For all integers a and b, if a | b and b | a, a = b.
Can something think of a proof?
I have done this:
Proof:
Suppose m and n are integers such that m|n and n|m, but n ≠ m. Let m = 1 and n = -1
We know that $m|n$ implies $n = mk$ where $k$ is an integer
1 = -1(-1) where (-1) is integer, therefore, $m|n$ and $n|m$ implies $m = nl$ where $l$ is an integer
-1 = 1 (-1) where (-1) is integer, therefore, $n|m$
however, $m \neq n$ since $1 \neq -1$.
This is not true. What we have is $a = \pm b$.
We have $a=k_1b$ because $b \mid a$ and $b= k_2a$ because $a \mid b$.
Multiplying gives $ab=k_1k_2ab$, thus $k_1k_2=1$. Therefore we have $k_1=k_2=1$ or $k_1=k_2=-1$. Thus $a=b$ or $a=-b$.