I try to understand and get a feeling which gaps p-adic numbers fill to complete $\mathbb{Q}$.
In the course of this I depicted (for $p = 2$) the "base" $\{p^k\}_{k\in\mathbb{Z}}$ with respect to which every p-adic number can be written
$$ n = \sum_{i = m}^{\infty}a_i p^i$$
with $a_i \in \{0,\dots p-1\}$ – first on the real line (black dots) and then its "projection" on the circle with center $i$ and radius $1$ (blue dots).
It appears as if the ever smaller p-adic numbers will fill "continuously" the circle near $0$ while the ever bigger p-adic numbers will fill "continuously" the circle near $\infty$.
There is an astonishing symmetry in this picture (at least for me): If we draw lines (yellow) through the points on the circle corresponding to $p^{1+k}$ and $p^{1-k}$ these lines are parallel:
I wonder what this might mean – and if such kind of visualization makes sense at all?
You may wish to compare the pictures above with the usual drawing of perspective (giving rise to real projective spaces):
[Side question: What does the original curve in the plane look like that gives rise – in projection – to the circle? Is it a straight line, a parabola or a hyperbola? Depending on what? (I didn't try to check, but what else?)]
And finally with some roots of unity:
This picture depicts the (maybe superficial and not "deep") symmetry between $0$ and $\infty$ as is discussed in the comments:
But the picture for the roots of unity is symmetric, too:
Best Answer
View the projection as a map from the real line to the circle in $\Bbb{R}^2$ with center $(0,1)$ and radius $1$. Then the point $r\in\Bbb{R}$ is mapped to the intersection of the line $2x+ry=2r$ and the circle $x^2+(y-1)^2=1$, which is $$(x,y)=\left(\frac{4r}{r^2+4},\frac{2r^2}{r^2+4}\right).$$ Now let $p$ be a prime number. The line through the projections of the points $p^{1+k}$ and $p^{1-k}$ has slope \begin{eqnarray*} \frac{\frac{4p^{1+k}}{(p^{1+k})^2+4}-\frac{4p^{1-k}}{(p^{1-k})^2+4}} {\frac{2(p^{1+k})^2}{(p^{1+k})^2+4}-\frac{2(p^{1-k})^2}{(p^{1-k})^2+4}} &=&\frac{4p^{1+k}((p^{1-k})^2+4)-4p^{1-k}((p^{1+k})^2+4)} {2(p^{1+k})^2((p^{1-k})^2+4)-2(p^{1-k})^2((p^{1+k})^2+4)}\\ &=&\frac{(4p^{3-k}+16p^{1+k})-(4p^{3+k}+16p^{1-k})}{(2p^4+8p^{2+2k})-(2p^4+8p^{2-2k})}\\ &=&\frac{(4-p^2)(p^k-p^{-k})}{2p(p^{2k}-p^{-2k})}=\frac{4-p^2}{2p(p^k+p^{-k})}. \end{eqnarray*} If $p=2$ then the slope equals $0$ for all $k$, so all these lines are parallel. However, if $p\neq2$ then the slope varies as $k$ varies, and the lines are not parallel.
In fact, your observation becomes a lot less astonishing after a projective transformation.
Consider the hyperbola in $\Bbb{R}^2$ given by $y=\frac{1}{x}$. Take two points $a,b\in\Bbb{R}$ such that $ab=1$. The vertical projections of $(a,0)$ and $(b,0)$ onto the hyperbola are given by $(a,\frac{1}{a})$ and $(b,\frac{1}{b})$. The line connecting these two points has slope $-1$. This is geometrically clear because the hyperbola is symmetric in the line $x=y$, and algebraically because $b=\frac{1}{a}$. The picture below shows the projection lines in black, and the connecting lines in yellow, for $a=2$ and $a=4$.
Applying the projective transformation $$(x:y:z)\ \longmapsto\ (2z:2y:x+y),$$ now yields your picture, and a simple scaling yields a similar picture for any real number; this is not a property of any $p$-adic valuation, but a property of (real) conics.