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Diffstat (limited to 'library/numbers.tex')
| -rw-r--r-- | library/numbers.tex | 62 |
1 files changed, 31 insertions, 31 deletions
diff --git a/library/numbers.tex b/library/numbers.tex index d3af3f1..a675ea5 100644 --- a/library/numbers.tex +++ b/library/numbers.tex @@ -93,8 +93,8 @@ For all $n,m \in \naturals$ we have $(n + m) \in \naturals$. \end{axiom} -\subsubsection{Natural numbers as ordinals} +\subsubsection{Natural numbers as ordinals} \begin{lemma}\label{nat_is_successor_ordinal} Let $n\in\naturals$. @@ -187,7 +187,7 @@ Follows by \cref{suc_eq_plus_one,naturals_addition_axiom_2,naturals_addition_axiom_1,naturals_inductive_set,one_is_suc_zero}. \end{proof} -\begin{proposition}\label{naturals_add_kommu} +\begin{proposition}\label{naturals_add_comm} For all $n \in \naturals$ we have for all $m\in \naturals$ we have $n + m = m + n$. \end{proposition} \begin{proof} @@ -235,17 +235,17 @@ \end{align*} \end{proof} -\begin{proposition}\label{naturals_add_remove_brakets} +\begin{proposition}\label{naturals_add_remove_parens} Suppose $n,m,k \in \naturals$. - Then $(n + m) + k = n + m + k = n + (m + k)$. + Then $(n + m) + k = n + m + k = n + (m + k)$. \end{proposition} -\begin{proposition}\label{naturals_add_remove_brakets2} +\begin{proposition}\label{naturals_add_remove_parens2} Suppose $n,m,k,l \in \naturals$. - Then $(n + m) + (k + l)= n + m + k + l = n + (m + k) + l = (((n + m) + k) + l)$. + Then $(n + m) + (k + l)= n + m + k + l = n + (m + k) + l = (((n + m) + k) + l)$. \end{proposition} -%\begin{proposition}\label{natural_disstro_oneline} +%\begin{proposition}\label{natural_distrib_oneline} % Suppose $n,m,k \in \naturals$. % Then $n \rmul (m + k) = (n \rmul m) + (n \rmul k)$. %\end{proposition} @@ -262,7 +262,7 @@ % $ \suc{n} \rmul (m' + k') = (n \rmul (m' + k')) + (m' + k') = ((n \rmul m') + (n \rmul k')) + (m' + k') = ((n \rmul m') + (n \rmul k')) + m' + k' = (((n \rmul m') + (n \rmul k')) + m') + k' = (m' + ((n \rmul m') + (n \rmul k'))) + k' = ((m' + (n \rmul m')) + (n \rmul k')) + k' = (((n \rmul m') + m') + (n \rmul k')) + k' = ((n \rmul m') + m') + ((n \rmul k') + k') = (\suc{n} \rmul m') + (\suc{n} \rmul k')$. %\end{proof} -\begin{proposition}\label{natural_disstro} +\begin{proposition}\label{natural_distrib} Suppose $n,m,k \in \naturals$. Then $n \rmul (m + k) = (n \rmul m) + (n \rmul k)$. \end{proposition} @@ -283,11 +283,11 @@ \suc{n} \rmul (m' + k') \\ &= (n \rmul (m' + k')) + (m' + k') \\% \explanation{by \cref{naturals_mul_axiom_2}}\\ &= ((n \rmul m') + (n \rmul k')) + (m' + k') \\%\explanation{by assumption}\\ - &= ((n \rmul m') + (n \rmul k')) + m' + k' \\%\explanation{by \cref{naturals_add_remove_brakets,naturals_add_kommu}}\\ - &= (((n \rmul m') + (n \rmul k')) + m') + k' \\%\explanation{by \cref{naturals_add_kommu,naturals_add_remove_brakets,naturals_add_assoc}}\\ - &= (m' + ((n \rmul m') + (n \rmul k'))) + k' \\%\explanation{by \cref{naturals_add_kommu}}\\ + &= ((n \rmul m') + (n \rmul k')) + m' + k' \\%\explanation{by \cref{naturals_add_remove_parens,naturals_add_comm}}\\ + &= (((n \rmul m') + (n \rmul k')) + m') + k' \\%\explanation{by \cref{naturals_add_comm,naturals_add_remove_parens,naturals_add_assoc}}\\ + &= (m' + ((n \rmul m') + (n \rmul k'))) + k' \\%\explanation{by \cref{naturals_add_comm}}\\ &= ((m' + (n \rmul m')) + (n \rmul k')) + k' \\%\explanation{by \cref{naturals_add_assoc}}\\ - &= (((n \rmul m') + m') + (n \rmul k')) + k' \\%\explanation{by \cref{naturals_add_kommu}}\\ + &= (((n \rmul m') + m') + (n \rmul k')) + k' \\%\explanation{by \cref{naturals_add_comm}}\\ &= ((n \rmul m') + m') + ((n \rmul k') + k') \\%\explanation{by \cref{naturals_add_assoc}}\\ &= (\suc{n} \rmul m') + (\suc{n} \rmul k') %\explanation{by \cref{naturals_mul_axiom_2}} \end{align*} @@ -353,7 +353,7 @@ &=(k' \rmul (n' \rmul m')) + (k' \rmul m') \\ &=k' \rmul ((n' \rmul m') + m') \\ &=k' \rmul (\suc{n'} \rmul m') \\ - &=(\suc{n'} \rmul m') \rmul k' + &=(\suc{n'} \rmul m') \rmul k' \end{align*} \end{proof} @@ -385,7 +385,7 @@ Commutivatiy of the standart operations \begin{axiom}\label{reals_axiom_kommu} For all $x,y \in \reals$ $x + y = y + x$ and $x \rmul y = y \rmul x$. \end{axiom} - + \begin{axiom}\label{reals_axiom_assoc} For all $x,y,z \in \reals$ we have $(x + y) + z = x + (y + z)$. \end{axiom} @@ -393,7 +393,7 @@ Commutivatiy of the standart operations Existence of one and Zero \begin{axiom}\label{reals_axiom_zero} - For all $x \in \reals$ $x + \zero = x$. + For all $x \in \reals$ $x + \zero = x$. \end{axiom} \begin{axiom}\label{reals_axiom_one} @@ -453,7 +453,7 @@ Laws of the order on the reals \end{axiom} \begin{axiom}\label{reals_dense} - For all $x,y \in \reals$ if $x < y$ then + For all $x,y \in \reals$ if $x < y$ then there exist $z \in \reals$ such that $x < z$ and $z < y$. \end{axiom} @@ -471,15 +471,15 @@ Laws of the order on the reals \end{axiom} \begin{axiom}\label{reals_postiv_mul_is_positiv} - For all $x,y \in \reals$ such that $\zero < x,y$ we have $\zero < (x \rmul y)$. + For all $x,y \in \reals$ such that $\zero < x,y$ we have $\zero < (x \rmul y)$. \end{axiom} \begin{axiom}\label{reals_postiv_mul_negativ_is_negativ} - For all $x,y \in \reals$ such that $x < \zero < y$ we have $(x \rmul y) < \zero$. + For all $x,y \in \reals$ such that $x < \zero < y$ we have $(x \rmul y) < \zero$. \end{axiom} \begin{axiom}\label{reals_negativ_mul_is_negativ} - For all $x,y \in \reals$ such that $x,y < \zero$ we have $\zero < (x \rmul y)$. + For all $x,y \in \reals$ such that $x,y < \zero$ we have $\zero < (x \rmul y)$. \end{axiom} @@ -603,12 +603,12 @@ Laws of the order on the reals %\begin{abbreviation}\label{gcd} % $g$ is greatest common divisor of $\{a,b\}$ iff -% $g$ is a integral divisor of $a$ -% and $g$ is a integral divisor of $b$ +% $g$ is a integral divisor of $a$ +% and $g$ is a integral divisor of $b$ % and for all $g'$ such that $g'$ is a integral divisor of $a$ % and $g'$ is a integral divisor of $b$ we have $g' \leq g$. %\end{abbreviation} -% TODO: Was man noch so beweisen könnte: bruch kürzung, kehrbruch eigenschaften, bruch in bruch vereinfachung, +% TODO: Was man noch so beweisen könnte: bruch kürzung, kehrbruch eigenschaften, bruch in bruch vereinfachung, \subsection{Order on the reals} @@ -664,8 +664,8 @@ Laws of the order on the reals \begin{lemma}\label{order_reals_lemma1} Let $x,y,z \in \reals$. - Suppose $\zero < x$. - Suppose $y < z$. + Suppose $\zero < x$. + Suppose $y < z$. Then $(y \rmul x) < (z \rmul x)$. \end{lemma} \begin{proof} @@ -675,15 +675,15 @@ Laws of the order on the reals % (z \rmul x) \\ % &= ((y + k) \rmul x) \\ % &= ((y \rmul x) + (k \rmul x)) \explanation{by \cref{reals_disstro2}} - %\end{align*} - %Then $(k \rmul x) > \zero$. + %\end{align*} + %Then $(k \rmul x) > \zero$. %Therefore $(z \rmul x) > (y \rmul x)$. \end{proof} \begin{lemma}\label{order_reals_lemma2} Let $x,y,z \in \reals$. - Suppose $\zero < x$. - Suppose $y < z$. + Suppose $\zero < x$. + Suppose $y < z$. Then $(x \rmul y) < (x \rmul z)$. \end{lemma} \begin{proof} @@ -693,8 +693,8 @@ Laws of the order on the reals \begin{lemma}\label{order_reals_lemma3} Let $x,y,z \in \reals$. - Suppose $\zero > x$. - Suppose $y < z$. + Suppose $\zero > x$. + Suppose $y < z$. Then $(x \rmul z) < (x \rmul y)$. \end{lemma} \begin{proof} @@ -808,7 +808,7 @@ Laws of the order on the reals \begin{definition}\label{m_to_n_set} - $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. + $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. \end{definition} \begin{axiom}\label{abs_behavior1} |