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Chapter 1 #8(a): Suppose that the integer $k$ is not a perfect $n$th power. Prove that $\sqrt[n]{k} \not\in \mathbb{Q}$.

*Proof*: Suppose that $\sqrt[n]{k} \in \mathbb{Q}$, hence $\sqrt[n]{k}=\dfrac{p}{q}$ where $p,q \in \mathbb{Z}$ with $q \neq 0$ and with $\dfrac{p}{q}$ in lowest terms (i.e. "reduced" all the way). Since we are assuming that $k$ is not a perfect $n$th power, we must conclude that $\sqrt[n]{k}$ is not an integer, and hence $q \neq 1$. We have $k = \dfrac{p^n}{q^n}$ and so $\dfrac{p^n}{q^n}$ is an integer. But this implies that $q^n$ divides evenly into $p^n$, which is impossible because $\dfrac{p}{q}$ was assumed to be reduced. This is a contradiction, hence we conclude that $\sqrt[n]{k} \not\in \mathbb{Q}$. $\blacksquare$

*Problem A*: Prove from the definition for convergence that $\displaystyle\lim_{n \rightarrow \infty} \dfrac{1}{n}=0$.

*Solution*: Let $\epsilon > 0$, let $N \in \mathbb{N}$ with $N > \dfrac{1}{\epsilon}$, and let $n \geq N$. For such $n$, we have $n > \dfrac{1}{\epsilon}$ and so $\dfrac{1}{n} < \epsilon$. Now compute
$$\left| \dfrac{1}{n} - 0 \right| = \dfrac{1}{n} < \epsilon,$$
completing the proof. $\blacksquare$

*Problem B*: Prove from the definition for convergence that $\displaystyle\lim_{n \rightarrow \infty} \dfrac{n}{2n+1} = \dfrac{1}{2}$.

*Solution*: Let $\epsilon > 0$, let $N \in \mathbb{N}$ with $N > \dfrac{1}{4\epsilon} - \dfrac{1}{2}$, and let $n \geq N$. For such $n$, we have $4n > \dfrac{1}{\epsilon}-2$ and so $4n+2 > \dfrac{1}{\epsilon}$ and so $\dfrac{1}{4n+2} < \epsilon$. Thus compute
$$\left| \dfrac{n}{2n+1} - \dfrac{1}{2} \right| = \left| \dfrac{1}{4n+2} \right| = \dfrac{1}{4n+2} < \epsilon,$$
completing the proof. $\blacksquare$

*Problem C*: Prove from the definition for convergence that $\displaystyle\lim_{n \rightarrow \infty} \dfrac{3n+2}{5n-1} = \dfrac{3}{5}$.

*Proof*: Let $\epsilon > 0$, let $N \in \mathbb{N}$ with $N> \dfrac{13}{25\epsilon} + \dfrac{1}{5}$, and let $n \geq N$. For such $n$, we have $25n > \dfrac{9}{\epsilon}+5$ and so $\dfrac{9}{25n-5} < \epsilon$. Now calculate
$$\left| \dfrac{3n+2}{5n-1} - \dfrac{3}{5} \right| = \left| \dfrac{13}{25n-5} \right| = \dfrac{13}{25n-5} < \epsilon,$$
as was to be shown. $\blacksquare$