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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.
Exercise 12c question number 24
The tenth term in the binomial expansion of (1-1/4)(1-1/5)(1-1/6)...(1-1/n+3) is equal to
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Hi, I don''t know how to solve 2(5x+3)
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