maihuna wrote:1. x^2 < 1 for -1<x<1 => |x| <1 Suff

2. |x| < 1/x
trick here is we dont know sign of x but we do know the sign of |x|

However, since we know that |x| must be positive (since x&#8800;0), we can deduce that x is positive as well.

Let's just replace |x| with +

1/x > +

well, if 1/x is greater than a positive number, 1/x must be positive, which means that x is also positive.

Now that we know that x is positive, we can multiply both sides by x to get:

x^2 < 1

and, if x^2 < 1 AND x > 0, we know that x must be a positive fraction.