does the p-adic shows the other end side of numbers? i.e, from right to left?

Question

Reading this scientopia link i pondered whether p-adic shows the numbers from the other side (right to left from infinity's side)? For example the last digits the last digits of $\sqrt 2$ in 10-adics?

Answer 1

In the real numbers, regardless of base, irrational numbers have no "last" digit. There simply is no such thing at all. It is also incorrect to think of algebraics in $p$-adic fields as identical to corresponding algebraics in the complex numbers, any more than in any other kind of field. For example in the finite field $\Bbb F_7$ the element $2:=1+1$ has a couple of square roots (namely $3$ and $-3=4$) but neither of these is the same as the numbers $\sqrt{2}$ or $-\sqrt{2}$ in the reals $\Bbb R$. They are different sets and their elements are different. Thus, in particular, the $10$-adic numbers are not there to "repair" the reals by seeing "the other" side of numbers' decimal expansions. That's not what they're there for and it's not what they do.

Note that, number-theoretically, it is much more natural to consider $p$-adics for prime numbers $p$ rather than composites like $10$. With composites there are zero divisors and hence no fraction fields, and in any case $\Bbb Z_n\cong\prod_{p\mid n}\Bbb Z_p$ (to abuse notation for $n$-adics). Henceforth I'll assume $p$ is prime.

The decimal expansion of a real number is set up in such a way that each succeeding digit provides an approximation to the given real to an order of magnitude more precision than the last digit. The $p$-adics are set up the same way, except there is a much different notion of precision which is afforded by the new topology, metric, and measure on $\Bbb Q_p$. (This topology is totally disconnected, like that of a tree, and the metric satisfies an ultrametric inequality, two features which are the root of some strange but in some ways simplifying topological behavior). In an integer expansion $a_0+a_1p+a_2p^2+\cdots$ each new digit provides more precision, because the powers $1,p,p^2,p^3,\cdots$ tend to $0$ in the limit.

Rather than being seen as fixing $\Bbb R$, or being "dual" to $\Bbb R$, one should think of $\Bbb Q_p$ as being auxiliary to it in number theory. The reals are ground support for geometric information, where as $p$-adics are the foundation for arithmetic information, and these types of information go hand-in-hand in NT. The reals and the $p$-adics are the local completions of $\Bbb Q$, and one can put all of these together to get the ring of adeles which encodes a lot. For example, as Tate's thesis pointed out, many general $L$-functions are multiplicative Fourier transforms of Gaussians (fixed points of additive Fourier transforms) in the adelic setting. In particular this explains the particular form of the functional equation for $\zeta(s)$ and explains how the "gamma" factor relates to the other "local" factors of the Euler product.

Answer 2

$2$ doesn't have a square root in the $10$-adics: because $2$ divides $10$, the $10$-adics "understand" divisibility by $2$... and $2$ is divisible by an odd power of $2$, therefore it can't be a square.

Or put differently, it "understands" reduction modulo powers of $2$, and $x^2 \equiv 2 \pmod{4}$ has no solutions.

$3$ also doesn't have a square root in the $10$-adics: because $5$ divides $10$, the $10$-adics "understand" reduction modulo powers of $5$, and the equation $x^2 \equiv 3 \pmod{5}$ has no solutions.

$41$, however, does have a square-root in the 10-adics. Hensel's lemma (a.k.a. Newton's algorithm) tells us that because $x^2 \equiv 41 \pmod{8}$ has solutions, it has solutions in the $2$-adics, and the fact $x^2 \equiv 41 \pmod{5}$ tells us it has solutions in the $5$-adics. Since these are the only primes dividing $10$, it also has a solution in the $10$-adics.

The trailing 5 digits, for example, can be found by solving $x^2 \equiv 41 \bmod 2 \cdot 100000$, and reducing the solutions modulo $100000$.

Why the extra factor of $2$? It's because solving $x^2 = a \pmod {2^n}$ is weird: you need the extra factor of two to eliminate the "spurious" solutions for $x$. e.g. $1,3,5,7$ are all solutions to $x^2 \equiv 41 \pmod{8}$, but the solutions to $x^2 \equiv 41 \pmod{16}$ are $3, 5, 11, 13$. These latter four values are all either $3$ or $5$ modulo $8$: the mod-8 solutions $1,7$ were "spurious" and do not correspond to a $2$-adic square root of $41$. In general, solving $x^2 \equiv a \pmod{2^n}$ only tells you the reductions modulo $\pmod{2^{n-1}}$ of the $2$-adic square root of $a$.

(in the above, I believe I am assuming $a$ is relatively prime to $p$: if $a$ is divisible by the $p$, then factor the number into a power of $p$ times a number relatively prime to $p$, and handle taking the square root of the power of $p$ separately)

To demonstrate Newton's algorithm, recall that it iterates $x \leftarrow x - f(x)/f'(x)$ to converge to a solution to $f(x) = 0$. In this case, we use $f(x) = x^2 - 41$, so our iteration is $x \leftarrow x - (x^2 - 41)/(2x) = (x^2 + 41)/(2x)$.

$x=3$ is a known solution correct to three $2$-adic digits. This happens to be enough for Newton's algorithm to converge in the $2$-adics: if you have $k$ digits right, the next iteration will have $2k-1$ digits right. For other primes, you only need to have one digit right and each iteration doubles the number of digits. A simplistic explanation of the weird difference behavior is because there is a $2$ in the denominator of the formula for iterating Newton's algorithm.

Thus, $(3^2 + 41)/(2 \cdot 3) = 25/3$ has $5$ digits correct. However, we can simplify this modulo $25/3 \equiv 19 \pmod{32}$, so we can simplify to see that $19$ is a $2$-adic square root of $41$ correct to $5$ digits -- i.e. correct modulo $32$. (and indeed $19^2 \equiv 41 \pmod{64}$. Again, the extra digit is because squaring is weird $2$-adically)