Exercise 1.3.2

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\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\be} {\begin{enumerate}} ewcommand{\ee} {\end{enumerate}} ewcommand{\deb}{\begin{eqnarray*}} ewcommand{\fin}{\end{eqnarray*}} ewcommand{\ssi} {si et seulement si } ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} Let $f(y) = y^3+py+q$. The purpose of Exercises 2 to 5 is to prove Theorem 1.3.1 geometrically using graphing techniques. The proof breaks up into three cases corresponding to $p>0,p=0$, and $p<0$. This exercise will consider the case $p>0$. \begin{enumerate} \item[(a)] Explain why $\Delta <0$. \item[(b)] Analyse the sign of $f'(y)$, and show that $f(y)$ is always increasing. \item[(c)] Explain why $f(y)$ has only one real root. \end{enumerate}

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow} \def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers \def\dcro{\mbox{]\hspace{-.15em}]}} ewcommand{\be} {\begin{enumerate}} ewcommand{\ee} {\end{enumerate}} ewcommand{\deb}{\begin{eqnarray*}} ewcommand{\fin}{\end{eqnarray*}} ewcommand{\ssi} {si et seulement si } ewcommand{\D}{\mathrm{d}} ewcommand{\Q}{\mathbb{Q}} ewcommand{\Z}{\mathbb{Z}} ewcommand{\N}{\mathbb{N}} ewcommand{\R}{\mathbb{R}} ewcommand{\C}{\mathbb{C}} ewcommand{\F}{\mathbb{F}} ewcommand{\U}{\mathbb{U}} ewcommand{ e}{\,\mathrm{Re}\,} ewcommand{\im}{\,\mathrm{Im}\,} ewcommand{\ord}{\mathrm{ord}} ewcommand{\Gal}{\mathrm{Gal}} ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}} \begin{proof} \begin{enumerate} \item[(a)] As $p>0$, $-4p^3<0$, and $-27q^2\leq 0$, thus $\Delta = -4p^3-27q^2<0$. \item[(b)] For all $y\in \mathbb{R}$, $f'(y) = 3y^2+p>0$, thus $f$ is strictly increasing on $\mathbb{R}$. \item[(c)] As $f$ is strictly increasing on $\mathbb{R}$, $f$ is injective (one to one). Moreover $\lim\limits_{y o +\infty} f(y) = +\infty$, and $\lim\limits_{y o -\infty} f(y) = -\infty$, so there exists $y_1$ such that $f(y_1)<0$ and there exists $y_2$ such that $f(y_2)>0$. As $f$ is a continuous function, the intermediate values theorem gives the existence of a real root of $f$. As $f$ is injective, this is the only real root. \end{enumerate} \end{proof}

Exercise 1.3.2

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Exercise 1.3.2

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