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- If x and y are elements of an ordered integral domain D, prove the following inequalities. a. x22xy+y20 b. x2+y2xy c. x2+y2xyFor an element x of an ordered integral domain D, the absolute value | x | is defined by | x |={ xifx0xif0x Prove that | x |=| x | for all xD. Prove that | x |x| x | for all xD. Prove that | xy |=| x || y | for all x,yD. Prove that | x+y || x |+| y | for all x,yD. Prove that | | x || y | || xy | for all x,yD.2. Prove the following statements for arbitrary elements of an ordered integral domain . a. If and then . b. If and then . c. If then . d. If in then for every positive integer . e. If and then . f. If and then .
- Prove f(x) = x2+4 is not uniformly continuous on R.Suppose that w and r are continuous functions on (−∞, ∞), W (x) is an invertible antiderivative of w(x), and R(x) is an antiderivative of r(x). Circle all of the statements that must be true.Consider the Cauchy Problem y 0 = a(x) arctan y, y(0) = 1, where a(x) is a continuous function defined on R, such that for every x it holds that |a(x)| ≤ 1. Using the Global Picard–Lindel¨of Theorem, show that there exists a unique solution y defined on R.
- Let f: [a, b] -> R be continuous except on a finite number of points in (a, b]. Is f Lebesgue measurable? Answer it Yes or No. True or False: If f: E - › R is continuous and E has a Lebesgue measure 0, then its image f(E) has a Lebesgue measure 0.1. Show (in terms of ε - δ) that a function f : R3 → R defined byf(x; y; z) = (2x + 3y + 4z) is uniformly continuous.Suppose that f1 : [0, 1] → R and f2 : [0, 1] → R are continuous everywhere and that f1(0) < f2(0) and f1(1) > f2(1). Show that there exists a point c ∈ (0,1) such that f1(c) − f2(c) = 0.
- (a) Find a bijection f : Z → Z such that f^n is not equal to 1z for everyn ∈ N.(b) Find an injection f : R → R that is neither strictly increasing norstrictly decreasing.Let f: R -> R be contnuous. If f(c) > 0, prove there exists a neighborhood N of c such that f(x) > 0 for all x ∈ N.Let x, y, z ∈ ℕ. Suppose gcd(x, y) = 1. Prove that if x | yz, then x | z.