Let E1 = {a}, E2 = {a, b} and N be the set of natural numbers. Using a bijective function, a. prove that | E,*| = |N|.
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A: The solution for the above given question is given below:
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A: Answer: The solutions of both the parts are given below-
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A: F(X,Y,Z) = X'Y'Z' + X'YZ + XY'Z + XYZ where, No. of variables is 3 i.e. x, y, z
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A: Answer: I have given answered in the handwritten format in brief explanation.
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Q: . Let Q, R and S be sets. Show that (R – Q) ∪ (S – Q) = (R ∪ S) – Q.
A: Given: Let Q, R and S be sets. Show that (R – Q) ∪ (S – Q) = (R ∪ S) – Q.
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- Suppose that the equation ax b .mod n/ is solvable (that is, d j b, whered D gcd.a; n/) and that x0 is any solution to this equation. Then, this equation has exactly d distinct solutions, modulo n, given by xi D x0 C i.n=d / fori D 0; 1; : : : ; d 1Let Z be the set of all integers. An integer a has f as a factor if a = integer j. An integer is even if it has 2 as a factor. An integer a is odd if it is not even. Prove by contradiction that an odd number cannot have an even number as a factor.Prove: Let a, b, and c be integers. If Suppose a, b, c are integers with (a (a - b) | c, then a | c. b) | c. Then c = (a b) for some integer k, so c = ]a, · so a c.
- Let Z be the set of all integers. An integer a has f as a factor if a = fj for some integer j. An integer is even if it has 2 as a factor. An integer a is odd if it is not even. Prove by contradiction that an odd number cannot have an even number as a factor.Determine whether each of these functions is a bijection from ℝ to ℝ. If it is, write the inverse function. f(x) = 3|x| − 4Let S and T be sets with: |S| = 5, |T| = 7 How many one-to-one functions are there from S to T? Α. 0 В. 57 С. 75 D. 7!/2 E. 25-7
- Find f(n) and BigOExplain the Wronskian determinant test. Using the Wronskian determinant test, write the program using NumPy to determine whether the functions f(x)=e^(- 3x), g(x)=cos2x and h(x)=sin2x are linearly independent in the range (-∞, + ∞). #UsePythonIf A = {1, 2, 6} and B = {2, 3, 5}, then the union of A and B is