Foundational Arithmetic¶
Recall (doc/postdoc). Not a first course.
Job: ring operations on \(\mathbb{Z}\) (later \(\mathbb{Q},\mathbb{R},\mathbb{C}\)); order of ops is a convention for parsing, not a theorem.
Core¶
| Op | In rings | Notes |
|---|---|---|
| \(+\) | abelian group law | identity \(0\), inverse \(-a\) |
| \(\cdot\) | monoid (or ring mult.) | identity \(1\); \(a\cdot a^{-1}=1\) only if unit |
| \(-\) | \(a-b:=a+(-b)\) | not associative |
| \(\div\) | \(a/b:=a\cdot b^{-1}\) | requires unit \(b\) |
Division algorithm (in \(\mathbb{Z}\)): \(\forall a\in\mathbb{Z},\,b>0\), \(\exists!\,q,r\) with \(a=bq+r\), \(0\le r<b\).
Parsing convention (PEMDAS): grouping \(\to\) exponents \(\to\) \(\times/\div\) L→R \(\to\) \(+/-\) L→R. PEMDAS is not a theorem.
Zero divisors / \(0^{-1}\) do not exist in integral domains; \(a/0\) undefined.
Used later for¶
- Number theory: Euclidean algorithm, \(\gcd\), remainders
- Rings and fields: units, domains, \(\operatorname{Frac}(R)\)
- Modular arithmetic: residue of \(a\bmod b\)
- Real analysis: completeness not available in \(\mathbb{Z}\) alone
Check¶
In \(\mathbb{Z}\), does \(a=bq+r\) with \(0\le r<b\) determine \(q\) uniquely for fixed \(b>0\)?
Yes — uniqueness is part of the division algorithm; \(q=\lfloor a/b\rfloor\).