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## 编译环境
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## 我的问题
```
\documentclass{article}
\usepackage{amsmath}
\begin{document}
Considering $z \in [{k_1},{k_2}]$ and $x = {k_2},y=0$, by the condition condition , we have that $U_1^T({k_1},z) = U_1^T({k_2},{S^{{U_2}}}(0,z)) = {S^{{U_2}}}(U_1^T({k_2},0),z) = {S^{{U_2}}}({k_1},z) = z$, which implies that $U_1^T({k_2},z) = \min ({k_2},z)$ for all $z \in [{k_1},{k_2}]$.Thus $T$ must be an ordinal sum of the form $T \equiv \left(\left\langle {0,\frac{k_2 - k_1}{1 - k_1},T_1} \right\rangle,\left\langle {\frac{k_} - k_1}{1 -k_1},1,T_2}\right\rangle } \right)$.
\end{document}
```
用这个代码编辑的数学公式T超出了页边距，如何处理?
1 回答
19
提交代码之前先看看能不能编译，还有以后要把代码放代码块
```
\documentclass{article}
\usepackage{amsmath}
\usepackage[showframe]{geometry}
\begin{document}
Thus $T$ must be an ordinal sum of the form sum of the form sum of the form $T \equiv \left(\left\langle {0,\frac{k_2 - k_1}{1 - k_1},T_1} \right\rangle,\left\langle \frac{k - k_1}{1 -k_1},1,T_2\right\rangle \right)$.
Thus $T$ must be an ordinal sum of the form sum of the form sum of the form $T \equiv \Bigl(\left\langle {0,\frac{k_2 - k_1}{1 - k_1},T_1} \right\rangle$,
$\left\langle \frac{k - k_1}{1 -k_1},1,T_2\right\rangle \Bigr)$.
\end{document}
```
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