answer:To add the random numbers into a set in Python, you can use the `set()` function and pass the list of random numbers as an argument. Here's how you can do it: ```python import random random_numbers = [random.randint(1, 100) for _ in range(10)] random_set = set(random_numbers) print(random_set) ``` The `set()` function converts the list of random numbers into a set, which automatically removes any duplicate numbers because sets in Python cannot contain duplicate elements. Alternatively, you can use a set comprehension to directly generate the set of random numbers: ```python import random random_set = {random.randint(1, 100) for _ in range(10)} print(random_set) ```

answer:Here's a Python program that generates a list of 20 pseudorandom numbers between 1 and 100, then separates the numbers into two lists for odd and even values, and calculates the sum of each list: ```python import random # Generate a list of 20 pseudorandom numbers between 1 and 100 numbers = [random.randint(1, 100) for _ in range(20)] print(Original List:, numbers) # Separate the numbers into two lists for odd and even values odd_numbers = [num for num in numbers if num % 2 != 0] even_numbers = [num for num in numbers if num % 2 == 0] print(Odd Numbers:, odd_numbers) print(Even Numbers:, even_numbers) # Calculate the sum of each list sum_odd = sum(odd_numbers) sum_even = sum(even_numbers) print(Sum of Odd Numbers:, sum_odd) print(Sum of Even Numbers:, sum_even) ``` This program uses list comprehensions to generate the list of pseudorandom numbers and to separate the numbers into two lists based on whether they are odd or even. The `sum()` function is used to calculate the sum of each list. Note that the `%` operator is used to check if a number is odd or even by finding the remainder of the division by 2. If the remainder is not zero, the number is odd; if the remainder is zero, the number is even.

answer:To determine if f has an inverse function, we need to check if f is both injective (one-to-one) and surjective (onto). For a function to be invertible, it must satisfy both properties. First, let's examine injectivity. Suppose f(n) = f(m) for some positive integers n and m. Then, by definition of f, we have 2n + 1 = 2m + 1. Subtracting 1 from both sides gives 2n = 2m. Dividing both sides by 2 yields n = m. This shows that f is injective, as different inputs (n and m) cannot map to the same output. Next, we'll consider surjectivity. We need to check if every positive integer in the codomain Z+ has a corresponding preimage in the domain Z+. Let y be an arbitrary positive integer. Can we find an x in Z+ such that f(x) = y? We want 2x + 1 = y. Subtracting 1 from both sides gives 2x = y - 1. Dividing both sides by 2, we get x = (y - 1)/2. However, for x to be a positive integer, (y - 1) must be even. This implies that y must be odd. Thus, f is not surjective, as not every positive integer y has a preimage. Specifically, all even positive integers are excluded from the range of f. Despite not being surjective, f is injective, and we can define an inverse function f^-1 on the range of f, which consists of all odd positive integers. Let g be the function g : {odd positive integers} → Z+ defined by g(y) = (y - 1)/2. Then, g is both injective and surjective on its domain and satisfies g(f(n)) = n for all n in Z+ and f(g(y)) = y for all odd positive integers y. Therefore, g is the inverse function of f, albeit with a restricted domain. So, while f does not have an inverse function defined on the entire codomain Z+, it does have an inverse function when the codomain is restricted to the range of f, which consists of all odd positive integers.

answer:This statement is indeed true. To see why, consider the expression n(n + 1). This can be factored into n^2 + n. Now, consider the possible parity (evenness or oddness) of n. If n is even, then n^2 is also even, and the sum n^2 + n is the sum of two even numbers, which is also even. On the other hand, if n is odd, then n^2 is also odd, but when you add an odd n to it, the result is the sum of an odd and an odd, which is always even. Thus, regardless of whether n is even or odd, the expression n(n + 1) always evaluates to an even number. Another way to look at this is to recognize that n(n + 1) is the product of two consecutive integers. One of these integers must be even (specifically, if n is even, then n is the even one, and if n is odd, then n + 1 is the even one). The product of an even number and any other number is always even, so n(n + 1) must be even. This result has an interesting application in mathematics, particularly in combinatorics and number theory. For example, it can be used to prove that the sum of the first n positive integers, given by the formula n(n + 1)/2, is always an integer (since the numerator is always even, the division by 2 always results in an integer).