}{1}=x·x^{-2}$$

Now (1) can be shown to be false because it contradicts (2) which follows directly from the field axioms. What is true and might have confused you is: $$\frac{x}{\sqrt{x}}=\frac{x}{\sqrt{x}}·1=\frac{x}{\sqrt{x}}·\frac{\sqrt{x}}{\sqrt{x}}=\frac{x·\sqrt{x}}{x}=\frac{x}{x}·\sqrt{x}=1·\sqrt{x}=\sqrt{x}$$

I hope this helps you!