Solution

The correct answer is A, B, C Only

Key Points

To determine which of the statements have truth values when the universe of discourse is all integers, we need to evaluate each statement:

• A. \( \forall n \exists m (n^2 < m)\)
  • This means: For every integer n, there exists an integer m such that n² < m.
  • This is always true because we can always choose m = n² + 1, which will always be greater than n².
• B. \(\exists n \exists m (n < m^2)\)
  • This means: There exist integers n and m such that n < m².
  • True: Example — let n = 0, m = 1, then 0 < 1² = 1.
• C. \(\exists n \exists m (n^2 + m^2 = 5)\)
  • This means: There exist integers n and m such that n² + m² = 5.
  • True: Example — n = 1, m = 2 → 1² + 2² = 1 + 4 = 5.
• D. \(\exists n \exists m (n^2 + m^2 = 6)\)
  • This is false because there are no integer values of n and m whose squares sum to 6.
• E. \(\exists n \exists m (m + n = 4 \land n - m = 1)\)
  • Solving:
    • From n - m = 1, get n = m + 1
    • Substitute in m + n = 4: m + (m + 1) = 4 → 2m + 1 = 4 → m = 1.5
    • Then n = 2.5. These are not integers.
  • So, no integer solution exists. ❌

The correct answer is: 1) A, B, C Only

NOTE:  Some confusion may arise by misinterpreting statement A as false. However, it is true since for every integer n, there always exists a greater integer m such that n² < m.