\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}

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\usepackage{amssymb}
% Exercise 11: Theorms

\usepackage{amsthm}

%\newtheorem{theorem}{Theorem}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}{Corollary}[theorem]
\newtheorem{lemma}[theorem]{Lemma}

\theoremstyle{remark}
\newtheorem*{remark}{Remark}

\theoremstyle{definition}
\newtheorem{definition}{Definition}[section]

\title{Theorems And Proofs}
\author{ShareLaTeX Team }
\date{May 2014}


\renewcommand\qedsymbol{$\blacksquare$}

\begin{document}

\maketitle

\section{Introduction}
Theorems can easily be defined

\begin{theorem}
Let $f$ be a function whose derivative exists in every point, then $f$ is a continuous function.
\end{theorem}

\begin{theorem}[Pythagorean theorem]
\label{pythagorean}
This is a theorema about right triangles and can be summarised in the next equation 
\[ x^2 + y^2 = z^2 \]
\end{theorem}

And a consequence of theorem \ref{pythagorean} is the statement in the next corollary.

\begin{corollary}
There's no right triangle whose sides measure 3cm, 4cm, and 6cm.
\end{corollary}

You can reference theorems such as \ref{pythagorean} when a label is assigned.

\begin{lemma}
Given two line segments whose lengths are $a$ and $b$ respectively there is a real number $r$ such that $b=ra$.
\end{lemma}

\begin{proof}
To prove it by contradiction try and assume that the statemenet is false, proceed from there and at some point you will arrive to a contradiction.
\end{proof}

Unnumbered theorem-like environments are also possible.

\begin{remark}
This statement is true, I guess.
\end{remark}

And the next is a somewhat informal definition

\theoremstyle{definition}
\begin{definition}{Fibration}
A fibration is a mapping between two topological spaces that has the homotopy lifting property for every space $X$.
\end{definition}
\end{document}