\begin{propriete}
	\begin{enumerate}
		\item Le produit scalaire est \MotDefinition{commutatif}{}, c'est-à-dire que $\vv{u}\cdot\vv{v}=\vv{v}\cdot\vv{u}$.
		\item Si $\vv{u}=\vv{0}$ ou $\vv{v}=\vv{0}$, alors $\vv{u}\cdot\vv{v}=0$.
        \item $\vv{u}\cdot\vv{u}$ est également noté $\vv{u}^2$, appelé \og \MotDefinition{carré scalaire}{} de $\vv{u}$ \fg{}.

        On a alors $\vv{u}^2=\left\Vert{}\vv{u}\right\Vert{}^2$.
	\end{enumerate}

\vspace{-\baselineskip}
\end{propriete}

\begin{preuve}
	\begin{enumerate}
		\item On a $\vv{u}\cdot\vv{v}=\dfrac{1}{2}\left(\left\Vert{}\vv{u}+\vv{v}\right\Vert{}^2-\left\Vert{}\vv{u}\right\Vert{}^2-\left\Vert{}\vv{v}\right\Vert{}^2\right)
				=\dfrac{1}{2}\left(\left\Vert{}\vv{v}+\vv{u}\right\Vert{}^2-\left\Vert{}\vv{v}\right\Vert{}^2-\left\Vert{}\vv{u}\right\Vert{}^2\right)=\vv{v}\cdot\vv{u}$.
		\item Si $\vv{u}=\vv{0}$, alors $\vv{u}\cdot\vv{v}=\dfrac{1}{2}\left(\left\Vert{}\vv{0}+\vv{v}\right\Vert{}^2-\left\Vert{}\vv{0}\right\Vert{}^2-\left\Vert{}\vv{v}\right\Vert{}^2\right)
				=\dfrac{1}{2}\left(\left\Vert{}\vv{v}\right\Vert{}^2-0^2-\left\Vert{}\vv{v}\right\Vert{}^2\right)=0$.

Il en va de même si $\vv{v}=\vv{0}$.
        \item $\vv{u}^2%=\vv{u}\cdot\vv{u}
        =\dfrac{1}{2}\left(\left\Vert{}\vv{u}+\vv{u}\right\Vert{}^2-\left\Vert{}\vv{u}\right\Vert{}^2-\left\Vert{}\vv{u}\right\Vert{}^2\right)
				=\dfrac{1}{2}\left(\left\Vert{}2\vv{u}\right\Vert{}^2-2\left\Vert{}\vv{u}\right\Vert{}^2\right)
                =\dfrac{1}{2}\left(\left(|2|\left\Vert{}\vv{u}\right\Vert{}\right)^2-2\left\Vert{}\vv{u}\right\Vert{}^2\right)$

                $\phantom{\vv{u}^2}=\dfrac{1}{2}\left(4\left\Vert{}\vv{u}\right\Vert{}^2-2\left\Vert{}\vv{u}\right\Vert{}^2\right)
                =\dfrac{1}{2}\left(2\left\Vert{}\vv{u}\right\Vert{}^2\right)=\left\Vert{}\vv{u}\right\Vert{}^2$.
	\end{enumerate}				
\end{preuve}
