imc

Let us denote $a=x+k$ where $k>0$ and $b=y+l$ where $l>0$. We can write that $x+y<x+y+k+l⟹x+y<a+b$. It is important to note that $1$ counterexample fully disproves a claim. Let's try substituting $x=-3,y=-4,a=1,b=4$. $\text{(II)}$ states that $x-y<a-b⟹-3-(-4)<1-4⟹1<-3$.Therefore, $\text{(II)}$ is false. $\text{(III)}$ states that $xy<ab⟹(-3)\cdot (-4)<1\cdot 4⟹12<4$. Therefore, $\text{(III)}$ is false. $\text{(IV)}$ states that $\frac{x}{y}<\frac{a}{b}⟹\frac{-3}{-4}<\frac{1}{4}⟹0.75<0.25$. Therefore, $\text{(IV)}$ is false. One of our four inequalities is true, hence, our answer is $\boxed{1}$