Proof of $a\mid b\ \wedge\ b\mid a\ \Rightarrow a=\pm b$ for integer numbers I have this proof for divisibility over integer numbers: $$a\mid b\ \wedge\ b\mid a\ \Rightarrow a=\pm b$$ _Proof._ : Hypothesis: $a\times d=b$, $b\times e=a$. Hence \begin{align}a \times (d\times e)&=^{\langle {\rm Associativity}\ \times \rangle} \\\\[0.1cm]&=(a\times d)\times e=^{\langle {\rm Hyp.}:\ a\times d=b\rangle}\\\\[0.1cm]&=b\times e=^{\langle {\rm Hyp.:}\ b\times e=a\rangle}\\\\[0.1cm]&=a\end{align} I understand almost the entire proof, but I don't where the $a * (d*e)$ expression comes from.

That's just a starting point; in effect, the proof is basically saying that since $a * (d * e) = a$, then $d * e = 1$ and the only possible values for them are either $d=e=1$ or $d=e=-1$. This in turn shows that $a = \pm b$, given the definitions of $d$ and $e$.

Let's formulate the entire proof in complete sentences, to make it easier to grasp:

**Theorem:** If $a$ and $b$ divide each-other evenly, then $a=\pm b$.

**Proof:**

1. Let $d$ and $e$ be the respective results of evenly dividing $a/b$ and vice versa, i.e. $a = b*d$, $b = a * e$.

2. Next, consider the product $a * (d * e)$. Since $a*(d*e)=a$ (as shown in the question), then we must have $d*e=1$.

3. $d*e=1$ implies that $d=e=\pm 1$, since there are no other numbers that would multiply to $1$.

4. In step 1. we defined $a=b*d$, which with $d=\pm 1$ proves the theorem.