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| author | Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> | 2024-05-07 14:50:35 +0200 |
|---|---|---|
| committer | Simon-Kor <52245124+Simon-Kor@users.noreply.github.com> | 2024-05-07 14:50:35 +0200 |
| commit | aeef2bd2dfc7e1a7f1865ee5455e934d9dedaa32 (patch) | |
| tree | 7ae3b0de772215c2caec882e7865be3ef4374c11 /library | |
| parent | 3795588d157864a411baf2fc3afb31f9f5184d93 (diff) | |
Clean up of Notation in numbers.tex
First notation of tupels in the relation set was swapped with the canonical <.
Diffstat (limited to 'library')
| -rw-r--r-- | library/numbers.tex | 26 |
1 files changed, 13 insertions, 13 deletions
diff --git a/library/numbers.tex b/library/numbers.tex index afb7d3f..df47d81 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -53,16 +53,16 @@ \end{abbreviation} \begin{axiom}\label{reals_axiom_dense} - For all $x,y \in \reals$ if $(x,y)\in \lt[\reals]$ then - there exist $z \in \reals$ such that $(x,z) \in \lt[\reals]$ and $(z,y) \in \lt[\reals]$. + For all $x,y \in \reals$ if $x < y$ then + there exist $z \in \reals$ such that $x < z$ and $z < y$. \end{axiom} \begin{axiom}\label{reals_axiom_order_def} - $(x,y) \in \lt[\reals]$ iff there exist $z \in \reals$ such that $(\zero, z) \in \lt[\reals]$ and $x + z = y$. + $x < y$ iff there exist $z \in \reals$ such that $\zero < z$ and $x + z = y$. \end{axiom} \begin{lemma}\label{reals_one_bigger_than_zero} - $(\zero,1) \in \lt[\reals]$. + $\zero < 1$. \end{lemma} @@ -111,22 +111,22 @@ \begin{lemma}\label{order_reals_lemma1} - For all $x,y,z \in \reals$ such that $(\zero,x) \in \lt[\reals]$ - if $(y,z) \in \lt[\reals]$ - then $((y \times x), (z \times x)) \in \lt[\reals]$. + For all $x,y,z \in \reals$ such that $\zero < x$ + if $y < z$ + then $(y \times x) < (z \times x)$. \end{lemma} \begin{lemma}\label{order_reals_lemma2} - For all $x,y,z \in \reals$ such that $(\zero,x) \in \lt[\reals]$ - if $(y,z) \in \lt[\reals]$ - then $((x \times y), (x \times z)) \in \lt[\reals]$. + For all $x,y,z \in \reals$ such that $\zero < x$ + if $y < z$ + then $(x \times y) < (x \times z)$. \end{lemma} \begin{lemma}\label{order_reals_lemma3} - For all $x,y,z \in \reals$ such that $(x,\zero) \in \lt[\reals]$ - if $(y,z) \in \lt[\reals]$ - then $((x \times z), (x \times y)) \in \lt[\reals]$. + For all $x,y,z \in \reals$ such that $x < \zero$ + if $y < z$ + then $(x \times z) < (x \times y)$. \end{lemma} \begin{lemma}\label{a} |
