prove if a|b and b|a then $a = \pm b$ Fairly basic I guess. 
Attempt:
$a\neq\pm b \Rightarrow a\nmid b \vee b \nmid a$
let $a = \pm b + d, d\in \mathrm{Z} \wedge d\neq 0$  
then $a\mid b \Rightarrow b\nmid a$ and $b\mid a \Rightarrow a\nmid b$
 A: $b|a \rightarrow a = kb$, and
$a|b \rightarrow b = na$. Thus:
$a = k(na) = kna$, and assume $a \neq 0$, so: $nk = 1$. Thus:
either $n = k = 1$ or $n = k = -1$ which means $a = b$ or $a = - b$
A: $a|b$ so $b=ak$ for some integer $k$. Similarly, $b|a$ implies $a=bl$ for some integer $l$. Putting these together you get $b=ak=(bl)k=bkl$, which implies $1=kl$. Which integers can $k$ and $l$ be such that their product is $1$?
A: (1) If $u$ divides $v$, then $|u|$ divides $|v|$.
(2) If $u$ divides $v$, with $u>0$ and $v>0$, then $u\le v$.
Suppose $a$ divides $b$ and $b$ divides $a$.
By (1), we have $|a|$ divides $|b|$ and $|b|$ divides $|a|$.
By (2), we have $|a| \le |b| \le |a|$, which implies $|a|=|b|$. This means that $a=\pm b$.
A: Since $b = qa$ for some integer $q \neq 0$ and $a = q'b$ for some integer $q' \neq 0,$ we have
$$b = qq'b,$$
so that 
$$1 = qq',$$
whence
$$q, q' = 1\ or\ q, q'= -1.$$
A: Hint $\,\ |a|\,\overset{\underset{}{\large a\,\mid\, b}}=\gcd(a,b)\overset{\underset{}{\large b\,\mid\, a}}=|b|$
