Solve the following short questions: a) Prove (A - B) (A B) = A using set identities. b) Given X = , b\ and Y = , v list four binary relations from X to Y that are not functions.

You're on a roll — Here are the solutions to the questions from the image: Q2. Solve the following short questions a) Step 1: Use the set identity A - B = A B^c. (A - B) (A B) = (A B^c) (A B) Step 2: Apply the distributive law for sets, X (Y Z) = (X Y) (X Z), or in this case, (Y X) (Z X) = (Y Z) X . (A B^c) (A B) = A (B^c B) Step 3: Use the identity B^c B = U (the universal set). A (B^c B) = A U Step 4: Use the identity A U = A. A U = A Thus, (A - B) (A B) = A. Proven b) The set X = \a, b\ and Y = \u, v\. The Cartesian product X × Y = \(a,u), (a,v), (b,u), (b,v)\. A binary relation is a subset of X × Y. A relation is not a function if either: 1. An element in X is not mapped to any element in Y. 2. An element in X is mapped to more than one element in Y. Here are four binary relations from X to Y that are not functions: 1. R_1 = \(a,u)\ (Element b is not mapped). 2. R_2 = \(a,u), (a,v)\ (Element a is mapped to two elements). 3. R_3 = \(a,u), (a,v), (b,u)\ (Element a is mapped to two elements). 4. R_4 = \(b,u)\ (Element a is not mapped). Four relations: R_1 = \(a,u)\, R_2 = \(a,u), (a,v)\, R_3 = \(a,u), (a,v), (b,u)\, R_4 = \(b,u)\ c) To prove that if f and g are one-to-one functions, then their composition g f is one-to-one, we must show that if (g f)(x_1) = (g f)(x_2), then x_1 = x_2. Step 1: Assume (g f)(x_1) = (g f)(x_2). g(f(x_1)) = g(f(x_2)) Step 2: Since g is a one-to-one function, if g(A) = g(B), then A = B. Applying this, we have: f(x_1) = f(x_2) Step 3: Since f is a one-to-one function, if f(x_1) = f(x_2), then x_1 = x_2. x_1 = x_2 Since we started with (g f)(x_1) = (g f)(x_2) and concluded x_1 = x_2, the composition g f is one-to-one. Proven d) The sequence is defined by b_j = 1 - 2^j for all integers j 0. Step 1: Calculate the first term for j=0. b_0 = 1 - 2^0 = 1 - 1 = 0 Step 2: Calculate the second term for j=1. b_1 = 1 - 2^1 = 1 - 2 = -1 Step 3: Calculate the third term for j=2. b_2 = 1 - 2^2 = 1 - 4 = -3 Step 4: Calculate the fourth term for j=3. b_3 = 1 - 2^3 = 1 - 8 = -7 The first four terms of the sequence are 0, -1, -3, -7. 0, -1, -3, -7 Q3. Solve the following Long Questions. a) To show that f(x) = 3x + 2 and g(x) = (x-2)/(3) are inverse functions, we need to prove that f(g(x)) = x and g(f(x)) = x. Step 1: Calculate f(g(x)). f(g(x)) = f((x-2)/(3)) Substitute (x-2)/(3) into f(x) = 3x + 2: f((x-2)/(3)) = 3((x-2)/(3)) + 2 = (x-2) + 2 = x Step 2: Calculate g(f(x)). g(f(x)) = g(3x+2) Substitute 3x+2 into g(x) = (x-2)/(3): g(3x+2) = ((3x+2)-2)/(3) = (3x)/(3) = x Since f(g(x)) = x and g(f(x)) = x, the functions f and g are inverses of each other. Proven b) Given two distinct rational numbers r and s with r < s. We need to prove that there is a rational number x such that r < x < s. Step 1: Consider the number x = (r+s)/(2). Step 2: Show that x is a rational number. Since r and s are rational numbers, they can be expressed as r = (a)/(b) and s = (c)/(d), where a, b, c, d are integers and b, d 0. x = (a)/(b) + (c)/(d)2 = (ad+bc)/(bd)2 = (ad+bc)/(2bd) Since a, b, c, d are integers, ad+bc is an integer, and 2bd is a non-zero integer (because b 0 and d 0). Therefore, x is a rational number. Step 3: Show that r < x < s. Since r < s, we can add r to both sides of the inequality: r+r < s+r 2r < r+s Divide by 2: r < (r+s)/(2) So, r < x. Similarly, since r < s, we can add s to both sides of the inequality: r+s < s+s r+s < 2s Divide by 2: (r+s)/(2) < s So, x < s. Combining these results, we have r < x < s. Thus, for any two distinct rational numbers r and s with r < s, there exists a rational number x = (r+s)/(2) such that r < x < s. Proven c) We need to prove by mathematical induction that for any integer n 1, x^n - y^n is divisible by x-y, where x and y are any two integers with x y. Base Case (n=1): For n=1, we need to show that x^1 - y^1 is divisible by x-y. x^1 - y^1 = x-y. Since x-y is clearly divisible by x-y (with a quotient of 1), the base case holds. Inductive Hypothesis: Assume that for some integer k 1, x^k - y^k is divisible by x-y. This means x^k - y^k = M(x-y) for some integer M. Inductive Step: We need to prove that x^k+1 - y^k+1 is divisible by x-y. Consider the expression x^k+1 - y^k+1: x^k+1 - y^k+1 = x · x^k - y · y^k We can rewrite this by adding and subtracting x · y^k: x^k+1 - y^k+1 = x · x^k - x · y^k + x · y^k - y · y^k Factor out x from the first two terms and y^k from the last two terms: x^k+1 - y^k+1 = x(x^k - y^k) + y^k(x - y) Now, apply the inductive hypothesis, substituting x^k - y^k = M(x-y): x^k+1 - y^k+1 = x(M(x-y)) + y^k(x - y) Factor out (x-y): x^k+1 - y^k+1 = (xM + y^k)(x - y) Since x, y, M, k are integers, xM + y^k is also an integer. Therefore, x^k+1 - y^k+1 is divisible by x-y. Conclusion: By the principle of mathematical induction, x^n - y^n is divisible by x-y for all integers n 1. Proven Got more? Send 'em 📸