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* Remover demonstração que antes estava em ex1.tex * Actualizar ano, e remover listagens de teoremas * Exemplo de referência * Update para a release a fazer
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@misc{uabefoliotemplate, | ||
author = {{Pinto Machado}, Carlos}, | ||
title = {uab-efolio-template}, | ||
year = {2023}, | ||
version = {0.0.2}, | ||
year = {2024}, | ||
version = {0.0.3}, | ||
url = {https://github.com/cpmachado/uab-efolio-template} | ||
} |
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\exercicio | ||
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\paragraph{} Não esquecer que mencionar este template | ||
\cite[Algo]{uabefoliotemplate}. | ||
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\alinea | ||
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\begin{theorem}[Somatório de números naturais] | ||
Para $n \in \mathbb{N}$, tem-se que $\sum_{k = 1}^n k = \frac{n(n+1)}{2}$. | ||
\end{theorem} | ||
\begin{prop}\label{prop1a} | ||
Seja $n \in \mathbb{N}$, então $n + 1 \in \mathbb{N}$. | ||
\end{prop} | ||
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\begin{proof} | ||
\; \\ | ||
Caso base $n = 1$, tem-se que 1 = 1, pelo que verifica o caso base.\\ | ||
Fixado $n \in \mathbb{N}$, vamos supor: | ||
\begin{align*} | ||
\sum_{k = 1}^n k &= \frac{n(n+1)}{2} &&\text{(Hipótese de Indução)} | ||
\intertext{Pretende-se provar:} | ||
\sum_{k = 1}^{n + 1} k &= \frac{(n + 1)(n+2)}{2} &&\text{(Tese de Indução)} | ||
\intertext{Passo de Indução:} | ||
\sum_{k = 1}^{n + 1} k &= n + 1 + \sum_{k = 1}^n k | ||
\overset{\text{\tiny passo de indução}}{=} | ||
n + 1 + \frac{n (n + 1)}{2} \\ | ||
&= \frac{n (n + 1) + 2(n + 1)}{2} | ||
= \frac{(n +2)(n + 1)}{2} | ||
\end{align*} | ||
\\ | ||
Por indução, etc. fica provada a proposição \ref{prop1a}. | ||
\end{proof} | ||
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