Leo Shabarov · Notes

Bases & Adics

A clearer feel for place value and the \(p\)-adic flip. Chapters numbered in binary: \(1_{\,2}, 10_{\,2}, 11_{\,2}, 100_{\,2}, 101_{\,2}\).

Chapter \(1_{\,2}\): what base 10 really is (and why \(2+2\) can look like \(10\))

We write numbers using places. In base \(10\) there are ten symbols: \(0,1,2,3,4,5,6,7,8,9\). Each place is a power of \(10\).

So \(745\) means 7 hundreds, 4 tens, 5 ones: \(7\cdot 100 + 4\cdot 10 + 5\).

The idea of a base is to pick a bundle size. In base \(10\), ten ones bundle into a ten, ten tens into a hundred. If you change the bundle size to \(b\), the allowed digits are \(0\) up to \(b-1\). Each place is a power of \(b\). That’s the only change.

When someone says “\(2+2 = 10\),” they’re talking about notation, not value. The value four is written “\(10\)” in base \(4\) (one group of four and zero ones). In base \(2\), four is “\(100\)”.

Rule: inside one line, stick to one base. In base \(2\) you can’t even write the symbol “2,” so “\(2+2\)” isn’t legal there. You’d write \(10+10=100\) to mean “two plus two equals four” in base 2.

How carrying and borrowing work in any base

Carrying means you hit the bundle size and trade. In base \(b\), when a column sums to \(\ge b\), write the remainder and carry \(1\) to the next column.

Base 5 example: \(24_5 + 33_5\). Ones: \(4+3=7\). In base \(5\), \(7\) is “\(12\),” so write \(2\) and carry \(1\). Fives: \(2+3+1=6\). In base \(5\), \(6\) is “\(11\),” so write \(1\) and carry \(1\). New left digit is \(1\). Result \(112_5\). Check: \(24_5=14,\ 33_5=18,\ 14+18=32,\ 112_5=32\).

Multiplication and division follow the same school algorithms; the carry/borrow threshold is \(b\) instead of \(10\).

Chapter \(10_{\,2}\): reading & writing across bases

Read a base-\(b\) string

Expand by powers of \(b\).

Write a base-\(b\) string for a whole number \(N\)

Keep dividing by \(b\) and record the remainders; read them bottom to top.

Example: write \(117\) in base \(7\).
\(117=7\cdot 16+5\) (rem \(5\))
\(16=7\cdot 2+2\) (rem \(2\))
\(2=7\cdot 0+2\) (rem \(2\))
So \(117=\mathbf{225_7}\).

Fractions

Multiply the fractional part by \(b\) repeatedly; each whole part is the next digit after the point.

Example: \(0.375\) to base \(2\): \(0.375\!\times\! 2=0.75\) (digit \(0\)); \(0.75\!\times\! 2=1.5\) (digit \(1\), keep \(0.5\)); \(0.5\!\times\! 2=1.0\) (digit \(1\)). So \(0.375=0.\!011_2\).

Rationals terminate or repeat in any base. If the denominator’s primes are all in the base, you get a terminating expansion (e.g. base \(10\): only \(2\)s and \(5\)s terminate); otherwise it repeats.

Quick conversions when bases are powers

Sanity checks

All of this is still regular place value. Now we flip the viewpoint.

Chapter \(11_{\,2}\): a new sense of “close,” and 10-adic intuition

In adic land, “small” means “divisible by a big power of the base.” Two integers are close if they agree in many trailing digits.

Define distance so more shared trailing digits means closer. Powers of \(10\) become tiny steps. A 10-adic integer pins down every finite tail of decimal digits—picture a left-infinite \(\cdots d_3 d_2 d_1 d_0\). School arithmetic works; carries run left, and that’s fine because each finite tail stabilizes.

Two quick 10-adic facts:

  • \(\cdots 9999 = -1\). Each truncation \(9,99,999,\dots\) is \(-1\) mod \(10^k\); hence \(\cdots 9999 + 1 = 0\).
  • \(1+10+10^2+\dots = \dfrac{1}{1-10} = -\dfrac{1}{9}\) 10-adically; powers of \(10\) shrink.

Catch: \(10\) isn’t prime, so 10-adics split into \(2\)-adics × \(5\)-adics (zero divisors). For clean algebra, use a prime \(p\): the \(p\)-adics.

Chapter \(100_{\,2}\): \(p\)-adics from the ground up

Divisibility count & size

Let \(v_p(n)\) be the exponent of \(p\) dividing nonzero \(n\). Bigger \(v_p\) ⇒ “smaller” in \(p\)-adic size. Completing \(\mathbb Q\) under this size gives \(\mathbb Q_p\); its “integers” are \(\mathbb Z_p\).

Three equivalent views of \(\mathbb Z_p\)

Arithmetic is school arithmetic on digits, carrying left.

Units & why prime matters

In \(\mathbb Z_p\), numbers not divisible by \(p\) are invertible (first digit \(a_0\neq 0\)). Prime \(p\) ⇒ no zero divisors; the fraction field \(\mathbb Q_p\) is a field.

Compact notation & ultrametric

\(|x|_p = p^{-v_p(x)}\) (and \(|0|_p=0\)), distance \(d_p(x,y)=|x-y|_p\). Triangle rule: \(|x+y|_p \le \max(|x|_p,|y|_p)\).

By hand

Example: \(1/3\) in \(\mathbb Z_5\). Mod \(5\), \(3^{-1}\equiv 2\Rightarrow a_0=2\). Lift to \(25\): \(a=2+5t\), \(3a\equiv 1\Rightarrow t\equiv 3\Rightarrow a\equiv 17\). Digits start \(2,3,\dots\).

Hensel root: \(x^2=2\) in \(\mathbb Z_7\): start \(x\equiv 3\pmod 7\); one lift gives \(x\equiv 10\pmod{49}\); continue for more precision.

Chapter \(101_{\,2}\): connecting pictures & quick practice

Three views, one object

Chinese remainder: composite bases (e.g. \(10=2\cdot 5\)) split into prime pieces, introducing zero divisors—hence the focus on \(p\) prime.

Units, valuation, and size

Negatives & repeating tails

When Hensel can fail

If \(f'(x_0)\equiv 0\pmod p\), the simple version doesn’t apply (there are refined forms).


Worked examples you can trace

Convert bases (integers)

Convert \(300\) to base \(7\):
\(300=7\cdot 42+6\) (rem \(6\)) → \(42=7\cdot 6+0\) (rem \(0\)) → \(6=7\cdot 0+6\) (rem \(6\)).
Read up: \(\mathbf{606_7}\). Check: \(6\cdot 49+0\cdot 7+6=300\).

Convert \(1402_7\) to base \(10\): \(1\cdot 7^3 + 4\cdot 7^2 + 0\cdot 7 + 2 = 541\).

Convert bases (fractions)

\(1/5=0.2_{10}\) to base \(2\): digits repeat “0011,” so \(1/5=0.\overline{0011}_2\).

\(0.3_{10}\) to base \(3\): \(0.3\times 3=0.9\)(0); \(0.9\times 3=2.7\)(2); \(0.7\times 3=2.1\)(2); \(0.1\times 3=0.3\)(0); loop → \(0.\overline{220}_3\).

Multiplication in base \(5\)

\(243_5\times 14_5\). Values: \(243_5=73\), \(14_5=9\), product \(657\). Encode in base \(5\) by division → \(\mathbf{10112_5}\).\br/> (Or do school multiplication with carries at \(5\) to land on the same digits.)

A first 10-adic identity

Show \(\cdots 9999 + 1 = 0\) 10-adically. Each finite tail is \(-1\bmod 10^k\); adding \(1\) zeros every tail.

Inverses & lifts in \(\mathbb Z_p\)

\(1/3\) in \(\mathbb Z_5\) to five digits: \(2,3,1,3,1\), so \(1/3=2+3\cdot 5+1\cdot 5^2+3\cdot 5^3+1\cdot 5^4+\cdots\).

Square root in \(\mathbb Z_7\): start \(x\equiv 3\pmod 7\), lift \(x\equiv 10\pmod{49}\), then \(x\equiv 108\pmod{343}\), etc.

Practice (tap to reveal answers)

1) Convert \(300\) to base \(7\).
\(300=606_7\).
2) Convert \(0.2_{10}\) to base \(2\) (≈ 8 places).
\(0.2 \approx 0.\!0011\,0011_2\) (repeating \(0011\)).
3) Add in base \(6\): \(452_6 + 345_6\).
\(452_6 + 345_6 = \mathbf{1141_6}\).
4) Show \(\cdots 4444_5 = -1\) in \(\mathbb Z_5\).
Each truncation is \(-1\bmod 5^k\); the 5-adic limit is \(-1\).
5) First two base-5 digits of \(1/3\) in \(\mathbb Z_5\).
\(3^{-1}\equiv 2\pmod 5\Rightarrow a_0=2\). Lift to \(25\): \(a\equiv 17\Rightarrow a_1=3\).
6) Lift a root of \(x^2\equiv 2\pmod 7\) to \(\bmod 49\).
Start \(x_0=3\). One Hensel/Newton step gives \(x_1\equiv 10\pmod{49}\).
Clean end picture. Bases pick a bundle size (notation, not value). Adic closeness = many shared trailing digits. With prime \(p\), \(\mathbb Z_p\) and \(\mathbb Q_p\) give a robust arithmetic/analysis world where power series and lifting shine.