Determine and sketch the pairs $(x, y)$ in $\mathbb{R} \times \mathbb{R}$ that satisfy: $|x\cdot y| \leq 2$ I am learning to work with inequalities.  For an example, let's say I want to find the set of $(x, y) \in \mathbb{R}^{2}$ that satisfy: $|x \cdot y| \leq 2$.
 A: My suggestion:
(A) Start with the definition of absolute value $\left\vert z\right\vert $, with $z=xy,x,y\in\mathbb{R}$
$$\left\vert xy\right\vert =xy\qquad \text{if}\ xy\geq 0\iff \left( x\geq 0\wedge y\geq 0\right) \vee \left( x\leq 0\wedge y\leq 0\right) $$
$$\left\vert xy\right\vert =-xy\qquad \text{if}\ xy<0\iff \left( x\geq 0\wedge y\leq 0\right) \vee \left( x\leq 0\wedge y\geq 0\right) $$
(B) Add the condition  $\left\vert xy\right\vert \leq 2\iff -2\le xy\leq 2$. 
(Why?)
(C) Observe that 
(i) $xy\gt 0$ 
in the 1st and 3rd quadrants; and 
(ii)  $xy\lt 0$ 
in the 2nd and 4th quadrants. 
(Why?)
A: By taking the seperate cases where $x = 0$ or $y = 0$ and showing that they are in the set, then we can assume $x,y \ne 0$  Therefore we can take $-2 \le xy \le 2$.
Therefore:


*

*$-2/y \le x \le 2/y$

*$-2/x \le y \le 2/x$


So for all x,y real numbers satisfying the above, and (0, y real number) (x real number, 0), x,y are in R x R.
A: HINT: When you look at the answer of Chandru1 you must think twice. Do you know any algebraic properties of the absolute value? How do you solve an inequality?
You might also like to think of different points $(x,y)=(5,5)$, $(x,y)=(5,-5)$, $(x,y)=(-5,5)$, $(x,y)=(-5,-5)$ -- are they in you set? $(x,y)=(1/5,5)$, $(x,y)=(1/5,-5)$, $(x,y)=(-1/5,5)$, $(x,y)=(-1/5,-5)$ -- are they in you set?
