Prove that if $\{7k:k\in{\mathbb Z}\}\subsetneq\{nm:m\in{\mathbb Z}\}$, then $n=1.$

Question

Let n be a natural number. Prove that if $\{7k:k\in{\mathbb Z}\}\subsetneq\{nm:m\in{\mathbb Z}\}$, then $n=1$.

I know that we must show $x\in{A}$ implies $x\in{B}$, and that there exists $x\in{B}$ such that $x\notin{A}$, which means that $x\in{A}\neq{x\in{B}}$. Any help is appreciated; thanks

score 1 · Answer 1 · answered Feb 02 '19 at 21:33

1

The inclusion provided says that if $x$ is a multiple of $7$, then $x$ is a multiple of $n$. Moreover, $n \neq 7$, else the two sets are equal, but it's specified that they are not. Use your knowledge of the unique prime factorisation of $7$ to conclude.

answered Feb 02 '19 at 21:33

DanLewis3264

2,321

Ah I see what you are saying! Thanks so much. – macy Feb 02 '19 at 21:47
Key thing to take away: to divide is to contain (as Bill shows in symbols in his answer). As for proof writing, I hope you tried different things before you posted on here. I know I started off my answer by assuming $x \in 7\mathbb{Z}$, and seeing what happened, and it didn't quite work out, so I thought about why and started again. You should go through a similar process. – DanLewis3264 Feb 02 '19 at 22:13

score 0 · Answer 2 · answered Feb 02 '19 at 22:08

0

Hint $\ $ Show $\ a\Bbb Z\supseteq b\Bbb Z\iff a\mid b.\ $ Hence $\,n\Bbb Z\supseteq 7\Bbb Z\iff n\mid 7$

answered Feb 02 '19 at 22:08

Bill Dubuque

272,048

Prove that if $\{7k:k\in{\mathbb Z}\}\subsetneq\{nm:m\in{\mathbb Z}\}$, then $n=1.$

2 Answers2