ttaiwo wrote:

Hi Bunuel,

I also plugged in 2 for P...and got pretty lost from there once I found out that P=2 as well.

I followed your explanation and understand P@ = 2 since when you plug in 2...the result is 2.

However, I don't follow what happens from there...how exactly is P@@=P?

As for solution 2: how did you solve this algebraically? In other words, how did you end up knowing that P@@ required you to dividing P/P-1 P/P-1 -1

Thanks

1. If \(P = 2\). Then \(P@ = \frac{2}{2 - 1} = 2\).

So, we got that for p = 2, \(P@ = P\), then \((P@)@ =P@ = P\).

2. \(P@ = \frac{P}{(P-1)}\).

\(P@@ = (\frac{P}{(P-1)})@=\frac{\frac{P}{(P-1)}}{\frac{P}{(P-1)} - 1}\)