Latex breaks under reading mode

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1

When going from editing to reading mode 2 unintended things happen:

  1. Inline math breaks
  2. Some bullet points are different style than others.
    Atached is editing, reading and source.

Have any of you already dealt with a similiar issue?

imagen
imagen827×305 11.7 KB

imagen
imagen832×339 14 KB

imagen
imagen823×373 31.5 KB

As I am new to Obsidian I appreciate all help.
TYIA

2

Can you paste the entire section in a

code block,

so we can see what's happening?

IMG_9137
IMG_91371179×704 85.1 KB

https://help.obsidian.md/Editing+and+formatting/Basic+formatting+syntax#Code+blocks

More often than not, the indentation is off or a new line is needed for Markdown to render correctly. Seeing the source in a code block can help. :crossed_fingers:

3 
Para comprobar espacio vectorial hay que verificar: 
- Suma
	- Asociativa 
$$
\begin{align}
(\bar{x}[+]\bar{y})[+]\bar{z}&=\left(\sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{x_{n}^3 +y_{n}^3}\right) [+]\bar{z} \\
&= \left(\sqrt[3]{\left(\sqrt[3]{\bar{x}_{1}^3+ \bar{y}_{1}^3 }\right)^3 +\bar{z}_{1}^3},\dots,\sqrt[3]{\left(\sqrt[3]{\bar{x}_{n}^3+ \bar{y}_{n}^3 }\right)^3 +\bar{z}_{n}^3} \right) \\
&= \left( \sqrt[3]{\bar{x}_{1}^3 +\bar{y}_{1}^3+\bar{z}_{1}^3 },\dots,\sqrt[3]{\bar{x}_{n}^3 +\bar{y}_{n}^3+\bar{z}_{n}^3 }\right) = (1) \\
\bar{x}[+](\bar{y}[+]\bar{z}) &= \bar{x} [+]\left(\sqrt[3]{y_{1}^3+z_{1}^3},\dots,\sqrt[3]{y_{n}^3 +z_{n}^3}\right) \\
&= \left(\sqrt[3]{\bar{x}_{1}^3+\left(\sqrt[3]{\bar{y}_{1}^3+ \bar{z}_{1}^3 }\right)^3},\dots,\sqrt[3]{\bar{x}_{n}^3+\left(\sqrt[3]{\bar{y}_{n}^3+ \bar{z}_{n}^3 }\right)^3} \right) \\
&= \left( \sqrt[3]{\bar{x}_{1}^3 +\bar{y}_{1}^3+\bar{z}_{1}^3 },\dots,\sqrt[3]{\bar{x}_{n}^3 +\bar{y}_{n}^3+\bar{z}_{n}^3 }\right) = (2) \\ \\
(1) &= (2) 
\end{align}
$$


	- Comutativa
$$
\begin{align}
\bar{x} [+] \bar{y} &= \left(\sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{x_{n}^3 +y_{n}^3}\right) \\
&= \left(\sqrt[3]{y_{1}^3+x_{1}^3},\dots,\sqrt[3]{y_{n}^3 +x_{n}^3}\right)&= \bar{y}[+]\bar{x}
\end{align}
$$
	- Elemento neutro
$$
\begin{align}
\bar{x} [+] \bar{0} &= (x_{1},\dots,x_{n})[+] (0,\dots,0) \\
&= \left( \sqrt[3]{x_{1}^3 +0 },\dots,\sqrt[3]{{x}_{n}^3+0}  \right) \\
&= \left( \sqrt[3]{{x}_{1}^3},\dots,\sqrt[3]{{x}_{n}^3 } \right) \\
&= \bar{x}
\end{align}
$$

	- Elemento opuesto $\forall \bar{x}\exists -\bar{x}|\bar{x}[+](-\bar{x})=\bar{0}$

$$
\begin{align}
(\bar{x}_{1},\dots,\bar{x}_{n}) [+] (-\bar{x}_{1},\dots,-\bar{x}_{n}) &= \left( \sqrt[3]{x_{1}^3+(-x_{1}^3)},\dots,\sqrt[3]{x_{n}^3 +(-x_{n}^3})\right) \\
&=(0,\dots,0) \\
&= \bar{0}
\end{align}
$$
- Multiplicacion
	- Distributiva segun vectores
$$
\begin{align}
\lambda[\cdot](\bar{x}[+]\bar{y}) &= \lambda [\cdot] \left( \sqrt[3]{x_{1}^3+y_{1}^{3}},\dots,\sqrt[3]{x_{n}^3+y_{n}^3}  \right) \\
&= \left( \sqrt[3]{\lambda}\cdot \sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{\lambda}\cdot \sqrt[3]{x_{n}^3+y_{n}^3}  \right) &=(1) \\
\left((\lambda[\cdot]\bar{x})[+](\lambda[\cdot]\bar{y})\right)&= \left(\sqrt[3]{\lambda}x_{1},\dots,\sqrt[3]{\lambda}x_{n}\right)[+]\left( \sqrt[3]{\lambda}y_{1},\dots,\sqrt[3]{\lambda}y_{n}  \right) \\
&= \left( \sqrt[3]{\lambda x_{1}^3+\lambda y_{1}^3},\dots,\sqrt[3]{\lambda x_{n}^3+\lambda y_{n}^3}  \right) \\
&= \left( \sqrt[3]{\lambda}\cdot \sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{\lambda}\cdot \sqrt[3]{x_{n}^3+y_{n}^3}  \right) &=(2)  \\
(1) &= (2)
\end{align}
$$
	- Distributiva segun escalares
$$
\begin{align}
(\lambda + \mu)[\cdot]\bar{x} &= (\sqrt[3]{\lambda+\mu}\cdot x_{1},\dots,\sqrt[3]{\lambda+\mu}\cdot x_{n}) &=(1)\\
\lambda[\cdot]\bar{x}[+]\mu[\cdot]\bar{x} &= \left(\sqrt[3]{\lambda}x_{1},\dots,\sqrt[3]{\lambda}x_{n}\right) [+]\left(\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\mu}x_{n}\right) \\
&=\left(\sqrt[3]{\left(\sqrt[3]{\lambda}x_{1}\right)^3+\left(\sqrt[3]{\mu}x_{1}\right)^3},\dots,\sqrt[3]{\left(\sqrt[3]{\lambda}x_{n}\right)^3+\left(\sqrt[3]{\mu}x_{n}\right)^3}\right) \\
&= \left( \sqrt[3]{(\lambda+\mu)x_{1}^3} ,\dots,\sqrt[3]{(\lambda+\mu)x_{n}^3} \right) \\
&= \left( \sqrt[3]{\lambda+\mu}\cdot x_{1},\dots,\sqrt[3]{\lambda+\mu}\cdot x_{n}  \right) &= (2) \\
(1) = (2)
\end{align}
$$
	- Asociativa segun producto de escalares
$$
\begin{align}
\left( \lambda \mu\right) [\cdot]\bar{x} &= \left(\sqrt[3]{\lambda \mu}x_{1},\dots,\sqrt[3]{\lambda \mu}x_{n}\right) \\
&=\left(\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{n}\right) &= (1) \\
 \lambda[\cdot](\mu[\cdot]\bar{x}) &= \lambda [\cdot] \left(\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\mu}x_{n}\right) \\
&=\left(\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{n}\right) &= (2) \\
(1) &= (2)
\end{align}
$$
	- Identidad
$$
\begin{align}
1 [\cdot] \bar{x} &= \left(\sqrt[3]{1}x_{1},\dots,\sqrt[3]{1}x_{n}\right) \\
&= \left(x_{1},\dots,x_{n}\right) \\
&= \bar{x}
\end{align}
$$
4

I added some new lines separating the math blocks and fixed the indentation in a few places. So, I think the main issue is that you had a lot of

- hi
$$
\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}
$$
    - hello

where the - hello isn’t part of the - hi list anymore (because of the math block in the middle).

Hopefully this is a bit better:

Para comprobar espacio vectorial hay que verificar: 
- Suma
	- Asociativa 

$$
\begin{align}
(\bar{x}[+]\bar{y})[+]\bar{z}&=\left(\sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{x_{n}^3 +y_{n}^3}\right) [+]\bar{z} \\
&= \left(\sqrt[3]{\left(\sqrt[3]{\bar{x}_{1}^3+ \bar{y}_{1}^3 }\right)^3 +\bar{z}_{1}^3},\dots,\sqrt[3]{\left(\sqrt[3]{\bar{x}_{n}^3+ \bar{y}_{n}^3 }\right)^3 +\bar{z}_{n}^3} \right) \\
&= \left( \sqrt[3]{\bar{x}_{1}^3 +\bar{y}_{1}^3+\bar{z}_{1}^3 },\dots,\sqrt[3]{\bar{x}_{n}^3 +\bar{y}_{n}^3+\bar{z}_{n}^3 }\right) = (1) \\
\bar{x}[+](\bar{y}[+]\bar{z}) &= \bar{x} [+]\left(\sqrt[3]{y_{1}^3+z_{1}^3},\dots,\sqrt[3]{y_{n}^3 +z_{n}^3}\right) \\
&= \left(\sqrt[3]{\bar{x}_{1}^3+\left(\sqrt[3]{\bar{y}_{1}^3+ \bar{z}_{1}^3 }\right)^3},\dots,\sqrt[3]{\bar{x}_{n}^3+\left(\sqrt[3]{\bar{y}_{n}^3+ \bar{z}_{n}^3 }\right)^3} \right) \\
&= \left( \sqrt[3]{\bar{x}_{1}^3 +\bar{y}_{1}^3+\bar{z}_{1}^3 },\dots,\sqrt[3]{\bar{x}_{n}^3 +\bar{y}_{n}^3+\bar{z}_{n}^3 }\right) = (2) \\ \\
(1) &= (2) 
\end{align}
$$

- Comutativa

$$
\begin{align}
\bar{x} [+] \bar{y} &= \left(\sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{x_{n}^3 +y_{n}^3}\right) \\
&= \left(\sqrt[3]{y_{1}^3+x_{1}^3},\dots,\sqrt[3]{y_{n}^3 +x_{n}^3}\right)&= \bar{y}[+]\bar{x}
\end{align}
$$

- Elemento neutro

$$
\begin{align}
\bar{x} [+] \bar{0} &= (x_{1},\dots,x_{n})[+] (0,\dots,0) \\
&= \left( \sqrt[3]{x_{1}^3 +0 },\dots,\sqrt[3]{{x}_{n}^3+0}  \right) \\
&= \left( \sqrt[3]{{x}_{1}^3},\dots,\sqrt[3]{{x}_{n}^3 } \right) \\
&= \bar{x}
\end{align}
$$

- Elemento opuesto $\forall \bar{x}\exists -\bar{x}|\bar{x}[+](-\bar{x})=\bar{0}$

$$
\begin{align}
(\bar{x}_{1},\dots,\bar{x}_{n}) [+] (-\bar{x}_{1},\dots,-\bar{x}_{n}) &= \left( \sqrt[3]{x_{1}^3+(-x_{1}^3)},\dots,\sqrt[3]{x_{n}^3 +(-x_{n}^3})\right) \\
&=(0,\dots,0) \\
&= \bar{0}
\end{align}
$$

- Multiplicacion
	- Distributiva segun vectores

$$
\begin{align}
\lambda[\cdot](\bar{x}[+]\bar{y}) &= \lambda [\cdot] \left( \sqrt[3]{x_{1}^3+y_{1}^{3}},\dots,\sqrt[3]{x_{n}^3+y_{n}^3}  \right) \\
&= \left( \sqrt[3]{\lambda}\cdot \sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{\lambda}\cdot \sqrt[3]{x_{n}^3+y_{n}^3}  \right) &=(1) \\
\left((\lambda[\cdot]\bar{x})[+](\lambda[\cdot]\bar{y})\right)&= \left(\sqrt[3]{\lambda}x_{1},\dots,\sqrt[3]{\lambda}x_{n}\right)[+]\left( \sqrt[3]{\lambda}y_{1},\dots,\sqrt[3]{\lambda}y_{n}  \right) \\
&= \left( \sqrt[3]{\lambda x_{1}^3+\lambda y_{1}^3},\dots,\sqrt[3]{\lambda x_{n}^3+\lambda y_{n}^3}  \right) \\
&= \left( \sqrt[3]{\lambda}\cdot \sqrt[3]{x_{1}^3+y_{1}^3},\dots,\sqrt[3]{\lambda}\cdot \sqrt[3]{x_{n}^3+y_{n}^3}  \right) &=(2)  \\
(1) &= (2)
\end{align}
$$

- Distributiva segun escalares

$$
\begin{align}
(\lambda + \mu)[\cdot]\bar{x} &= (\sqrt[3]{\lambda+\mu}\cdot x_{1},\dots,\sqrt[3]{\lambda+\mu}\cdot x_{n}) &=(1)\\
\lambda[\cdot]\bar{x}[+]\mu[\cdot]\bar{x} &= \left(\sqrt[3]{\lambda}x_{1},\dots,\sqrt[3]{\lambda}x_{n}\right) [+]\left(\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\mu}x_{n}\right) \\
&=\left(\sqrt[3]{\left(\sqrt[3]{\lambda}x_{1}\right)^3+\left(\sqrt[3]{\mu}x_{1}\right)^3},\dots,\sqrt[3]{\left(\sqrt[3]{\lambda}x_{n}\right)^3+\left(\sqrt[3]{\mu}x_{n}\right)^3}\right) \\
&= \left( \sqrt[3]{(\lambda+\mu)x_{1}^3} ,\dots,\sqrt[3]{(\lambda+\mu)x_{n}^3} \right) \\
&= \left( \sqrt[3]{\lambda+\mu}\cdot x_{1},\dots,\sqrt[3]{\lambda+\mu}\cdot x_{n}  \right) &= (2) \\
(1) = (2)
\end{align}
$$

- Asociativa segun producto de escalares

$$
\begin{align}
\left( \lambda \mu\right) [\cdot]\bar{x} &= \left(\sqrt[3]{\lambda \mu}x_{1},\dots,\sqrt[3]{\lambda \mu}x_{n}\right) \\
&=\left(\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{n}\right) &= (1) \\
 \lambda[\cdot](\mu[\cdot]\bar{x}) &= \lambda [\cdot] \left(\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\mu}x_{n}\right) \\
&=\left(\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{1},\dots,\sqrt[3]{\lambda}\sqrt[3]{\mu}x_{n}\right) &= (2) \\
(1) &= (2)
\end{align}
$$

- Identidad

$$
\begin{align}
1 [\cdot] \bar{x} &= \left(\sqrt[3]{1}x_{1},\dots,\sqrt[3]{1}x_{n}\right) \\
&= \left(x_{1},\dots,x_{n}\right) \\
&= \bar{x}
\end{align}
$$