If the universe of disclosure is set of integers, then which of the followings are TRUE?

(A) ∀n ∃m(n² < m)

(B) ∃n ∀m(n < m²)

(C) ∃n ∀m(nm = m)

(D) ∃n ∃m(n² + m² = 6)

(E) ∃n ∃m(n + m = 4 ∧ n - m = 1)

Choose the correct answer from the options given below:

The correct answer is: Option 1) (A), (B), and (C) only

Key Points

Given: Universe of discourse = Integers

(A) ∀n ∃m (n² < m)
For every integer n, there exists an integer m such that n² < m.
✔️ Since integers are unbounded above, for any n², we can find m = n² + 1 (or higher).
✅ TRUE

(B) ∃n ∀m (n < m²)
There exists an integer n such that for every integer m, n < m².
✔️ Try n = -1. Since m² ≥ 0 for all integers m, -1 < m² always holds true.
✅ TRUE

(C) ∃n ∀m (n·m = m)
There exists an integer n such that for all m, n·m = m.
✔️ Try n = 1. Then 1·m = m for all m.
✅ TRUE

(D) ∃n ∃m (n² + m² = 6)
Check all integer combinations:
n = 1, m = 2 → 1² + 2² = 1 + 4 = 5 ❌
n = 2, m = 2 → 4 + 4 = 8 ❌
No integer solution satisfies n² + m² = 6.
❌ FALSE

(E) ∃n ∃m (n + m = 4 ∧ n - m = 1)
Solving:
n + m = 4
n - m = 1
⇒ 2n = 5 ⇒ n = 2.5, m = 1.5
 Not integers ⇒ no integer solution ❌ FALSE

Final Conclusion:
Only statements (A), (B), and (C) are TRUE.
✅ Correct Answer: Option 1