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Prove that the Poset has a unique least element
Prove that if (A, <) has a least element, then (A,≤) has a unique least element.
Ans: Let (A, ≤) be a poset. Suppose the poset A has two least elements x and y. Since x is the least element, it implies that x ≤ y. Using the same argument, we can say that y ≤ x, since y is supposed to be another least element of the same poset. ≤ is an anti-symmetric relation, so x ≤ y and y ≤ x ⇒ x = y. Thus, there is at most one least element in any poset.
Find the middle term of the AP 1, 8, 15....505. A ns: Middle terms a + (n-1)d = 505 a + (n-1)7 = 505 n - 1 = 504/7 n = 73 ∴ 37th term is middle term a 37
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