NOTES ON STATISTICS, PROBABILITY and MATHEMATICS

Motivation:

The p-adic expression of a number is akin to placing a filter on the number: the filter of the prime \(p\). In the case of \(5\)-adics for example, the valuation and the size of the \(5\)-adic expression of a number could be seen as an attempt at extracting the “five-ness” out of the number in question.

Consider a monetary system in which there were coins representing fractions of \(5,\) as well as bills with values multiples of \(5\). The p-adic twist would dictate that these coins had much larger value than the bills - the inverse of the monetary value, almost like the coins were actually made of precious metals. Imagine that at a shop an attendant is returning \(\$\,450\). The \(5\)-adic expression is:

\[450 = 0\cdot 5^0 + 0\cdot 5^1 + 3\cdot 5^2 + 3\cdot 5^3\]

Yes, the employee could hand over to the customer \(\$\,5\) bills, but this would be bulky. In fact, the most efficient way of giving out this quantity is with \(3\) bills of \(\$\,25\) and \(3\) bills of \(\$\,125.\) This is exactly what the expression is telling us.

The valuation is \(2\), because this measures the “five-ness” of \(450.\) If the coefficient for \(5^2\) had been \(0,\) and the expression had started at with a non-zero coefficient at \(5^3,\) the number would have had a valuation of \(3\) - a higher “five-ness”.

On the flip side, the distance to zero would be smaller the higher the valuation:

\[|450|_5=\frac1{5^{v(x)}}=\frac 1{5^2}\]

where \(v(x)=2\) is the valuation of \(450\) in this case.

This distance formula \(|x|_p = 1/p^v\) reveals that the \(5\)-adic world is not a continuous line, but a hierarchy. The higher the valuation, the deeper the number is nested within ‘folders’ of \(5\). This creates a tree-like topology where two numbers are close only if they share a long path of common digits. The \(p\)-adics are a Cantor Set. If you try to plot them on a 2D plane, they look like dust. Every “point” is actually the tip of an infinite branch. The reason you don’t see this in real math is that in real math, we carry. If you add \(1\) to \(0.444_5\), it rolls over and changes everything to the left. In \(p\)-adics, the “folders” are rigid. Addition happens “deep” in the tree and only affects the branches above it. If you add \(1\) to the \(5\)-adic number \(4\) (\(...0004\)), you get \(5\), which is written as \(...0010\). So it would be incorrect to say that we only carry in real numbers. The “carry” happens exactly as it does in base 5. The difference isn’t that we don’t carry; it’s where the carry goes and how it affects the “closeness” of the numbers. In \(p\)-adic math, a carry moves toward higher powers of \(p\) (\(5, 25, 125\)), which are smaller and closer to zero. On the other hand real carry pushes the number away from where you started, as opposed to pushing the number deeper into the tree, actually making it “smaller” in terms of distance to zero.

If the employee had had to return \(\$\, 1/5\), the expansion would have been:

\[ 1\cdot 5^{-1}+0\cdot 5^{0}+0\cdot 5^{1}+\dots \]

with a valuation of \(-1\) and a distance to zero of \(5.\) This change is return in some “giant coin”. In this system, \(1/5\) is actually larger (further from zero) than \(1\), which has an expansion of

\[1 = \mathbf{1} \cdot 5^0 + 0 \cdot 5^1 + 0 \cdot 5^2 + \dots\]

And hence a valuation of \(0,\) and a distance to zero of \(1.\) In the \(5\)-adic world, \(5\) is small

\[5 = 0 \cdot 5^0 + \mathbf{1} \cdot 5^1 + 0 \cdot 5^2 + \dots\]

with a valuation of \(1,\) and a distance to \(0\) of \(1/5.\)

\(25\) is smaller, and \(1/5\) is a giant.

If you only look at a number through the “\(5\)-adic prism,” you only see its relationship to the prime \(5.\) To truly “know” the number in its entirety—its “global” identity—you have to look at it through every single prime prism (\(2, 3, 5, 7, 11 \dots\)) plus the “standard” prism of the real numbers (\(\mathbb{R}\)).

It is important to point out that the digits in the 5-adic expansion of a real number like \(450\) is identical to the expression in base \(5\).

In fact, the relationship is also extended to fractions. For instance, in the case of \(1/3\):

only the first digit is different. The connection between the first digit of the Base \(5\) expansion (\(1\)) and the first digit of the \(5\)-adic expansion (\(2\)) is hidden in a simple piece of arithmetic: \(1 + 2 = 3\) (the denominator). This isn’t a coincidence. In many cases, the \(p\)-adic expansion and the standard fractional expansion act like “partners” trying to reach the same target from different directions. In base \(5\) (\(0.1...\)) this \(1\) tells you that \(1/3\) is at least \(1/5\). It is a “floor” or an estimate from below. In \(5\)-adic (\(...2\)) this \(2\) is the solution to \(3x \equiv 1 \pmod 5\). It is the unique integer that, when multiplied by the denominator, gets you as close as possible to a “reset” (a multiple of \(5\)). The Base 5 version is trying to build \(0.333...\) (in decimal). The \(5\)-adic version is trying to “undo” the \(3.\) Because \(3 \times 2 = 6\), and \(6\) is one more than \(5,\) that \(2\) is the perfect “starting piece” to begin the \(p\)-adic cancellation process. There is a famous \(p\)-adic identity that explains why the digits look like reflections. If you look at the 5-adic expansion of \(-1\), it is an infinite string of \(4\)s: \[-1 = 4 + 4(5) + 4(25) + 4(125) + \dots\] When you calculate a positive fraction like \(1/3\) in \(p\)-adics, the math often forces the digits to be the “complements” (relative to \(p-1\)) of the standard base expansion.

Notice the math in the calculation:

In base \(5,\) the number \(1/3\) is written as:

\[\frac{1}{3} = (d_1 \times 5^{-1}) + (d_2 \times 5^{-2}) + (d_3 \times 5^{-3}) + \dots\]

We want to find \(d_1\) (the first digit). To isolate \(d_1\), we multiply the entire equation by \(5\):

\[\left(\frac{1}{3}\right) \times 5 = \left(d_1 \times 5^{-1} + d_2 \times 5^{-2} + \dots\right) \times 5\]

\[\frac{5}{3} = 1 + \frac 2 3 = d_1 + (d_2 \times 5^{-1} + d_3 \times 5^{-2} + \dots)\]

Now, look at the left side: \(5/3\) is \(1\) plus a remainder (\(2/3\)). On the right side, everything in the parentheses is now less than \(1.\) Therefore, \(d_1\) must be \(1\). For the next step, we take the remainder, \(5/3 - d_1 = 2/3,\) which is equivalent to moving the newly calculated \(d_1\) to the LHS:

\[\frac{5}{3} - 1 = \frac 2 3 = d_2 \times 5^{-1} + d_3 \times 5^{-2} + \dots\]

and multiply by \(5\) both sides to isolate \(d_2\):

\[\frac{10}{3} =3 + \frac 1 3 = d_2 + d_3 \times 5^{-1} + \dots\implies d_2 = 3\] Etc.:

Now let’s compare it to the calculation of \(1/3\) in \(5\)-adics:

\[\frac{1}{3} = d_0 + d_1(5^1) + d_2(5^2) + d_3(5^3) + \dots\] In modular arithmetic, any multiple of \(5\) becomes zero. \(d_1(5)\) is a multiple of \(5 \rightarrow 0\); \(d_2(25)\) is a multiple of \(25 \rightarrow 0\). The entire “tail” is made of multiples of 5, so it all vanishes. So, the equation \(\frac{1}{3} = d_0 + d_1(5) + \dots\) simplifies to:v\(\frac{1}{3} \equiv d_0 \pmod 5\). Multiplying by \(3\) to get rid of the fraction: \(1 \equiv 3d_0 \pmod 5\) we get to solve \(3x \equiv 1 \pmod 5\), and \(3 \times \mathbf{2} = 6 \equiv 1\). So \(d_0 = 2\).

Next we calculate (\(d_1\)): Subtract your progress and divide by \(5:\)

\[\frac{\frac 13 - 2 } 5 =\frac{-1}3 = d_1 + d_2(5^1) + d_3(5^2) + \dots\]

Solve \(3x \equiv -1 \pmod 5\).\(3 \times \mathbf{3} = 9 \equiv -1\). So \(d_1 = 3\). Here is the sequence:

The big picture: When you collect all \(p\)-adic “local” views and stitch them together into one massive mathematical object, you get the Adeles (short for “additive ideles”).If each \(p\)-adic field is a single prism, the Adele Ring (\(\mathbb{A}\)) is the entire crystal chandelier.1. What is an Adele, intuitively?Think of an Adele as an infinite string or a “vector” where every slot corresponds to a different way of measuring a number. If you were to represent the number \(10\) as an Adele, it would look like this:

\[\text{Adele}(10) = (10_{\mathbb{R}}, 10_2, 10_3, 10_5, 10_7, 10_{11}, \dots)\]

Calculations with p-adics:

Brief introduction:

P-adic numbers are power series:

\[s =\sum_{i=k}^\infty a_ip^i= a_kp^k + a_{k+1}p^{k+1} +a_{k+2}p^{k+2} + \cdots\] where \(k\) is an integer and \(0\leq a_i < p.\) These coefficients are the residue field (see below).

From this YouTube video:

The definition of the absolute value of the elements in a field determines the metric, which maps two elements of the field to a positive value.

The p-adic absolute value is defined as (for \(\alpha \in \mathbb Q\))

\[\left| \alpha \right|_p = \left| p^n \, \frac g h \right|_p = p^{-n}\] This takes only discrete values \(\left|\cdot\right|\in\{p^n:n\in\mathbb Z\}\cup\{0\}\), since \(|0|_p=0\).

For a p-adic number represented as a power series, the absolute value is determined by the lowest exponent term with a non-zero coefficient. The term with the lowest exponent of \(p\) signifies the highest power of \(p\) that divides the number.

Examples:

For \(p=11\) and \(\alpha =968/9,\)

\[\left| 968/9\right|_{11} = \left| 11^2 \, \frac{8}{9} \right|_{11} = 11^{-2}\]

In SageMath,

x = Qp(11)(968/9) 
print(abs(x))
1/121 = 11^2

This results in puzzling results for the metric. For instance, with \(p=7\) the distance between \(2\) and \(3\) is greater than between \(28814\) and \(2\):

\[\left| 28814-2\right|_{7}= \left| 28812\right|_{7}= \left| 7^4 \, 13 \right|_{7} = 7^{-4}=1/2401\]

whereas

\[\left| 3-2\right|_{7}= \left| 1\right|_{7}= \left| 7^0 \right|_{7} = 7^{0}=1\]

Or in 3-adic distance:

\(1\) and \(3\): Their difference \((2)\) is not divisible by \(3\), so their \(3\)-adic distance is relatively large. \(1\) and \(10\): Their difference \((9)\) is divisible by \(3^2\), making them closer in the \(3\)-adic sense.

This metric determines other seemingly puzzling results, such as the fact that in the \(2\)-adic sense

\[1 + 2 + 2^2 + 2^3 + 2^4 + \cdots=\frac{1}{1-2}=-1\]

makes sense, based on the fact that the added values (in \(2\)-adic) are getting smaller and smaller. This is presented by 3B1B in here.

The p-adic norm \(\left|x \right|_p = p^{-\nu}\) has power law behavior. Because the norm is defined as a power of \(p\), the addition of two p-adic numbers cannot result in a norm larger than the larger of the two individual norms. This is a characteristic feature of non-Archimedean spaces, where the usual additive properties of norms are replaced by the strong form of the triangle inequality, \(\left|x+y \right|_p \leq \max(\left|x \right|_p, \left|y \right|_p)\).

Visualization of the p-adic norm as an infinity tree, corresponding to the power series. Each branch represents an element of the residue field: This is the set of numbers that remain after dividing all integers by \(p\). For example, if \(p=3\), the residue field would be \(\{0, 1, 2\}\) because any integer divided by \(3\) leaves a remainder of \(0, 1, or 2\). The p-adic numbers are down at the boundary of this infinite tree. The p-adic norm can be interpreted as geodesic distances on this tree.

Here is a similar representation as a tree from here:

A p-adic integer is a walk up the tree from root to leaf, although one would never get to the leaf since p-adics are infinite expressions.

Examples of p-adic numbers in \(\mathbb Q_5\) (from here):

\[ \begin{align} (1)_{\mathbb Q_5} &=1\\ (5)_{\mathbb Q_5} &= (0+ 1\times 5)_{\mathbb Q_5}=10\\ (-1)_{\mathbb Q_5} &=(4 + 4 \cdot 5 + 4 \cdot5^2+\cdots)_{\mathbb Q_5}=...4444 \end{align} \] because adding \(...4444\) to \(1\), which is \(...000001\) will result in \(0\) (additive inverse of one).

Fractions:

Simple algorithm compared to long division based on a couple of examples:

Based on this YT presentation, let’s look first at the decimal expansion of \(1/3\):

\[\frac 1 3 = \frac {c_0}{10^0} + \frac {c_1}{10^1} + \frac {c_2}{10^2} + \frac {c_3}{10^3}+ \dots \]

Multiplying \(\times 3\)

\[1= 3\left(\frac {c_0}{10^0} + \frac {c_1}{10^1} + \frac {c_2}{10^2} + \frac {c_3}{10^3}+ \dots \right) \]

\(c_0\), in \(\frac{c_0}{10^0}\), multitplied times \(3\) doesn’t fit into \(1\), and hence \(c_0\) is \(0.\cdots\), and this gets us to the point when we have to add a zero \(0\) to the right of \(1\) to find \(c_1\) (i.e. multiply \(\times 10\)). Naturally,

\[10 = 3\left( c_1 + \frac {c_2}{10^1} + \frac {c_3}{10^2}+ \dots \right) \]

Now \(3 c_1=10 \implies c_1 = 3\) and

\[1=10 - 9 = 3\left( \frac {c_2}{10^1} + \frac {c_3}{10^2}+ \dots \right) \]

Again we have to multiply by ten:

\[10 = 3\left(c_2 + \frac{c_3}{10^1}+ \dots \right)\]

and we see that \(C_2=3.\) Et cetera.

This is mirrored in the p-adics as follows.

Calculating p-adic expansions for fractions (\(1/3\) in \(5\)-adic):

Example: calculate what \(1/3\) is in \(\mathbb Q_5:\)

We want to express \(1/3\) as a power series:

\[\frac 1 3 = a_0 + a_1\, p + a_2 \, p^2+ \dots \]

Rearranging:

\[1 = 3\left(a_0 + a_1\, p + a_2 \, p^2+ \dots \right) \]

If we express it \(\mod 5\) the only term that will survive is the first

\[1 = 3\,a_0\]

and

\[a_0= 2\]

because

\[3\times 2 = 6 \equiv 1 \mod(5)\]

But we want finer refinement (inverse limit). Since the calculated \(6\) is congruent to the LHS (i.e. \(1\)), the difference will be a multiple of \(5\):

\[1 - 6=-5= 3\left(a_1\, p + a_2 \, p^2+ \dots \right) \] Now dividing by \(5\) on both sides, and since \(p=5\):

\[ -1 = 3\left(a_1 + a_2 \, p^1+ \dots \right)\]

Given that \(-1 \equiv 4 \mod(5)\), if we look at that equation \(\mod 5\), all the powers of \(p\) will be zero, allowing us to calculate that \[a_1 = 3\]

since \(3 \times 3 = 9\equiv 4 \mod(5).\) Note that the fact that the terms containing powers of \(p\) go to zero doesn’t imply that the coefficients are zero.

Rinse and repeat:

\[-1 - 9= -10 = 3\left(a_2\, p + a_3 \, p^2+ \dots \right)\]

Divide by \(5\):

\[3 \mod(5) \equiv -2 = 3\left(a_2 + a_3 \, p+ \dots \right)\]

yielding

\[a_2=1\]

Rinse and repeat:

\[-2-3=-5 = 3\left( a_3\,p+ a_4\,p^2\dots \right)\]

Divide by \(5\):

\[4 \mod(5)\equiv -1 = 3\left( a_3+ a_4\,p\dots \right)\]

so \[a_3=3\]

Et cetera.

\[ \require{enclose} \begin{array}{rll} & \color{blue}{2}\,\color{aqua}{3}\,\color{magenta}{1}\,\color{lime}{3}\dots \\[3pt] 3 &\enclose{longdiv}{\phantom{0}\color{brown}{\bf 1\equiv 1}}\kern-.2ex \\[3pt] & \underline{-\,\color{orange}{\bf 6\equiv 1}} && \hbox{($\color{blue}{2} \times 3 =\color{orange}{\bf{6}}\equiv \color{brown}{\bf{ 1}}\mod(5)$)}\quad\hbox{$\color{brown}{\bf 1}=3\,(\color{blue}2\,p^0 +x_1 \,p^1 + x_2 \,p^2+ x_3 \,p^3+\cdots)$} \\[3pt] &\color{red}{\bf{-5}}\phantom{0}\to \color{brown}{\bf{-1\equiv 4}} && \hbox{($\frac{\color{red}{\bf{-5}}}{p} = \color{brown}{\bf{-1}}\equiv 4\mod(5)$)}\quad\hbox{$\frac{\color{red}{\bf{-5}}}{p}=3\,(x_1 + x_2\,p^1+ x_3\,p^2+\cdots)$} \\[3pt] &\phantom{0000000}\underline{-\,\color{orange}{\bf 9\equiv 4}\phantom{0}} && \hbox{($\color{aqua}{3} \times 3 =\color{orange}{\bf 9} \equiv 4 \mod(5)$)} \quad\hbox{$\color{brown}{\bf{-1}}=3\,(\color{aqua}3\,p^0 + x_2\,p^1+ x_3\,p^2+\cdots)$}\\[3pt] &\phantom{000000}\color{red}{\bf{-10}} \phantom{0}\to \color{brown}{\bf{-2\equiv 3}} && \hbox{($\frac{\color{red}{\bf{-10}}}{p} = \color{brown}{\bf{-2}}\equiv 3\mod(5)$)}\quad\hbox{$\frac{\color{red}{\bf{-10}}}{p}=3\,(x_2+ x_3\,p+\cdots)$} \\[3pt] &\phantom{00000000000000}\underline{-\,\color{orange}{\bf{3\equiv 3}}} && \hbox{($\color{magenta}{1} \times 3 = \color{orange}{\bf{3}}\equiv 3\mod(5)$)}\quad\hbox{$\color{brown}{\bf{-2}}=3\,(\color{magenta}1\,p^0 + x_3\,p^1+\cdots)$} \\[3pt] &\phantom{00000000000000}\color{red}{\bf{-5}} \to \color{brown}{\bf{-1\equiv 4}} && \hbox{($\frac{\color{red}{\bf{-5}}}{p} = \color{brown}{\bf{-1}}\equiv 4\mod(5)$)}\quad\hbox{$\frac{\color{red}{\bf{-5}}}{p}=3\,(x_3+\cdots)$} \\[3pt] &\phantom{00000000000000000000}\underline{-\,\color{orange}{\bf{9\equiv 4}}} && \hbox{($\color{lime}{3} \times 3 = \color{orange}{\bf{9}}\equiv 4\mod(5)$)}\quad\hbox{$\color{brown}{\bf{-1}}=3\,(\color{lime}3\,p^0 +\cdots)$}\\[2pt] &\phantom{00000000000000000000}\color{red}{\bf{-10}}\\[2pt] &&\phantom{000000000}\vdots \end{array} \]

\[\left(1/3 \right)_{\mathbb Q_5}= 2 + 3 \cdot 5 + 1 \cdot 5^2 + 3 \cdot 5^3 + \cdots\] Notice that the algorithm keeps reducing the remainder modulo \(5\), as we get the coefficients for higher and higher \(p^n\) values. This is akin to adding a \(0\) to the remainder to keep dividing \(1\) into \(3\) in the process of getting the decimal expression of \(1/3,\) with each \(0\) added to the remainder equivalently contributing to higher and higher \(10^{-n}\) values in the decimal expression.

In SageMath:

padic_printing.mode('digits')
print(Qp(5)(1/3))
...31313131313131313132
padic_printing.mode('series')
print(Qp(5)(1/3))
2 + 3*5 + 5^2 + 3*5^3 + 5^4 + 3*5^5 + 5^6 + 3*5^7 + 5^8 + 3*5^9 + 5^10 + 3*5^11 + 5^12 + 3*5^13 + 5^14 + 3*5^15 + 5^16 + 3*5^17 + 5^18 + 3*5^19 + O(5^20)

\[...31313131313131313132 = 2 + 3\cdot 5 + 5^2 + 3\cdot 5^3 + 5^4 + 3\cdot 5^5 + 5^6 + 3\cdot 5^7 + 5^8 + 3\cdot 5^9 + 5^{10} + 3\cdot 5^{11} + O(5^{12})\]

Another example: \(2/3\) in \(5\)-adic:

In Sagemath:

padic_printing.mode('digits')
print(Qp(5)(2/3))
...13131313131313131314
padic_printing.mode('series')
print(Qp(5)(2/3))
4 + 5 + 3*5^2 + 5^3 + 3*5^4 + 5^5 + 3*5^6 + 5^7 + 3*5^8 + 5^9 + 3*5^10 + 5^11 + 3*5^12 + 5^13 + 3*5^14 + 5^15 + 3*5^16 + 5^17 + 3*5^18 + 5^19 + O(5^20)

\[2 = 3\left(a_0 + a_1\, p + a_2 \, p^2+ \dots \right) \]

\[a_0=4\]

because \(3 \times 4 = 12\equiv2 \mod(5)\)

Now

\[2-3 \times 4= -10 = 3\left(a_1\, p + a_2 \, p^2+ \dots \right) \]

Divide by \(5\)

\[3 \mod(5)\equiv -2 = 3\left(a_1 + a_2 \, p+ \dots \right) \]

yields

\[a_1 = 1\]

and therefore

\[-2 -3=-5= 3\left(a_1 + a_2 \, p+ \dots \right) \]

divide by \(5\)

\[4 \mod(5)\equiv -1 = 3\left( a_2 + a_3\, p+ \dots \right) \]

and

\[a_2=3\]

Subtract

\[-1 - 9=-10= 3\left(a_3\, p+ \dots \right) \]

Divide by \(5\)

\[3\mod(5)\equiv-2= 3\left(a_3\, p+ \dots \right) \]

yielding

\[a_3=1\]

And so on.

Notice that this would break if we tried \(1/24\) in \(2\)-adics:

\[1=24\, (a_0+ a_1 \,p + a_2 \,p^2+\dots)\]

because there is no number \(n\) such that \(24 \, n\equiv 1\mod(2).\) Hence we we’ll need the factorization trick explained below with \(1/24= 1/3\times 2^{-3}.\)

Wikipedia example of \(1/5\) in \(3\)-adics:

\[ \require{enclose} \begin{array}{rll} & \color{blue}{2}\,\color{red}{0}\,\color{magenta}{1}\,\color{lime}{2}\dots \\[-3pt] 5 &\enclose{longdiv}{\phantom{0}1}\kern-.2ex \\[-3pt] & \underline{-10} && \hbox{($\color{blue}{2} \times 5 = 10\equiv 1\mod(3)$)} \\[-3pt] &-9\phantom{0}\to -3 && \hbox{($-9/p = -3\equiv 0\mod(3)$)}\,\hbox{($-9/p=5(x_1 + x_2p^1+ x_3p^2+\cdots)$)} \\[-3pt] &\phantom{0000000}\underline{-0\phantom{0}} && \hbox{($\color{red}{0} \times 5 =0 \equiv 0 \mod(3)$)} \\[-3pt] &\phantom{000000}-3 &\to -1 & \hbox{($-3/p = -1\equiv 2\mod(3)$)}\,\hbox{($-3/p=5(x_2+ x_3p+\cdots)$)} \\[-3pt] &&\phantom{000}\underline{-5} & \hbox{($\color{magenta}{1} \times 5 = 5\equiv 2\mod(3)$)} \\[-3pt] &&\phantom{00}-6 \to -2 & \hbox{($-6/p = -2\equiv 1\mod(3)$)}\,\hbox{($-6/p=5(x_3+\cdots)$)} \\[-3pt] &&\phantom{00000000}-10 & \hbox{($\color{lime}{2} \times 5 = 10\equiv 1\mod(3)$)}\\[-3pt] &&&\vdots \end{array} \]

In SageMath the \(3\)-adic expansion is:

padic_printing.mode('series')
print(Qp(3)(1/5))
2 + 3^2 + 2*3^3 + 3^4 + 3^6 + 2*3^7 + 3^8 + 3^10 + 2*3^11 + 3^12 + 3^14 + 2*3^15 + 3^16 + 3^18 + 2*3^19 + O(3^20)
padic_printing.mode('digits')
print(Qp(3)(1/5))
...21012101210121012102

whereas base \(3\) base expression is

\[0.012101210121\dots_3\]

Notice that the expansion of \(5\) in \(3\)-adic may look similar in structure:

padic_printing.mode('series')
print(Qp(3)(5))
2\cdot 3^0 + 1\cdot 3^1 + O(3^20)

but it ends: after a certain coefficient, the rest are all zeros:

padic_printing.mode('digits')
print(Qp(3)(5))
...00000000000000000012

This is the difference between the p-adic expression of an integer and a fraction: The p-adic representation of a fraction that is not an integer always requires an infinite number of non-zero terms in its expansion in powers of p. It cannot be represented by a finite sum of powers of p. This is equivalent to saying that the p-adic expansion will never terminate with zeros to the left.

p-adic expansion of integers:

P-adics expand to infinity on to the left as in \(\mathbb Z_3,\) corresponding to the \(3\)-adics. For example, in this system the number \(3\) is

\[...00000000000000000010 = 0 \cdot 3^0 + 1 \cdot 3^1 \]

In Sage

padic_printing.mode('digits')
print(Zp(3)(3))
...000000000000000000010
padic_printing.mode('series')
print(Zp(3)(3))
3 + O(3^21)

For natural numbers, the p-adic representation and the base-p representation are essentially the same. For natural numbers, the p-adic representation will have only a finite number of non-zero terms, effectively reducing to the base-p representation.

For example, compare the \(5\)-adic expansion of \(233\)

\[...00000000000000001413 = 3 \cdot 3^0 + 1 \cdot 5^1 + 4\cdot 5^2 + 1\cdot 5^3+\cdots\]

padic_printing.mode('series')
print(Qp(5)(233))
3 + 5 + 4*5^2 + 5^3 + O(5^20)
padic_printing.mode('digits')
print(Qp(5)(233))
...00000000000000001413

to the expression mod \(5\):

\[233 \mod(5)= 3 + 1 \cdot 5^1 + 4 \cdot 5^2 + 1\cdot 5^3 \]

a = 233
base = 5
print(" + ".join([f"{digit}*{base}^{i}" if i > 0 else str(digit) for i, digit in enumerate(a.digits(base)) if digit]))
3 + 1*5^1 + 4*5^2 + 1*5^3

The negative reals are simply the additive inverses of their positive counterparts, so \(-1 = ...11111111\)

Because

\[...11111111 + ...00000001 = ...000000\]

since in base \(2\) and summing from right to left, \(1 + 1 = 0\) and we carry \(1\) to the left to add it to the \(1\) in the second position in the expression of \(-1.\)

Negative integers:

A negative integer can be expressed as the difference between \(0\) and a positive integer. The coefficients will not be all zeros beyond a certain point as in positive integers, because they represent the “borrowing” that occurs during the subtraction. However, the representation will still be finite.

Example:

\(2\)-adic representation of \(-33\): We want to find a \(2\)-adic number \(x\) such that \(x + 33 = 0\) (in the \(2\)-adic field). We can compute this by finding the inverse of \(33\) in the \(2\)-adic field:

padic_printing.mode('digits')
print(Qp(2)(33))
...00000000000000100001
padic_printing.mode('digits')
print(Qp(2)(-33))
...11111111111111011111

Negative exponents in p-adics:

Negative exponents typically come into play when dealing with fractions or representing very small numbers in the p-adic number system.

In the p-adic number system, fractions that involve negative exponents emerge when the denominator of the fraction is divisible by \(p\), the base of the system. This is because the p-adic expansion needs to represent the fractional part appropriately by incorporating terms with negative exponents. Let’s consider an example in the \(5\)-adic system: Let’s take \(1/25\). Convert to \(5\)-adic Expansion: \(1/25=5^{−2}\).

padic_printing.mode('digits')
print(Qp(5)(1/25))
...000000000000000000.01
padic_printing.mode('series')
print(Qp(5)(1/25))
5^-2 + O(5^18)

Here is another example of negative exponents, which correspond to digits to the right of the period. When you see a p-adic number written with digits to the right of a “p-adic point” (or sometimes represented with negative powers of p explicitly), it indicates that the number is not a p-adic integer. It is a p-adic number that lies in the fraction field of the p-adic integers. In this case it is the \(2\)-adic expression of \(\frac{1}{2^5 \cdot33}:\)

padic_printing.mode('digits')
print(Qp(2)(1/(2^5*33)))
...111110000011111.00001
padic_printing.mode('series')
print(Qp(2)(1/(2^5*33)))
2^-5 + 1 + 2 + 2^2 + 2^3 + 2^4 + 2^10 + 2^11 + 2^12 + 2^13 + 2^14 + O(2^15)

Digits to the right of the period in \(p\)-adic expressions (e.g. \(.011\)):

How are the finite digits (\(.011\)) to the right of the period calculated in an expression such as \(...10101010101010101.011\), and where do they come from?

The issue comes up in the introduction of \(\mathbb Q_p\) numbers. For instance, in the case of the 2-adic expression of \(\frac 1{24}\). P-adic numbers can be factored as

\[n = p^k \, \text{unit}\] where \(k\) is an exponent measuring the valuation of the p-adic as \(\frac 1{p^k}\), and the \(\text{unit}\) is… well quick review:

A unit in a ring is an element that has a multiplicative inverse. In other words, if \(a\) is a unit, there exists an element \(b\) in the ring such that \(a \cdot b = 1\). A zero divisor is a non-zero element \(a\) in a ring such that there exists a non-zero element \(b\) where \(a \cdot b = 0.\)

Therefore, in the case of a fraction:

\[\frac m n = \frac m{p^k} \frac 1{\text{unit}}= \frac m{\text{unit}}p^{-k}\]

Applying it to the example of \(1/24,\) the denominator can be factorized as \(2^3 \cdot 3:\)

\[\frac 1{24}=\frac 1 3 \cdot 2^{-3}\]

\(1/3\) is just the multiplicative inverse of \(3,\) which can be obtained in sagemath as:

from sage.all import *
from sage.rings.padics.padic_printing import pAdicPrinter
padic_printing.mode('digits')
print(Qp(2)(1/3))
# ...10101010101010101011

and easily confirmed to be accurate:

third = Qp(2)(1/3)
three = Qp(2)(3)
print(third * three)
# ...00000000000000000001

Therefore,

\[1/24 = ...10101010101010101011 \cdot 2^{-3}\]

but since we are operating in base \(2,\) the factor \(2^{-3}\) just moves the period to the right, like \(10^{-3}\) would in base \(10.\) In this way, we get the final result:

\[1/24=...10101010101010101.011=1\cdot2^{-3} + 1\cdot2^{-2}+0\cdot2^{-1}+1\cdot2^{0} + 0\cdot2^{1}+ 1\cdot2^{2}+0\cdot2^{3}+1\cdot2^{4}\]

or in sagemath

padic_printing.mode('digits')
print(Qp(2)(1/24))
...10101010101010101.011
padic_printing.mode('series')
print(Qp(2)(1/24))
2^-3 + 2^-2 + 1 + 2^2 + 2^4 + 2^6 + 2^8 + 2^10 + 2^12 + 2^14 + 2^16 + O(2^17)

p-adic use in computers:

Suppose our computer stores the following 5-adic number: \(...4321\) (where the digits extend infinitely to the left). However, the computer can only store the last four digits: \(4321\).

  1. Interpretation as a Base-5 Number:

We treat the stored digits, \(4321\), as a base-\(5\) number.

  1. Conversion to Base-10 (Decimal):

We convert this base-5 number to its decimal (base-10) equivalent:

\(4 * 5³ + 3 * 5² + 2 * 5¹ + 1 * 5⁰ = 4 * 125 + 3 * 25 + 2 * 5 + 1 * 1 = 500 + 75 + 10 + 1 = 586\)

Computers typically use floating-point representation for real numbers. This allows them to represent a wide range of numbers, both very small and very large, with reasonable precision. Computer hardware is designed to perform arithmetic on binary numbers, which are used to represent integers and floating-point numbers. There is no direct hardware support for p-adic arithmetic. Any p-adic calculations would have to be implemented in software, which would be slower.

Hensel’s lemma:

A simple root of a polynomial is a root where the polynomial crosses the x-axis and does not touch or bounce off it. A root \(r\) of a polynomial \(f(x)\) is considered simple if \(f(r) = 0\) and the derivative \(f'(r)\neq 0\). This means that at \(r\), the polynomial has a non-zero slope, indicating that it crosses the x-axis at that point. In contrast, a multiple root (or repeated root) would satisfy \(f(r) = 0\) and \(f’(r) = 0\), meaning the polynomial touches or bounces off the x-axis at \(r\) but does not cross it.

If a polynomial has a simple root modulo a prime \(p\), Hensel’s theorem allows us to lift this root to a root modulo \(p^k\) for any \(k\). This process can be extended to find roots in the p-adic integers \(\mathbb{Z}_p\).

\[a_{n+1}\equiv a_n \mod{p^{n+1}}\]

and

\[f(a_n)\equiv 0 \mod {p^{n+1}}\]

will allow the construction of a sequence of entries in the p-adic expression.

For instance (see here), \(f(x) = x^2 + 1\) has two solutions \(\mod{5}\): \(3\) with multiplicity \(1\), and \(2\) with multiplicity \(1.\)

R.<x> = GF(5)[]
p = x^2 + 1
p.roots()
# [(3, 1), (2, 1)]

These will start the sequence in the p-adic (of \(5\)-adic) expansion. So for \(r = 2,\) this will be \(a_0=2.\) For \(a_1,\) the equation above is

\[\begin{align} a_1 &\equiv a_0 \mod{5^1}\\ a_1 &\equiv 2 \mod{5} \\ a_1 &= 2 + 5t \end{align}\]

and

\[\begin{align} f(a_1) &\equiv 0 \mod{5^2}\\ (2 + 5t)^2 + 1 &\equiv 0 \mod{5^2} \\ t &\equiv 1 \mod {5^2} \end{align}\]

yielding

\[\begin{align} a_1 = 2 + 5 \cdot 1 =7 \end{align}\]

And at this point we are ready for

\[\begin{align} a_2 &\equiv a_1 \mod{5^2}\\ a_2 &\equiv 7 \mod{25}\\ a_2 &= 7 +25t \end{align}\]

and \[(7 +25t)^2+1 \equiv 0 \mod{5^3}\]

which results in \(a_2 =57.\)

Progressing to

\[\begin{align} a_3 &\equiv a_2 \mod{5^3}\\ a_3 &\equiv 57 \mod{125}\\ a_3 &= 57 +125t \end{align}\]

which can be plugged into the function:

\[(57+125t)^2 + 1\equiv 0 \mod{5^4}\]

resulting in \(t=1\) and \(a_3 =57 + 125 = 182\)

In sagemath,

import numpy as np

def hensel(roots, p):
    first_root = roots[0][0]
    scnd_root = roots[1][0]
    first_root_str = str(first_root).replace('...', '')
    first_root_list = [int(digit) for digit in first_root_str]
    first_root_list_reversed = first_root_list[::-1]
    first_result = [digit * (p ** i) for i, digit in enumerate(first_root_list_reversed)]
    scnd_root_str = str(scnd_root).replace('...', '')
    scnd_root_list = [int(digit) for digit in scnd_root_str]
    scnd_root_list_reversed = scnd_root_list[::-1]
    scnd_result = [digit * (p ** i) for i, digit in enumerate(scnd_root_list_reversed)]

    return [np.cumsum(first_result), np.cumsum(scnd_result)]
padic_printing.mode('series')
R.<x>  = Qp(5,8)[]
p = x^2 + 1
p.roots()
#[(2 + 5 + 2*5^2 + 5^3 + 3*5^4 + 4*5^5 + 2*5^6 + 3*5^7 + O(5^8), 1),
 (3 + 3*5 + 2*5^2 + 3*5^3 + 5^4 + 2*5^6 + 5^7 + O(5^8), 1)]
padic_printing.mode('digits')
R.<x> = Qp(5,8)[]
p = x^2 + 1
p.roots()
#[(...32431212, 1), (...12013233, 1)]
roots = p.roots()
hensel(roots, 5)
#[array([     2,      7,     57,    182,   2057,  14557,  45807, 280182]),
 array([     3,     18,     68,    443,   1068,   1068,  32318, 110443])]

Notice how this makes sense:

2 + 5 + 2*5^2 = 57

p-adics and base p:

There is a “mirror-image” relationship between standard base-\(3\) expansions, for example and \(3\)-adic numbers. The most fundamental difference is the “direction” of the infinite digits. Base-\(3\) are real Numbers, expanding to the right into increasingly smaller fractions (\(3^{-1}, 3^{-2}, \dots\)). On the other hand, \(3\)-adics expand to the left into increasingly higher powers of \(3\) (\(3^1, 3^2, \dots\)).

The mechanical steps for calculating a fraction like \(5/29\) are inverted:

Base-3 Algorithm:

In base 3, we only use the digits \(0\), \(1\), and \(2\). We are looking for

\[5/29 = d_{-1}3^{-1} + d_{-2}3^{-2} + \dots$Since $5 < 29\]

the expansion starts with \(0\). To find the digits, we multiply the remainder by \(3\) and see how many times \(29\) fits into it.

  • \(5 \times 3 = 15\). (\(15\) is less than \(29\)). Digit: \(0\)

  • \(15 \times 3 = 45\). (\(29\) goes into \(45\) \(1\) time, remainder \(16\)). Digit: \(1\)

  • \(16 \times 3 = 48\). (\(29\) goes into \(48\) \(1\) time, remainder \(19\)). Digit: \(1\)

  • \(19 \times 3 = 57\). (\(29\) goes into \(57\) \(1\) time, remainder \(28\)). Digit: \(1\)

  • \(28 \times 3 = 84\). (\(29\) goes into \(84\) \(2\), remainder \(26\)). Digit: 2

Hence the base-3 result is \(5/29 \approx 0.01112\dots_3\)

You multiply the remainder by \(3\) at each step and keep the quotient (how many times \(29\) “fits”). This zooms into the number line.

\(3\)-adic algorithm:

You subtract the residue and divide the error by \(3\) at each step to find the next remainder.

For the 3-adic version, we want to find digits \(d_0, d_1, d_2 \dots\) such that:

\[5/29 = d_0 + d_1(3) + d_2(9) + d_3(27) + \dots\]

This is equivalent to solving the congruence \(29x \equiv 5 \pmod{3^k}\) for increasing values of \(k\).

  • Step 1: Find \(d_0\): Solve \(29d_0 \equiv 5 \pmod 3\). Simplify 29 and 5 modulo 3: \(29 \equiv 2 \pmod 3\) and \(5 \equiv 2 \pmod 3\). The equation becomes: \(2d_0 \equiv 2 \pmod 3\). \(d_0 = 1\).

  • Step 2: Find \(d_1\): We know the expansion looks like \(1 + 3d_1 + \dots\). Substitute this into the original fraction: \[\frac{5 - 29(1)}{3} = \frac{-24}{3} = -8\] Now solve for \(d_1\): \(29d_1 \equiv -8 \pmod 3\). Simplify: \(2d_1 \equiv 1 \pmod 3\) (since \(-8 \equiv 1 \pmod 3\)). \(d_1 = 2\) (because \(2 \times 2 = 4 \equiv 1 \pmod 3\)).

  • Step 3: Find \(d_2\): Take the previous “leftover” (\(-8\)) and subtract the new term (\(29 \times 2\)): \[\frac{-8 - 29(2)}{3} = \frac{-8 - 58}{3} = \frac{-66}{3} = -22\]

Now solve for \(d_2\): \(29d_2 \equiv -22 \pmod 3\). Simplify: \(2d_2 \equiv 2 \pmod 3\).\(d_2 = 1\).

Therefore the \(3\)-adic result is \(5/29 = \dots 121_3\)

Explained slightly differently:

The goal is to find \(...d_2d_1d_0\). To pick the digit \(d_i\) we have to answer what \(d_i \in \{0, 1, 2\}\) makes the numerator of \((\text{Value} - d_i)\) divisible by \(3\)? For \(5/29\), \(d_0 = 1\) because \(5-29(1) = -24\), which is divisible by \(3.\) Subtract that digit: \(5/29 - 1 = -24/29\). Divide by \(3\): \((-24/29) \div 3 = -8/29\). Repeat: What \(d_1\) makes \(-8/29 - d_1\) divisible by 3? \(-8 - 29(2) = -66\), which is divisible by \(3\). So \(d_1 = 2\).

In general terms the algorithm works as follows:

  1. Set the initial “residual value” to your fraction:

\[x_0 = x\]

  1. Extract the residue: Find the digit \(d_i \in \{0, 1, \dots, p-1\}\) such that:

\[x_i \equiv d_i \pmod p\]

Technically, if \(x_i = \frac{a}{b}\), you solve the linear congruence \(b \cdot d_i \equiv a \pmod p\). Since \(p\) does not divide \(b\), a unique solution for \(d_i\) always exists.

  1. Calculate the next residual value \(x_{i+1}\) by removing the digit and performing a standard algebraic division by \(p\):\[x_{i+1} = \frac{x_i - d_i}{p}\]

This division is not a modular operation. It is a literal division in the field of rational numbers \(\mathbb{Q}\). Because we chose \(d_i\) such that \(x_i - d_i\) is a multiple of \(p\), the division “clears” a factor of \(p\) from the numerator, effectively shifting the \(p^{i+1}\) term into the “units” position for the next iteration. In modular arithmetic, it would always be \(0.\)

In p-adics we keep the residue \(d_i\): the unique digit in \(\{0, \dots, p-1\}\) that, when subtracted from the current value, creates a difference divisible by \(p\). This ensures that when we subsequently divide by \(p\) to find the next digit, we are performing an exact algebraic shift that moves the next power of \(p\) into the units position.

Base-\(3\) approximations (\(0.17\)) get closer to a value by reducing the physical distance on a ruler. \(3\)-adic approximations get closer by making the error more divisible by \(3\). This is why we can represent fractions as “infinite integers” in p-adics: we are adding terms that are so highly divisible by \(3\) that they eventually become “invisible” or zero in the limit. While regular division feels like chasing an approximation, p-adic steps feel exact. This is because each p-adic digit is a perfect solution to a modular equation (\(29x \equiv 5 \pmod{3^n}\)). Once a digit is found, it is “locked in” and never changes.

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NOTE: These are tentative notes on different topics for personal use - expect mistakes and misunderstandings.