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\title{MATH221604: Intro. to Abstract Mathematics: Homework ???} % Enter the correct homework number
\author{Your Name???} % Enter your name
\date{Date: due date???} % Enter the due date
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%%%%%%%%%%%%%%%%%%%%%% Your Solutions %%%%%%%%%%%%%%%%%%%%%%
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\begin{enumerate}
\item $\sqrt{2}$ is an irrational number.
\begin{proof}
Suppose $\sqrt{2}$ is rational. Then we may write
$$\sqrt{2}=\dfrac{p}{q},$$
for some $p,q\in\Z$ with $q\neq0$. Since $\sqrt{2}\neq0$, we also have that $p\neq0$. We may assume that $p$ and $q$ are co-prime. Then we have
$$(\sqrt{2})^2=\left(\dfrac{p}{2}\right)^2$$
$$2=\dfrac{p^2}{q^2}$$
$$2q^2=p^2.$$
Note that $m=q^2$ is an integer. Since $p^2=2m$ for an integer $m$, we must have that $p^2$ is even. Since $2$ is prime, $p$ must be even. So $p=2n$ for some $n\in\Z$. So we have
$$2q^2=p^2$$
$$2q^2=(2n)^2$$
$$2q^2=4n^2$$
$$q^2=2n^2.$$
Note that $k=n^2$ is an integer. Since $q^2=2k$ for an integer $k$, we must have that $q^2$ is even. Since $2$ is prime, $q$ must be even. So $q=2j$ for some $j\in\Z$. Then $p$ and $q$ have a common factor of $2$, i.e., $p$ and $q$ are not co-prime, a contradiction.
Therefore, $\sqrt{2}$ is an irrational number.
\end{proof}
\end{enumerate}
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