Prove $a|b \wedge b|a \implies a=\pm b$

Question

Let $a,b\in\mathbb{Z} \backslash \{0\}$. Show that

$a\mid b$, $b\mid a$ $\implies a=\pm b$.

I can see why this is true, but not write it down.

Hint: $a\mid b$ implies that $b=c_1a$ for some integer $c_1$. Similarly you get $a=c_2b$. what can you deduce from $c_1c_2=1$? — Jyrki Lahtonen, Aug 22 '15 at 18:37

score 2 · Accepted Answer · answered Aug 22 '15 at 18:36

2

If $a|b, b=an$ where $n$ is an integer

Similarly, $b|a\implies a=bm$ where $m$ is an integer

$\implies ab=abmn\iff mn=1$

But $m,n$ are integers, so what are the possible values of $m,n?$

answered Aug 22 '15 at 18:36

lab bhattacharjee

score 1 · Answer 2 · answered Aug 22 '15 at 18:37

1

$a=tb$ and $b=sa$ gives $a=tsa$, which means $ts=1$. Now, $t$ and $s$ are both integers, so...

answered Aug 22 '15 at 18:37

Arthur

score 0 · Answer 3 · answered Aug 22 '15 at 19:00

0

If $a|b$ then $|a|\leq |b|$. Similarly $b|a$ implies $|b|\leq |a|$. So $a=b$ or $a=-b$.

answered Aug 22 '15 at 19:00

Euler88 ...

3 Answers3