Introduction to $p$-adic Numbers I was watching a video on youtube.com by Dr. Norman Wildberger (sp?) in relation to the formulation of $p$-adic numbers.  He starts off by asking (or saying that Euler was interested in) the following sum
$$...+x^3+x^2+x+1+\frac1{x}+\frac1{x^2}+\frac1{x^3}+...$$
Calling it $A$, we can multiply it by $x$ and obtain $$Ax=A\rightarrow A(x-1)=0$$
We have $x\neq 0$.  But for nonzero term $(x-1)$, this forces $A=0$.  However, if $x=1$, doesn't this mean for this value of $x$ that $A$ itself can be nonzero?  This is my confusion.  Does that mean then that an infinite sum of 1's is equal to $0$ or some other constant?
It's probably something easy... but I'm confused.
 A: For simplicity, first consider positive $x>0$. Then the expression 
$$Ax-A$$
is a subtraction of two divergent infinite sum each tending to $\infty$.  
Now in your example you have shown that by allowing "usual arithmetic" and arranging the sums in a particular way, you get a conclusion that $A=0$. This contradicts the divergence of $A$ and hence you concluded that something is wrong.  
The problem here is such a subtraction, of two infinite divergent sums, is not well defined. In this case, you cannot assume that
$$
Ax-A=0
$$
even though it is intuitive to do so. (Perhaps it may be more convincing to see that you have shown it can lead to the absurdity of $A=0$.) It's the same thing for $x<0$ due to divergent series.  
Also see this example for a comprehensive discussion. Another often mentioned related problem is this, where the author also showed a rearrangement of the terms leads to a conclusion that
$$
0 = S = 1+2+3+\cdots
$$
A: I think a better way to present the formula to give it meaning is to write a purely periodic rational number's representation in the p-adics as,
$$\frac{a}{b}=\sum_{n=0}^\infty m p^{kn} $$
$m$ is a $k$ digit natural number (not necessarily excluding leading zeroes), so for instance in the 5-adics we could write in multiple ways as,
$$-\frac{1}{24} = \frac{1}{1-25} = \sum_{n=0}^\infty 1 * 5^{2k} = ...01010101$$
Similarly we can write the negative of this rational number as a real number in base p with,
$$-\frac{a}{b} = \sum_{n=1}^\infty m p^{-kn}$$
So for instance,
$$\frac{1}{24}= \frac{\frac{1}{25}}{1-\frac{1}{25}} = \sum_{n=1}^\infty 1* 5^{-2k} = .01010101...$$
So you could naively add the perfectly legitimate rational numbers on the left hand side here, and do the totally wrong thing on the right side by dipping your left foot in the p-adics and right foot in the reals,
$$\frac{a}{b}-\frac{a}{b}=\sum_{n=0}^\infty m p^{kn} + \sum_{n=1}^\infty m p^{-kn} $$
$$0=\sum_{n=-\infty }^\infty m p^{kn} $$
Then you can imagine factoring out that natural number and dividing it out and letting $x=p^k$ a special case,
$$0 = \sum_{n=-\infty}^\infty x^n$$
Which is the formula we were looking at. Now the reason $x=1$ would not work from this perspective is that would have $k=0$ which, in this already broken situation, represent the even more broken scenario of having m be a zero digit number and so if it didn't represent anything before it definitely doesn't represent anything now.
I'm not really sure what to say past the fact that it's more like an interesting consequence of the fact that a rational number x not equal to 1 can be represented in these two equivalent ways, except the left side can be expanded as a p-adic number while the right side can be expanded as a real number.
$$\frac{1}{x-1} = \frac{x^{-1}}{1-x^{-1}}$$
A: The only way to write the expression $...+x^{-3}+x^{-2}+x^{-1}+1+x+x^2+x^3+...$ (¤) without worrying about convergence would be to put it in a (minimal) ring of formal power series $\mathbf Z[[x,x^{-1}]]$, which should be a quotient of $\mathbf Z[[x,y]]$, but is necessarily trivial (see the remark by @Lubin).  Having no patience to look at the entire video, I don't know the motivation behind Euler's (or Dr. Wildberger's) mind. But notice that Euler was fond of "divergent calculation", which sometimes helped him to obtain miraculous intuitive formulas. A typical example is the value of the Riemann zêta function at $s=-1/2$ which he got like this: let $S=1+2+3+4+5+6...$, then $4S=4+8+12+...$ and $-3S=S-4S=1+(2-4)+3+(4-8)+5+(6-12)+...=1-2+3-4+5-6+...$ But $\frac 1{1+x}=1-x+x^2-x^3+...$, which gives by (formal) derivation $\frac 1{(1+x)^2}=1-2x+3x^2-...$ Taking $x=1$, Euler concluded that $-3S=1/4$, hence the formula $\zeta (-1) =-1/12$... which is correct !
Coming back to numbers and the introduction to $p$-adic numbers, convergence must be worried about. Recall the astonishing Ostrowski theorem which states that the only absolute values on $\mathbf Q$ (in fact on any number field) which are compatible with the field operations are the usual archimedean absolute value $\mid .\mid_{\infty}$ and the $p$-adic ones $\mid .\mid_p$. Completing $\mathbf Q$ w.r.t. these metrics, one gets resp. $\mathbf R$ and the $p$-adic fields $\mathbf Q_p$. The point in the expression (¤) is that, in the parlance of  @Steven,  it has one foot in the $p$-adic world and another in the archimedean world. The geometric series converges $p$-adically iff $\mid x\mid_p <1$ (which implies that $\mid x\mid_{\infty}$ is large), and it converges archimedean-ly iff $\mid x\mid_{\infty}<1$ (which implies that $\mid x\mid_p$ is large). So (¤) has no sense in either world, archimedan or $p$-adic. This phenomenon actually expresses an incompatibility between two topologies : the archimedean world is connected whereas the $p$-adic world is totally disconnected. 
But this is a fertile incompatibility. The global-local principle which has taken root in in algebraic number theory and arithmetic geometry postulates that the true comprehension of a global arithmetic property should - as often as possible - come from studying and putting together its archimedean and (all) $p$-adic variants obtained by taking completions. A striking  illustration is the so-called "Zêta element": its archimedean avatar is the usual complex meromorphic zêta function (Riemann or Dedekind, or more generally L-functions attached to algebraic varieties), its $p$-adic incarnations are $p$-adic zêta functions which "interpolate" the archimedean one at "special values". For example, it is well-known that the values of the Riemann zêta function at odd negative integers are rational, more precisely $\zeta(1-2m)=-B_{2m}/2m$, where $B_{2m}$ is the classical Bernoulli number. Since this "special value" is a rational number, it can be viewed as a $p$-adic number, at it happens that it is also (up to some adjustment) a "special value" of an adequate $p$-adic zêta function. The gain in understanding is the subsequent (very deep) formula: for any odd integer $m>0$, up to powers of $2$, one has $\mid \zeta(-m)=\mid K_{2m}(\mathbf Z)\mid / \mid K_{2m+1}(\mathbf Z)\mid$, where the Quillen groups $ K_i (.)$ are topologically defined objects !
