x' is just a point, not a length. If you want to think about lengths, you need to be a little careful.
Let's imagine two points at opposite ends of a rod of length 2L, at rest parallel to the x-axis and centered at the origin. We measure the positions of the ends of the rods at some time. Choosing t=0 makes the maths simpler, so let's choose that. End 1 of the rod is at x1=L and end 2 is at x2=-L. Now we transform those measurements into a frame moving at speed u in the +x direction using the Lorentz transform ##x'=\gamma(x-ut)##:
<br />
\begin{eqnarray}<br />
x'_1&=&\gamma(x_1-0)\\<br />
&=&\gamma L\\<br />
x'_2&=&\gamma(x_2-0)\\<br />
&=&-\gamma L<br />
\end{eqnarray}<br />
That makes the length in the new frame x'_1-x'_2=2γL and, as you note, that is greater than 2L. We've apparently got length dilation, not length contraction. What's going on?
We've been bitten by the relativity of simultaneity. It is very important to remember that two things that happen simultaneously in one frame don't generally happen simultaneously in another. You can see this here by Lorentz transforming the times of the measurements using ##t'=\gamma(t-ux/c^2)##:
<br />
\begin{eqnarray}<br />
t'_1&=&\gamma(0-ux_1/c^2)\\<br />
&=&-\gamma uL/c^2\\<br />
t'_2&=&\gamma(0-ux_2/c^2)\\<br />
&=&\gamma uL/c^2<br />
\end{eqnarray}<br />
Those measurements of the positions of the ends of the rod weren't made simultaneously in the new frame. This is significant because, in this frame, the rod is moving at speed -u. Imagine trying to measure the length of a beetle. If it sits still on the ruler, you can just note down the ruler measurement at each end and take the difference. It it's walking, however, you need to make certain that you note down the measurements at the same instant or you'll get nonsense because you conflate the beetle's motion with its size. In this case, we measured simultaneously in the rest frame of the rod (when it didn't actually matter), but not simultaneously in the frame where it's moving (when it did matter).
We need to answer the question: in the moving frame, where is end 2 of the rod at the time ##t'_1##? This is easy enough to answer - we know ##x_2=-L## and we've just chosen ##t'_2=-\gamma uL/c^2##. We can just sub these into the Lorentz transforms to get what we don't know. It turns out to be easiest to use the inverse Lorentz transform for x, ##x=\gamma(x'+ut')##:
<br />
\begin{eqnarray}<br />
x_2&=&\gamma(x'_2+ut'_2)\\<br />
-L&=&\gamma(x'_2-\gamma\frac{u^2}{c^2}L)<br />
\end{eqnarray}<br />
which, with a bit of work, yields
<br />
x'_2=\gamma L-2\frac{L}{\gamma}<br />
That in turn makes the length (correctly measured with simultaneous measurements) in the moving frame:
<br />
\begin{eqnarray}<br />
x'_1-x'_2&=&\gamma L-\left(\gamma L-2\frac{L}{\gamma}\right)\\<br />
&=&2\frac{L}{\gamma}<br />
\end{eqnarray}<br />
...which is length contraction.
To summarise, then, you added some of the rod's velocity to its length because you didn't measure the length in one go. You can fix this by making measurements simultaneously in the frame you are interested in.
