Homotopy Type Theory UMyn8W7b (Rev #54, changes)

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Euclidean semirings

Given a additively cancellative commutative semiring $R$ , a term $e:R$ is left cancellative if for all $a:R$ and $b:R$ , $e \cdot a = e \cdot b$ implies $a = b$ .

\mathrm{isLeftCancellative}(e) \coloneqq \prod_{a:R} \prod_{b :R}(e \cdot a = e \cdot b) \to (a = b)

A term $e:R$ is right cancellative if for all $a:R$ and $b:R$ , $a \cdot e = b \cdot e$ implies $a = b$ .

\mathrm{isRightCancellative}(e) \coloneqq \prod_{a:R} \prod_{b :R}(a \cdot e = b \cdot e) \to (a = b)

An term $e:R$ is cancellative if it is both left cancellative and right cancellative.

\mathrm{isCancellative}(e) \coloneqq \mathrm{isLeftCancellative}(e) \times \mathrm{isRightCancellative}(e)

The multiplicative submonoid of cancellative elements in $R$ is the subset of all cancellative elements in $R$

\mathrm{Can}(R) \coloneqq \sum_{e:R} \mathrm{isCancellative}(e)

A Euclidean semiring is a additively cancellative commutative semiring $R$ for which there exists a function $d \colon \mathrm{Can}(R) \to \mathbb{N}$ from the multiplicative submonoid of cancellative elements in $R$ to the natural numbers, often called a degree function, a function $(-)\div(-):R \times \mathrm{Can}(R) \to R$ called the division function, and a function $(-)\;\%\;(-):R \times \mathrm{Can}(R) \to R$ called the remainder function, such that for all $a \in R$ and $b \in \mathrm{Can}(R)$ , $a = (a \div b) \cdot b + (a\;\%\; b)$ and either $a\;\%\; b = 0$ or $d(a\;\%\; b) \lt d(g)$ .

Non-cancellative and non-invertible elements

Given a ring $R$ , an element $x \in R$ is non-cancellative if: if there is an element $y \in \mathrm{Can}(R)$ with injection $i:\mathrm{Can}(R) \to R$ such that $i(y) = x$ , then $0 = 1$ . An element $x \in R$ is non-invertible if: if there is an element $y \in R^\times$ with injection $j:R^\times \to R$ such that $j(y) = x$ , then $0 = 1$ .

To On the real numbers and square roots

Q-vector space

free Q-vector spaces?

strictly ordered Q-vector space?

Archimedean Q-vector space?

There is a significant difference between square roots and $n$ -th roots. Square roots are the inverse operation of the diagonal $f(x, x)$ for any binary operation $f$ , while $n$ -th roots are inverse operations of the $n$ -dimensional diagonals $g(x, x, \ldots, x)$ for $n$ -ary operations, which we typically do not formally talk about in typical practice for rings, etc…

Algebra

Given a commutative ring $R$ , there is a commutative ring $A$ where $R$ is a subring of $A$ , with a function $(-)\circ(-):A \times A \to A$ called composition, a term $x:A$ called the composition identity, a function $S_{(-)}:R \times A \to A$ called the shift, and a function $\partial:A \to A$ called the derivative such that

rules for composition:

for all $a:R$ , $a \circ f = a$
for all $f:A$ , $f \circ x = f$
for all $f:A$ , $x \circ f = f$
for all $f:A$ , $g:A$ , and $h:A$ , $f \circ (g \circ h) = (f \circ g) \circ h$
for all $f:A$ , $g:A$ , and $h:A$ , $(f + g) \circ h = (f \circ h) + (g \circ h)$
for all $f:A$ , $g:A$ , and $h:A$ , $(f g) \circ h = (f \circ h) (g \circ h)$

rules for shifts:

for all $a:R$ and $f:A$ $S_a f = f \circ (x - a)$

rules for derivatives:

$\partial(x) = 1$
for all $a:R$ , $\partial(a) = 0$
for all $f:A$ and $g:A$ , $\partial(f + g) = \partial(f) + \partial(g)$
for all $f:A$ and $g:A$ , $\partial(f g) = \partial(f) g + f \partial(g)$

Symbolic representations of formal smooth functions on the entire domain.

Revision on June 3, 2022 at 20:49:15 by Anonymous?. See the history of this page for a list of all contributions to it.