This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Here are the solutions to the problems. Problem 1: Let m be an odd integer. Show that the number (m^4 + 4m^2 + 11)/(16) is an integer. Step 1: Express m as an odd integer. Let m = 2k+1 for some integer k. Step 2: Calculate m^2. m^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1 Since k and k+1 are consecutive integers, their product k(k+1) is always an even number. Let k(k+1) = 2j for some integer j. m^2 = 4(2j) + 1 = 8j + 1 Step 3: Calculate m^4. m^4 = (m^2)^2 = (8j+1)^2 = (8j)^2 + 2(8j)(1) + 1^2 = 64j^2 + 16j + 1 Step 4: Substitute m^2 and m^4 into the given expression. (m^4 + 4m^2 + 11)/(16) = ((64j^2 + 16j + 1) + 4(8j + 1) + 11)/(16) Step 5: Simplify the numerator. = (64j^2 + 16j + 1 + 32j + 4 + 11)/(16) = (64j^2 + (16j + 32j) + (1 + 4 + 11))/(16) = (64j^2 + 48j + 16)/(16) Step 6: Divide each term in the numerator by 16. = (64j^2)/(16) + (48j)/(16) + (16)/(16) = 4j^2 + 3j + 1 Since j is an integer, 4j^2, 3j, and 1 are all integers. The sum of integers is an integer. Therefore, the expression is an integer. The expression is an integer. Problem 2: If a, b, and c are positive real numbers, then show that: sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1 Step 1: Combine the terms under a single square root using the property sqrt(x) × sqrt(y) × sqrt(z) = sqrt(xyz). sqrt((b)/(a) × (c)/(b) × (a)/(c)) Step 2: Multiply the fractions inside the square root. sqrt((b · c · a)/(a · b · c)) Step 3: Cancel out common terms in the numerator and denominator. sqrt((abc)/(abc)) = sqrt(1) Step 4: Evaluate the square root. sqrt(1) = 1 Thus, sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1. sqrt((b)/(a)) × sqrt((c)/(b)) × sqrt((a)/(c)) = 1 Problem 3: Suppose sqrt(n) = (3)/(2), find -sqrt(n). Step 1: Substitute the given value of sqrt(n) into the expression -sqrt(n). -sqrt(n) = -((3)/(2)) Step 2: Simplify the expression. -(3)/(2) -(3)/(2) Problem 4: Let a and b be positive real numbers. Show that (a)/(b) + (b)/(a) 2. Step 1: Start with the known inequality that for any real numbers x and y, (x-y)^2 0. Expand this inequality: x^2 - 2xy + y^2 0 Step 2: Rearrange the terms. x^2 + y^2 2xy Step 3: Let x = sqrt(a) and y = sqrt(b). Substitute these into the inequality. (sqrt(a))^2 + (sqrt(b))^2 2sqrt(a)sqrt(b) a + b 2sqrt(ab) This is the AM-GM inequality for a and b. Step 4: