finding the interval of the product $x\cdot y$ using logical operations If we have:
$-1< x < 1$ and $-1<y<1$, then find the interval of $x\cdot y$.
Here is my method to solve it: 
I used the fact that the first inequality means simply that $\left\lvert x\right\rvert<1$, the second also : $\left\lvert y\right\rvert<1$ then we can do muliplication term by term to find that $\left\lvert x\cdot y\right\rvert<1$, then transform it to: $-1< x\cdot y<1$
My question is: how to use a pure logical operations to achieve the same result? In other words, how to transform the problem to logical assertions and manipulate it using (and, or) operations or sets operations (union, intersection) ?
Thank you.
 A: I think you have to use some properties of the real numbers. You cannot do it just by pure logical operations. E.g. you have to know that when $-1 \leq x \leq 0$ and $0 \leq y \leq 1$ then $-1 \leq xy \leq 0$. At least I don't see how you can make it just through pure logical operations (also I don't really see what exactly you mean by that).  
A: You have for premisses four inequalities: $-1<x$, $-1<y$, $x<1$, and $y<1$. You want to derive some bounds on the product $x\cdot y$ - using your notation. You are free to use any results pertaining to the ordered field $\mathbb{R}$ - assuming x and y element of $\mathbb{R}$. 
The argument is easier if we split it along the sign of y for example: $y \gt 0$ or $y \lt 0$ or $y=0$.


*

*$y \gt 0 \wedge -1<x$ implies $-y \lt xy$, hence $-1 \lt -y \lt xy$. Likewise $y \gt 0 \wedge x<1$ implies $xy \lt y$, hence $xy \lt y \lt 1.$ Conclusion: $-1 \lt -y \lt xy \lt y \lt 1$, hence $-1 \lt xy \lt 1.$

*$y \lt 0 \wedge -1<x$ implies ... . Conclusion:  $-1 \lt xy \lt 1.$

*y=0 case. ... . Conclusion:  $-1 \lt xy \lt 1.$

*Invoking a bit more logic, you have just proved (under the stated assumptions) $-1 \lt xy \lt 1.$

