Question

Which of the following expressions will definitely be true if the expression \( R > O = A > S < T \) is definitely true?

• A. \( O > T \)
• B. \( S < R \)
• C. \( T < A \)
• D. \( S = O \)

Hint:

Split the chain into pairwise relations and use transitivity through the term that is common to two of them.

Answer 1

The Correct Option is B

Instead of reasoning with symbols alone, this can be checked quickly by substituting a set of numbers that satisfies \( R > O = A > S < T \), and then testing which option survives across different valid choices.

1. Pick one valid set of numbers: let \( S = 1 \), \( T = 5 \), \( A = 2 \), \( O = 2 \) and \( R = 3 \). Checking the original condition: \( 3 > 2 = 2 > 1 < 5 \), which holds true, so this is a valid assignment.
2. Test each option with this set: \( O > T \) becomes \( 2 > 5 \), false. \( S < R \) becomes \( 1 < 3 \), true. \( T < A \) becomes \( 5 < 2 \), false. \( S = O \) becomes \( 1 = 2 \), false. Only \( S < R \) survives here.
3. Try a second valid set to be sure it is not a coincidence: let \( S = 1 \), \( T = 1.5 \), \( A = 2 \), \( O = 2 \), \( R = 3 \). This also satisfies \( 3 > 2 = 2 > 1 < 1.5 \). Now \( O > T \) becomes \( 2 > 1.5 \), true this time, and \( T < A \) becomes \( 1.5 < 2 \), also true this time. Since these two options flip between true and false depending on the numbers chosen, neither of them is definitely true. \( S < R \) still gives \( 1 < 3 \), true, and \( S = O \) still gives \( 1 = 2 \), false.
4. Conclude: \( S < R \) is the only option that stays true across every valid assignment, because it follows directly from \( R > O = A > S \) by transitivity of the greater than relation, so it is the one that is definitely true.

The expression that is definitely true is \( S < R \). $\boxed{S < R}$