0
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ (1) No. (2) Yes. $\endgroup$ – arctic tern Oct 6 '16 at 21:59
  • $\begingroup$ 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 $\endgroup$ – Ron Oct 6 '16 at 22:13
2
$\begingroup$

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:

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

$\endgroup$
2
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.