Given $x>0, y>0, x>y$ and $z \neq 0$, find the only false inequality.
(a) $x+z>y+z$
(c) $x z>y z$
(e) $x z^{2}>y z^{2}$
(b) $x-z>y-z$
(d) $\frac{x}{z^{2}}>\frac{y}{z^{2}}$
The correct option is (c).

In this question, we will use the following properties of inequalities. We can add the same number to both sides of an inequality without changing its direction. We can multiply both sides of an inequality by a positive number without changing its direction. Thus,

$$
x>y \Rightarrow\left\{\begin{array}{l}
x+z>y+z \text { (adding } z \text { to both sides) } \\
x z>y z \text { (multiplying by } z>0 \text { on both sides) }
\end{array}\right.
$$

Therefore, (a) and (b) are correct because $z$ and $-z$ were added to both sides, and (d) and (e) are correct because both sides were multiplied by $1 / z^{2}$ and $z^{2}$, both positive since $z \neq 0$. Option (c) is false because $z$ can be negative. For example, if $x=5, y=3$ and $z=-2$, we have $5>3$ and yet,

$$
\underbrace{5 \times 2}_{x z}=-10<-6=\underbrace{3 \times(-2)}_{y z}
$$