Proof that dividing irrational number by an irrational number can result in an integer?

By giving an example. For instance, $2\sqrt{2}$ divided by $\sqrt{2}$?

Note well: To prove that something can happen, all you need to do is exhibit a single instance of it happening. To prove that something must happen you need to show that it happens always. So to show that dividing an irrational number by an irrational number other than itself can be an integer, all you need to do is cook up a single example of it happening. If you wanted to prove that, say, a rational divided by rational other than zero must be a rational, an example would not suffice; you would need to show that for any rational $q$ and any rational $r\neq 0$, you would have $\frac{q}{r}$ a rational.

The basic idea works much more generally: it amounts to how a subgroup acts on its complement.

$\rm If\: \ a\in A \subset B\ni b\ \: then\ \ ab \in A \!\iff\! b \in A\ \, $ when $\rm\, \ a^{-1}\in A\, $ and $\rm\, A,B\, $ are monoids.

Thus $\rm\ \, b\not\in A\, \Rightarrow \ ab\not\in A,\, \ and\ \ (ab)/b = a\in A\ $ as you desire.

Your case is simply $\rm\ \ A = \mathbb Q,\ \ B = \mathbb R,\ \ a\in \mathbb Z$

The key property above is essentially this "complementary" view of a subgroup:

Theorem $\ \ $ A nonempty subset $\rm\:A\:$ of abelian group $\rm\:B\:$ comprises a subgroup $\rm\iff\ A\ \bar A\ = \bar A\ $ where $\rm\: \bar A\:$ is the complement of $\rm\:A\:$ in $\rm\:B$

Instances of this are ubiquitous in concrete number systems, e.g.

                                 transcendental 
     algebraic * nonalgebraic  =  nonalgebraic  if  nonzero 
      rational * irrrational   =   irrational   if  nonzero 
          real *   nonreal     =    nonreal     if  nonzero 

         even  +     odd       =      odd          additive example
       integer + noninteger    =   noninteger

A trivial example would be $2\pi/\pi=2$. In fact every example would be of this form (one irrational would be an integer multiple of the other), for if $a,b$ are distinct irrational numbers such that $\frac{a}{b}=c$ where $c$ is an integer, then, $a=bc$