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authorJuan Marin Noguera <juan@mnpi.eu>2022-12-04 22:49:17 +0100
committerJuan Marin Noguera <juan@mnpi.eu>2022-12-04 22:49:17 +0100
commitc34b47089a133e58032fe4ea52f61efacaf5f548 (patch)
tree4242772e26a9e7b6f7e02b1d1e00dfbe68981345 /ga/n2.lyx
parent214b20d1614b09cd5c18e111df0f0d392af2e721 (diff)
Oops
Diffstat (limited to 'ga/n2.lyx')
-rw-r--r--ga/n2.lyx38
1 files changed, 19 insertions, 19 deletions
diff --git a/ga/n2.lyx b/ga/n2.lyx
index caf4b8a..6617d41 100644
--- a/ga/n2.lyx
+++ b/ga/n2.lyx
@@ -610,7 +610,7 @@ Ambos son subanillos de
\end_inset
sería cuadrado de racional, pero si llamamos
-\begin_inset Formula $\frac{p}{q}:=\frac{a}{b}$
+\begin_inset Formula $\frac{p}{q}\coloneqq \frac{a}{b}$
\end_inset
como fracción irreducible,
@@ -2549,7 +2549,7 @@ equivalentes
\end_inset
de
-\begin_inset Formula $\mathbb{N}_{n}:=\{1,\dots,n\}$
+\begin_inset Formula $\mathbb{N}_{n}\coloneqq \{1,\dots,n\}$
\end_inset
tal que para
@@ -3305,7 +3305,7 @@ Demostración:
.
Sea
-\begin_inset Formula $I:=(a_{1},a_{2},\dots)=\bigcup_{n\in\mathbb{N}}(a_{n})$
+\begin_inset Formula $I\coloneqq (a_{1},a_{2},\dots)=\bigcup_{n\in\mathbb{N}}(a_{n})$
\end_inset
, como
@@ -3376,7 +3376,7 @@ euclídea
\end_layout
\begin_layout Enumerate
-\begin_inset Formula $\forall a\in D,b\in D\setminus\{0\},\exists q,r\in D\mid (a=bq+r\land(r=0\lor\delta(r)<\delta(b)))$
+\begin_inset Formula $\forall a\in D,b\in D\setminus\{0\},\exists q,r\in D:(a=bq+r\land(r=0\lor\delta(r)<\delta(b)))$
\end_inset
.
@@ -3416,7 +3416,7 @@ status open
\begin_layout Plain Layout
Si
-\begin_inset Formula $x:=a+bi$
+\begin_inset Formula $x\coloneqq a+bi$
\end_inset
con
@@ -3424,7 +3424,7 @@ Si
\end_inset
,
-\begin_inset Formula $\delta(x):=|x|^{2}=a^{2}+b^{2}\in\mathbb{N}$
+\begin_inset Formula $\delta(x)\coloneqq |x|^{2}=a^{2}+b^{2}\in\mathbb{N}$
\end_inset
.
@@ -3438,11 +3438,11 @@ Si
, de donde se obtiene la primera condición.
Sean ahora
-\begin_inset Formula $a:=a_{1}+a_{2}i$
+\begin_inset Formula $a\coloneqq a_{1}+a_{2}i$
\end_inset
y
-\begin_inset Formula $b:=b_{1}+b_{2}i\neq0$
+\begin_inset Formula $b\coloneqq b_{1}+b_{2}i\neq0$
\end_inset
con
@@ -3450,7 +3450,7 @@ Si
\end_inset
,
-\begin_inset Formula $x:=x_{1}+x_{2}i:=\frac{a}{b}$
+\begin_inset Formula $x\coloneqq x_{1}+x_{2}i\coloneqq \frac{a}{b}$
\end_inset
con
@@ -3474,11 +3474,11 @@ Si
\end_inset
respectivamente,
-\begin_inset Formula $q:=q_{1}+q_{2}i$
+\begin_inset Formula $q\coloneqq q_{1}+q_{2}i$
\end_inset
y
-\begin_inset Formula $r:=a-bq$
+\begin_inset Formula $r\coloneqq a-bq$
\end_inset
.
@@ -3810,7 +3810,7 @@ Sean
\end_inset
un dominio y
-\begin_inset Formula $X:=D\times(D\setminus\{0\})$
+\begin_inset Formula $X\coloneqq D\times(D\setminus\{0\})$
\end_inset
, definimos la relación binaria
@@ -3864,7 +3864,7 @@ status open
\begin_layout Standard
Llamamos
-\begin_inset Formula $a/s:=\frac{a}{s}:=[(a,s)]\in Q(D):=X/\sim$
+\begin_inset Formula $a/s\coloneqq \frac{a}{s}\coloneqq [(a,s)]\in Q(D)\coloneqq X/\sim$
\end_inset
, y las operaciones
@@ -4155,7 +4155,7 @@ Demostración:
\end_inset
dada por
-\begin_inset Formula $u(a):=a/1$
+\begin_inset Formula $u(a)\coloneqq a/1$
\end_inset
es un homomorfismo inyectivo, por lo que podemos ver a
@@ -4191,7 +4191,7 @@ Propiedad universal del cuerpo de fracciones:
\end_inset
dada por
-\begin_inset Formula $u(a):=a/1$
+\begin_inset Formula $u(a)\coloneqq a/1$
\end_inset
:
@@ -4300,7 +4300,7 @@ status open
\begin_layout Plain Layout
El homomorfismo
-\begin_inset Formula $f:=g\circ u=h\circ u$
+\begin_inset Formula $f\coloneqq g\circ u=h\circ u$
\end_inset
es inyectivo por serlo
@@ -4530,7 +4530,7 @@ Demostración:
\end_inset
, sea
-\begin_inset Formula $t:=(c+d\sqrt{m})(c-d\sqrt{m})$
+\begin_inset Formula $t\coloneqq (c+d\sqrt{m})(c-d\sqrt{m})$
\end_inset
, entonces
@@ -4633,7 +4633,7 @@ Demostración:
\end_inset
dado por
-\begin_inset Formula $f(n):=n1$
+\begin_inset Formula $f(n)\coloneqq n1$
\end_inset
es un homomorfismo inyectivo y la propiedad universal nos da un homomorfismo
@@ -4647,7 +4647,7 @@ Demostración:
.
Es claro entonces que
-\begin_inset Formula $K':=\tilde{f}(\mathbb{Q})$
+\begin_inset Formula $K'\coloneqq \tilde{f}(\mathbb{Q})$
\end_inset
es isomorfo a