Question 6

B: rt < rs

The statement "rt < rs" cannot be true given the inequality 0 < r < s < t. Since s is greater than r, multiplying both sides of the inequality by r (which is positive) means that rt must be greater than or equal to rs.

A) r ^ 2 < s ^ 2 < t ^ 2

This statement can be true because squaring the values of r, s, and t maintains the order of the inequalities since all values are positive. Thus, r² will indeed be less than s², which in turn will be less than t².

B) rt < rs

This statement cannot be true. Given that s is greater than r, multiplying both sides of the inequality by r (which is positive) yields rt > rs, contradicting the statement.

C) 1/t < 1/s < 1/r

This statement can be true. Since t is the largest among the three variables, 1/t is the largest reciprocal, followed by 1/s, and finally 1/r, satisfying the inequality.

D) s/r < t/r

This statement can be true as well. Dividing both sides of the inequality by r (a positive number) preserves the inequality, confirming that s/r is less than t/r since s < t.

E) r/t < r/s

This statement can also be true. Since t > s, the fraction r/t will be larger than r/s, as dividing r by a larger denominator (t) yields a smaller result compared to dividing it by a smaller denominator (s).

Conclusion

The statement "rt < rs" is definitively incorrect, as it contradicts the established order of the values r and s. All other options logically follow from the relationships defined in the inequalities, demonstrating their validity. Thus, option B is the only choice that cannot be true.