Question 308661
I. Clearly this is false. If {{{a<0}}}, then the inequality sign should flip, but it does not. If the sign did flip, then it would only be true for negative values of 'a', but what if 'a' was positive? Since we have this uncertainty, statement I is false.



II. If {{{a<>0}}} and {{{b<>0}}}, then {{{a^2}}} and {{{b^2}}} are both positive numbers. Recall that if {{{x<y}}} and 'x' and 'y' are both positive, then {{{1/x>1/y}}} which shows us that statement II is true.



III.



{{{a^2 < b^2}}} Start with the given inequality.



{{{a^2 - b^2 < 0}}} Subtract {{{b^2}}} from both sides.



{{{(a+b)(a-b) < 0}}} Factor the left side using the difference of squares.



So statement III is true.