Question

If $a$ and $b$ are nonzero integers such that $a|b$ and role="math" localid="1646123689270" $b|a$, prove that $a=\pm b$.

Short answer

It is proved that if $a$ and $b$ are nonzero integers, such that $a|b$ and $b|a$, then $a=\pm b$.

Step-by-step solution

Apply divisibility definition 

As $a|b$ then by definition of divisibility, we have $b=ax$ for some integer $x$ and $a\ne 0$.

As $b|a$ then by definition of divisibility, we have $a=by$ for some integer $y$ and $b\ne 0$.

Prove that if a|b  and b|a  then a=±b

Substitute $b=ax$into $a=by$ as:

$$\begin{gathered} a=by \\ =\left(ax\right)y \\ =a\left(xy\right) \end{gathered}$$

As $a\ne 0$, divide both sides of the obtained equation as:

$\begin{array}{rcl}\frac{a}{a}& =& \frac{a\left(xy\right)}{a}\\ 1& =& xy\end{array}$

From the obtained equation as $x=\pm 1$ and $y=\pm 1$, this implies that $a=\pm b$.

Hence, it is proved that if $a$ and $b$ are nonzero integers, such that $a|b$ and $b|a$ , then $a=\pm b$.