Fraction in a fraction powers: 4V/ V/2A

[Asmayus]

Hello, just wondering how you solve the following problem to get S = Equation:

$\displaystyle S = 4A + 4V/ (V/2A)$

If you bring the fraction above the line:

$\displaystyle 4A + 4V (V/2A)^-1$

And again:

$\displaystyle 4A + 4V (V^-1)/(2A^-1)$

$\displaystyle 4A + 4V (V^-1) (2A^-2)$

And then Cancel to get:

$\displaystyle 4A + 4 (2A^-2)$


This Gives us
$\displaystyle S = 4A + 8A^-2$

Is what I've done correct?

 

[galactus]

No, not quite. It simplifies more than that.

$\displaystyle \frac{4V}{\frac{V}{2A}}=4V\cdot\frac{2A}{V}=8A$

Add this to 4A and we get S=12A.

 

[Deleted member 4993]

Hello, just wondering how you solve the following problem to get S = Equation:

$\displaystyle S = 4A + 4V/ (V/2A)$

If you bring the fraction above the line:

$\displaystyle 4A + 4V (V/2A)^-1$

And again:

$\displaystyle 4A + 4V (V^-1)/(2A)^-1$ <<< Corrected

$\displaystyle 4A + 4V (V^-1) (2A^-2)$ <<< why is that 2A[sup:h2wvhb35]-2[/sup:h2wvhb35] ... should be 2A

And then Cancel to get:

$\displaystyle 4A + 4 (2A)$ <<< Corrected


This Gives us
$\displaystyle S = 4A + 8A^-2$

Is what I've done correct? NO

 

[Asmayus]

Thank you for the replies so far.

I realise I'm incorrect now but the logic was if you bring it above the line once it becomes 2A[sup:1s71xi9y]-1[/sup:1s71xi9y] so twice should give you 2A[sup:1s71xi9y]-2[/sup:1s71xi9y].

So would I be correct in stating that if you bring a fraction (to the power of +1) above a line once, it's to the power of -1, and if you raise it above the line again it becomes +1 once more?

 

[Aladdin]

So would I be correct in stating that if you bring a fraction (to the power of +1) above a line once, it's to the power of -1, and if you raise it above the line again it becomes +1 once more?

You change signs...

 

[Deleted member 4993]

Thank you for the replies so far.

I realise I'm incorrect now but the logic was if you bring it above the line once it becomes 2A[sup:2j0qarc2]-1[/sup:2j0qarc2] so twice should give you 2A[sup:2j0qarc2]-2[/sup:2j0qarc2]. <<< Do not count like that - it would be confusing. All the powers in denominator are multiplied by (-1), if denominator is transformed to numerator.

for example:

$\displaystyle \frac{x}{y} \, = \frac{x^1}{y^1} \, = \, \, x^1\cdot y^{-1} \, = \, x\cdot y^{-1}$


So would I be correct in stating that if you bring a fraction (to the power of +1) above a line once, it's to the power of -1, and if you raise it above the line again it becomes +1 once more?

Please show an example for this proposed action.

Remember what I said above -

All the powers in denominators are multiplied by (-1), if denominators are transformed to numerator.

 

[Asmayus]

Ah I see! Thank you!

@Subhotosh Khan

A worked example would be

\(\displaystyle {\frac{x}{\frac{y}{z}}\)

\(\displaystyle x{\frac{y^-^1}{z^-^1}}\) (sign changes to negitive)

\(\displaystyle x \cdot y^-^1 \cdot z\) (sign changes to positive)

 

[Denis]

Not sure what the purpose of that is, but if you're simply looking for gymnastic moves:

4v / (v / (2a)) = 4 / ((v^-1)(v^1) / (2a)) = 4 / (1 / (2a)) = 8a