Making something appear below center

redsoxfan325 · #1 $Old$ September 23rd, 2009, 08:02 PM

$\frac{\displaystyle\frac{{\color{red}P^x}~\displaystyle\frac{\Gamma}{(P\Rightarrow Q)}}{Q}~\displaystyle\frac{\Gamma}{(Q\Rightarrow R)}}{\displaystyle\frac{R}{(P\Rightarrow R)}~x}$

Is there any way to push down that $P^x$ so that it's sitting right on top of the line (while keeping everything else the same)?

Thanks.

Stefan_K · #2 $Old$ October 10th, 2009, 08:34 AM

A quick workaround using \llap:

$\frac{\displaystyle\frac{\qquad\displaystyle\frac{\Gamma}{\llap{$\color{red}P^x$~}(P\Rightarrow Q)}}{Q}~\displaystyle\frac{\Gamma}{(Q\Rightarrow R)}}{\displaystyle\frac{R}{(P\Rightarrow R)}~x}$

Stefan

redsoxfan325 · #3 $Old$ October 10th, 2009, 10:45 AM

Quote:

Originally Posted by Stefan_K $View Post$

A quick workaround using \llap:

$\frac{\displaystyle\frac{\qquad\displaystyle\frac{\Gamma}{\llap{$\color{red}P^x$~}(P\Rightarrow Q)}}{Q}~\displaystyle\frac{\Gamma}{(Q\Rightarrow R)}}{\displaystyle\frac{R}{(P\Rightarrow R)}~x}$

Stefan

One more question: what if I want to adjust something on the right. I tried qquad and rlap, but I got this:

$\frac{\displaystyle\frac{\Gamma}{(Q\Rightarrow P)\rlap{~~$Q$}}}{P}$

Stefan_K · #4 $Old$ October 10th, 2009, 10:49 AM

This way for instance:
$\frac{\displaystyle\frac{\Gamma}{(Q\Rightarrow P)\rlap{~~$Q$}}\qquad}{P}$

Stefan