Exercise 1.6

Question

Determine whether or not $S$ is a subring of $R$, when
\begin{enumerate}[label=(\alph*), leftmargin=*,align=left] 
	\item $S$ is the set of all rational numbers $a / b$, where $b$ is not divisible by $3$, and $R = \mathbb Q$.
	\item $S$ is the set of functions which are linear combinations with integer coefficients of the functions $\{1, \cos nt, \sin nt\}$, $n \in \mathbb Z$, and
		$R$ is the set of all real valued functions of $t$.
\end{enumerate}

Abdullah Zubair · Accepted Answer

\begin{enumerate}[label=(\alph*), leftmargin=*,align=left] 
	\item We claim that $S$ is a subring of $R$. Indeed, first note that $1 = 1/1 \in S$ since $3$ does not divide $1$. Next,
		let $r = a / b$ and $s = c / d$ be elements of $S$ where $3 \nmid d$ and $3 \nmid b$. We show that $a / b - c / d \in S$. Since 
		$$
			\dfrac{a}{b} - \dfrac{c}{d} = \dfrac{ad - cb}{bd}
		,$$
		it suffices to check that $3 \notdiv bd$. But this is true since if $3 \mid bd$ then $3 \mid d$ or $3 \mid b$ (Euclid's Lemma). The contrapositve gives $r - s \in
		S$. Now, we show that $rs \in S$ for any $r,s \in S$. Let $r,s \in S$ from before. Then $rs = (ac)(bd)$, and using the same argument we have that
		$3 \nmid bd$, so that $rs \in S$. QED

\item We show that $S$ is not subring of $R$. Indeed, note that $\cos (t) \sin (t) = \dfrac{1}{2}\sin(2t)$, and $\dfrac{1}{2}\sin(2t)$ cannot be written as a linear
		combination of the functions in $S$ (this would contradict the linear independence of $\{1,\cos(t),\sin(t), \cos(2t), \sin(2t),\dots\}$).

\end{enumerate}

Exercise 1.6

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Exercise 1.6

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