Section 4: Problem 11 Solution

Working problems is a crucial part of learning mathematics. No one can learn topology merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.

James R. Munkres

Given $m\in\mathbb{Z}$ , we say that $m$ is even if $m/2\in\mathbb{Z}$ , and $m$ is odd otherwise.

(a) Show that if $m$ is odd, $m=2n+1$ for some $n\in\mathbb{Z}$ . [Hint: Choose $n$ so that $n<m/2<n+1$ .]

(b) Show that if $p$ and $q$ are odd, so are $p\cdot q$ and $p^{n}$ , for any $n\in\mathbb{Z}_{+}$ .

(c) Show that if $a>0$ is rational, then $a=m/n$ for some $m,n\in\mathbb{Z}_{+}$ where not both $m$ and $n$ are even. [Hint: Let $n$ be the smallest element of the set $\{x|x\in\mathbb{Z}_{+}\mbox{ and }x\cdot a\in\mathbb{Z}_{+}\}$ .]

(d) Theorem. $\sqrt{2}$ is irrational.

(a) According to Exercise 9(b), $m/2\notin\mathbb{Z}$ implies there is $n\in\mathbb{Z}$ such that $n<m/2<n+1$ . Then, $2n<m<2n+2$ , so that $m=2n+1$ (this last point requires some clarification, namely, $2n+2=(2n+1)+1$ , and there is no $z\in\mathbb{Z}$ such that $2n<z<2n+1$ or $2n+1<z<(2n+1)+1$ , see page 32, so the only possibility for $m\in\mathbb{Z}$ is $m=2n+1$ ).

(b) In addition to (a), we need its converse, that is if $m=2n+1$ for some $n\in\mathbb{Z}$ , then $m$ is odd. Indeed, in this case, $n<m/2<n+1$ , and, according to page 32, there are no integer number between $n$ and $n+1$ . Now, if $p$ and $q$ are odd, then for some $m,n\in\mathbb{Z}$ , $p=2m+1$ and $q=2n+1$ , hence, $p\cdot q=2(2mn+m+n)+1$ is odd. From this, by induction, if $p$ and $p^{n}$ is odd, then $p^{n+1}=p^{n}p$ is odd.

(c) Using the hint, $n$ is the smallest positive integer such that $na\in\mathbb{Z}_{+}$ . Let $m=na$ , then $a=m/n$ . If both $m$ and $n$ are even, then $n/2\in\mathbb{Z}_{+}$ , $(n/2)a=m/2\in\mathbb{Z}_{+}$ and $n/2<n$ , contradiction.

(d) By definition, $\sqrt{2}\ge0$ , and $\sqrt{2}\neq0$ as $0^{2}=0$ . Assuming $\sqrt{2}$ is rational, by using (c), $\sqrt{2}=m/n$ , where at least one of $m$ and $n$ is odd. Then $m^{2}/n^{2}=2$ , $2n^{2}=m^{2}$ , and $m^{2}$ is even, as $m^{2}/2=n^{2}$ . Now, from (b) it follows that $m$ is even, as otherwise $m^{2}$ would be odd, hence, $m=2k$ for some $k\in\mathbb{Z}_{+}$ . Therefore, $2n^{2}=4k^{2}$ , and $n^{2}=2k^{2}$ . Using the same argument, $n^{2}$ is even, and, hence, $n$ must be even as well. Contradiction.

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Section 4: Problem 11 Solution

Section 4: Problem 11 Solution