Listing Elements And Intervals Of Sets: A Math Guide

Hey guys! Today, we're diving into the exciting world of sets and real numbers. We're going to explore how to list elements or define intervals for different sets based on given conditions. It might sound a bit complex, but trust me, we'll break it down step by step. So, grab your thinking caps, and let's get started!

a) A = {x ∈ R ||x| = 1}

Let's kick things off with set A. In this case, we're looking for all real numbers x such that the absolute value of x is equal to 1. Remember, the absolute value of a number is its distance from zero on the number line. So, we need to find the numbers that are exactly 1 unit away from zero.

The absolute value |x| = 1 means that x can be either 1 or -1. Why? Because |1| = 1 and |-1| = 1. There are no other real numbers that satisfy this condition. Therefore, we can list the elements of set A as follows:

A = {-1, 1}

That's it! Set A contains just two elements: -1 and 1. Understanding absolute values is crucial here. They always give you the magnitude of a number, irrespective of its sign. This simple example sets the stage for more complex scenarios, so make sure you're comfortable with this concept before moving on.

Understanding set notation is also essential. The notation {x ∈ R ||x| = 1} tells us that we're dealing with a set of elements x that belong to the set of real numbers (R), and these elements satisfy the condition |x| = 1. In simpler terms, we're collecting all the real numbers whose absolute value is 1.

Now, let's move on to the next set and see how we can apply similar principles to solve it.

b) B = {x ∈ R ||x + 2| = 3}

Next up, we have set B, which is defined as the set of all real numbers x such that the absolute value of x + 2 is equal to 3. This means we need to find all real numbers x that, when you add 2 to them, the absolute value of the result is 3.

To solve this, we consider two possibilities:

1. x + 2 = 3
2. x + 2 = -3

Let's solve each equation separately. For the first equation, x + 2 = 3, we subtract 2 from both sides to get x = 3 - 2, which simplifies to x = 1. So, 1 is one of the elements in set B.

Now, let's tackle the second equation, x + 2 = -3. Again, we subtract 2 from both sides to get x = -3 - 2, which simplifies to x = -5. Therefore, -5 is another element in set B.

Thus, the elements of set B are 1 and -5. We can write this as:

B = {-5, 1}

To double-check our work, let's plug these values back into the original equation |x + 2| = 3. For x = 1, we have |1 + 2| = |3| = 3, which is correct. For x = -5, we have |-5 + 2| = |-3| = 3, which is also correct. This confirms that both 1 and -5 are indeed the solutions.

The key here is to understand that the absolute value equation |x + 2| = 3 essentially gives us two separate equations to solve. By considering both the positive and negative possibilities, we can find all the real numbers that satisfy the given condition. This approach is fundamental to solving various types of absolute value problems.

c) C = {x ∈ R ||2x + 1| = 1}

Alright, let's move on to set C, where we're looking for all real numbers x that satisfy the equation |2x + 1| = 1. Similar to the previous example, we need to consider two cases due to the absolute value:

1. 2x + 1 = 1
2. 2x + 1 = -1

First, let's solve the equation 2x + 1 = 1. Subtracting 1 from both sides gives us 2x = 0. Dividing both sides by 2, we find that x = 0. So, 0 is one of the elements in set C.

Now, let's solve the second equation, 2x + 1 = -1. Subtracting 1 from both sides gives us 2x = -2. Dividing both sides by 2, we find that x = -1. Thus, -1 is another element in set C.

Therefore, the elements of set C are 0 and -1. We can write this as:

C = {-1, 0}

Let's verify our solutions by plugging them back into the original equation |2x + 1| = 1. For x = 0, we have |2(0) + 1| = |1| = 1, which is correct. For x = -1, we have |2(-1) + 1| = |-2 + 1| = |-1| = 1, which is also correct.

In this example, we had to deal with a slightly more complex expression inside the absolute value, but the approach remains the same. We split the problem into two separate equations and solve each one independently. This method is universally applicable to any absolute value equation, regardless of the complexity of the expression inside the absolute value bars.

Remember, always check your solutions to ensure they satisfy the original equation. This will help you avoid mistakes and build confidence in your problem-solving skills.

d) D = {x ∈ R | x ≤ 2}

Now, let's switch gears a bit and look at set D, which is defined as the set of all real numbers x such that x is less than or equal to 2. Unlike the previous sets, this one involves an inequality rather than an equation. This means that set D will contain an infinite number of elements, as there are infinitely many real numbers less than or equal to 2.

To represent this set, we use interval notation. In interval notation, we use brackets and parentheses to indicate whether the endpoints are included in the set or not. Since x can be equal to 2, we use a square bracket to include 2 in the set. And since x can be any real number less than 2, we extend the interval to negative infinity, which is always represented with a parenthesis because infinity is not a specific number.

Thus, we can represent set D as follows:

D = (-∞, 2]

This notation means that set D includes all real numbers from negative infinity up to and including 2. For example, -10, -5, 0, 1, 1.5, and 2 are all elements of set D.

Understanding interval notation is crucial when dealing with inequalities. It allows us to represent a continuous range of numbers in a concise and precise manner. In this case, the interval (-∞, 2] clearly defines the set of all real numbers less than or equal to 2. When visualizing this on a number line, you would shade everything to the left of 2, including 2 itself.

The concept of infinity can be a bit tricky, but remember that it simply represents an unbounded quantity. In this context, negative infinity means that there is no lower limit to the numbers in the set. This makes set D an infinite set, as it contains an unlimited number of elements.

e) E = {x ∈ R ||x-2| < 3}

Moving on to set E, we have the set of all real numbers x such that the absolute value of x - 2 is less than 3. This is another inequality involving absolute values, but don't worry, we'll tackle it step by step.

The inequality |x - 2| < 3 means that the distance between x and 2 on the number line is less than 3. In other words, x must be within 3 units of 2.

To solve this inequality, we can rewrite it as a compound inequality:

-3 < x - 2 < 3

Now, we can solve for x by adding 2 to all parts of the inequality:

-3 + 2 < x - 2 + 2 < 3 + 2

This simplifies to:

-1 < x < 5

So, set E includes all real numbers x that are greater than -1 and less than 5. In interval notation, we represent this as:

E = (-1, 5)

Notice that we use parentheses instead of brackets because x must be strictly greater than -1 and strictly less than 5. The endpoints -1 and 5 are not included in the set.

To verify this, let's pick a few numbers within this interval and see if they satisfy the original inequality. For example, let's try x = 0. We have |0 - 2| = |-2| = 2, which is less than 3. Now, let's try x = 4. We have |4 - 2| = |2| = 2, which is also less than 3. Both of these numbers satisfy the inequality, which confirms that our solution is correct.

The key to solving absolute value inequalities like this is to rewrite them as compound inequalities. This allows you to isolate x and determine the range of values that satisfy the given condition. Always remember to pay attention to whether the endpoints are included or excluded based on the inequality symbol.

f) F = {x ∈ R | |x| ≥ 2}

Last but not least, we have set F, which is defined as the set of all real numbers x such that the absolute value of x is greater than or equal to 2. This means we're looking for all real numbers that are at least 2 units away from zero on the number line.

To solve this inequality, we again consider two cases:

1. x ≥ 2
2. x ≤ -2

The first case, x ≥ 2, simply means that x can be any real number greater than or equal to 2. In interval notation, this is represented as [2, ∞).

The second case, x ≤ -2, means that x can be any real number less than or equal to -2. In interval notation, this is represented as (-∞, -2].

Since x can satisfy either of these conditions, set F is the union of these two intervals:

F = (-∞, -2] ∪ [2, ∞)

This notation means that set F includes all real numbers from negative infinity up to and including -2, as well as all real numbers from 2 up to positive infinity. In other words, it includes all real numbers except those strictly between -2 and 2.

To verify this, let's pick a few numbers and see if they satisfy the original inequality. For example, let's try x = -3. We have |-3| = 3, which is greater than or equal to 2. Now, let's try x = 3. We have |3| = 3, which is also greater than or equal to 2. Both of these numbers satisfy the inequality.

Understanding the union of intervals is crucial in this case. The union of two sets is simply the set containing all the elements from both sets. In this context, it means that any number that satisfies either x ≥ 2 or x ≤ -2 is an element of set F. This type of problem often arises when dealing with absolute value inequalities involving