Truth Value Of Propositions: A Math Analysis

Hey guys! Let's dive into determining the truth value of some mathematical propositions. We're going to evaluate each statement to see if it holds water. It's like being a detective, but with numbers and intervals! So, grab your magnifying glasses, and let's get started!

Analyzing the Propositions

a) -2 ∈ (-3; 2)

When determining if -2 ∈ (-3; 2), we need to verify whether -2 is included within the open interval (-3, 2). An open interval means that the endpoints, -3 and 2, are not included in the interval itself. The numbers in the interval (-3, 2) are all the real numbers strictly greater than -3 and strictly less than 2. In other words, x ∈ (-3, 2) if and only if -3 < x < 2. Here, we're checking if -2 fits into this criterion. Since -3 < -2 < 2, -2 indeed falls within this range. So, the proposition is true.

To further illustrate, let’s think of a number line. Place -3 and 2 on the number line, and consider all the numbers between them but not including -3 and 2. You will clearly see that -2 resides comfortably inside this open interval. Therefore, we can confidently say that the statement -2 ∈ (-3; 2) is true. This is a fundamental concept in understanding intervals and set membership, which are essential in calculus, real analysis, and many other areas of mathematics. Intervals define a range of values, and checking if a number belongs to an interval is a common task in problem-solving. Understanding the nuances of open and closed intervals is crucial. Remember, open intervals like (-3, 2) exclude the endpoints, while closed intervals include them, which will be relevant as we move on to the subsequent propositions.

b) 1 ∈ (-∞; 0)

In the proposition 1 ∈ (-∞; 0), we are asking whether the number 1 is an element of the interval that extends from negative infinity up to, but not including, 0. The interval (-∞, 0) includes all real numbers that are strictly less than 0. That is, x ∈ (-∞, 0) if and only if x < 0. Now, let’s consider the number 1. Is 1 less than 0? Absolutely not! 1 is a positive number, and it's definitely greater than 0. Therefore, 1 does not belong to the interval (-∞, 0).

Graphically, on the number line, (-∞, 0) represents all numbers to the left of 0, not including 0 itself. The number 1 is to the right of 0. Thus, it's evident that 1 cannot be in the interval (-∞, 0). This understanding is critical because intervals involving infinity are frequently used in defining the domain and range of functions in calculus and analysis. For instance, when dealing with functions like f(x) = 1/x, understanding that x cannot be 0 is essential, and intervals help express such constraints precisely. So, the statement 1 ∈ (-∞; 0) is false. It is vital to grasp the concept of intervals involving infinity because they pop up frequently in various mathematical contexts, especially when defining limits, continuity, and other fundamental concepts.

c) 10.2 ∈ (10; +∞)

For the proposition 10.2 ∈ (10; +∞), we need to determine if the number 10.2 belongs to the open interval that starts at 10 and extends to positive infinity. The interval (10, +∞) consists of all real numbers that are strictly greater than 10. In other words, x ∈ (10, +∞) if and only if x > 10. Now, let's evaluate whether 10.2 fits this criterion. Is 10.2 greater than 10? Yes, it is! 10.2 is indeed larger than 10, so it falls within the specified interval.

Visualizing this on the number line can be helpful. Imagine a number line starting at 10 and extending to the right indefinitely. Since 10.2 is to the right of 10, it is clearly part of the interval (10, +∞). The concept of intervals extending to infinity is particularly useful when dealing with unbounded sets and limits in calculus. For example, when considering the limit of a function as x approaches infinity, we are essentially examining the function's behavior within an interval of the form (a, +∞) for some value a. Therefore, the proposition 10.2 ∈ (10; +∞) is true. Understanding intervals with infinity is crucial for advanced mathematical topics like limits, integrals, and asymptotic behavior of functions.

d) -2 ∈ [-2, 5]

Moving on to proposition -2 ∈ [-2, 5], we are checking if -2 is an element of the closed interval [-2, 5]. A closed interval means that the endpoints are included in the interval. The interval [-2, 5] includes all real numbers greater than or equal to -2 and less than or equal to 5. So, x ∈ [-2, 5] if and only if -2 ≤ x ≤ 5. Now, let’s consider -2 itself. Is -2 greater than or equal to -2? Yes, it is, because it is equal to -2. Since -2 is one of the endpoints of the closed interval and closed intervals include their endpoints, -2 is indeed part of the interval [-2, 5].

On the number line, the interval [-2, 5] includes both -2 and 5, and all numbers in between. The fact that -2 is part of this interval makes the statement true. Understanding the difference between open and closed intervals is critical here. While -2 would not be part of the open interval (-2, 5), it is certainly part of the closed interval [-2, 5]. This distinction is important when discussing continuity, differentiability, and the behavior of functions at the boundaries of their domains. Thus, the statement -2 ∈ [-2, 5] is true. Closed intervals play a vital role in defining the domain and range of functions, especially when dealing with inequalities and boundary conditions.

e) 3 ∈ [-3, 3]

Now, let's evaluate the proposition 3 ∈ [-3, 3]. We need to check if the number 3 is an element of the closed interval [-3, 3]. As we've discussed, a closed interval includes its endpoints. The interval [-3, 3] consists of all real numbers that are greater than or equal to -3 and less than or equal to 3. In other words, x ∈ [-3, 3] if and only if -3 ≤ x ≤ 3. Now, let’s consider the number 3. Is 3 less than or equal to 3? Yes, it is, because it is equal to 3. Therefore, 3 is one of the endpoints of the closed interval, making it a part of the interval.

Graphically, on the number line, [-3, 3] includes -3, 3, and all numbers between them. Since 3 is one of the endpoints, it's clear that 3 belongs to the interval. The concept of closed intervals is essential in real analysis, especially when dealing with completeness and compactness. The inclusion of endpoints can significantly affect the properties of a set, particularly in the context of limits and continuity. Therefore, the statement 3 ∈ [-3, 3] is true. Remember, the key aspect here is that closed intervals contain their endpoints, which is a fundamental concept in understanding set membership and interval notation.

f) 1/2 ∉ (0, 1)

In the proposition 1/2 ∉ (0, 1), we are asserting that the number 1/2 is not an element of the open interval (0, 1). The open interval (0, 1) includes all real numbers strictly greater than 0 and strictly less than 1. In other words, x ∈ (0, 1) if and only if 0 < x < 1. To evaluate this proposition, we must determine whether 1/2 falls within this interval. Is 1/2 greater than 0? Yes, it is. Is 1/2 less than 1? Yes, it is. Since 0 < 1/2 < 1, the number 1/2 is indeed an element of the interval (0, 1).

However, the proposition claims that 1/2 is not in the interval. This contradicts our finding that 1/2 is indeed in (0, 1). Therefore, the proposition is false. This exercise reinforces the importance of understanding the meaning of