If the universe of disclosure is set of integers, then which of the followings…
2023
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 (n m = 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 :
- A.
(A), (B) and (C) Only
- B.
(B) and (C) Only
- C.
(C), (D) and (E) Only
- D.
(C) and (E) Only
Attempted by 32 students.
Show answer & explanation
Correct answer: A
Correct answer: the three true statements are the ones asserting (1) for every integer n there exists an integer m with n^2 < m, (2) there exists an integer n such that n < m^2 for all integers m, and (3) there exists an integer n such that n·m = m for all integers m.
Statement: For every integer n there exists an integer m with n^2 < m. True. Proof: take m = n^2 + 1, which is an integer greater than n^2.
Statement: There exists an integer n such that for every integer m, n < m^2. True. Proof: choose n = -1. Since m^2 ≥ 0 for all integers m, we have -1 < m^2 for every m.
Statement: There exists an integer n such that for every integer m, n·m = m. True. Proof: n = 1 satisfies 1·m = m for all integers m.
Statement: There exist integers n and m with n^2 + m^2 = 6. False. Explanation: integer squares are 0,1,4,9,...; no pair of these sums to 6.
Statement: There exist integers n and m with n + m = 4 and n - m = 1. False. Explanation: solving gives n = 5/2 and m = 3/2, which are not integers.
Therefore the correct choice is the one listing the first three statements (the inequalities and the multiplicative identity case); the other two statements are false.