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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.
Write the quadratic equation whose roots are real and non conjugate Ans) x^2-x+6=0 ...roots are real and non conjugate
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( a+2b)x + (2a - b)y = 2, (a - 2b)x + (2a +b)y = 3 (Ans: 5b - 2a/10ab , a + 10b/10ab ) Ans: 2ax + 4ay = y , we get 4bx - 2by = -1 2ax+ 4ay = 5 4bx- 2by = - 1
Theorem, from Definition of Derivative If f(x) is differentiable at x = a then f(x) is continuous at x =a. Proof : Since f(x) is differentiable at x = a we know, f'(a
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