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 \(10\).
- The rightmost place means “ones.”
- Next left means “tens.”
- Next left means “hundreds,” and so on.
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.
- Base \(4\) uses digits \(0,1,2,3\) and places \(1,4,16,64,\dots\).
- Base \(2\) uses digits \(0,1\) and places \(1,2,4,8,\dots\).
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
- Binary → octal: group bits in 3s from the point.
- Binary → hex: group bits in 4s.
Sanity checks
- Value check: convert small samples to base \(10\).
- Digit range: digits must be \(0,\dots,b-1\).
- Length check: huge values shouldn’t have suspiciously few digits.
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.
- Match last digit ⇔ difference is multiple of \(10\).
- Match last two digits ⇔ multiple of \(100\).
- Match last \(k\) digits ⇔ multiple of \(10^k\).
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 \(\mathbb Z_p\)
- Limits of integers agreeing on longer base-\(p\) tails.
- Inverse limit of \(\mathbb Z/p^k\mathbb Z\).
- Power series \(a_0+a_1p+a_2p^2+\cdots\) with \(a_i\in\{0,\dots,p-1\}\).
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
- Negatives: \(-1=\cdots (p-1)\).
- Geometric: \(1+p+p^2+\cdots=\frac{1}{1-p}\).
- Inverses by lifting (when \(p\nmid m\)): solve one digit at a time via modular inverses.
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
- Digits (great for hand arithmetic)
- Residues mod \(p^k\) (great for lifting)
- Metric/limits (great for convergence/analysis)
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
- \(v_p(x+y)\ge \min(v_p(x),v_p(y))\) (strict if leading digits cancel).
- \(|p|_p=1/p\): powers of \(p\) shrink.
- \(a/b\in\mathbb Z_p \iff v_p(a)\ge v_p(b)\).
Negatives & repeating tails
- In base \(p\), \(\cdots (p-1) = -1\).
- More generally, left repeats encode negatives plus a finite correction.
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.