Sets As Intervals: Solving Modulus Inequalities

In this article, we're going to dive into expressing sets defined by inequalities involving absolute values as intervals. This is a fundamental concept in mathematics, especially when dealing with real numbers. We'll tackle several examples step-by-step, making sure you understand how to convert these sets into interval notation. So, let's get started, guys!

Converting Sets to Intervals

Example 1: A = {x ∈ R | |x| ≤ sqrt(2)}

Okay, so our first set is defined as all real numbers x such that the absolute value of x is less than or equal to the square root of 2. What does this actually mean? Well, the absolute value of a number is its distance from zero. So, we're looking for all numbers that are within a distance of sqrt(2) from zero. Mathematically, this translates to:

-sqrt(2) ≤ x ≤ sqrt(2)

Therefore, we can express this set as the interval:

A = [-sqrt(2), sqrt(2)]

This interval includes all numbers from -sqrt(2) to sqrt(2), including -sqrt(2) and sqrt(2) themselves. Remember, the square brackets indicate that the endpoints are included in the interval. Visualizing this on a number line can be super helpful. Imagine a number line with zero in the middle. Then, mark -sqrt(2) and sqrt(2) on either side. The interval represents everything in between those two points, including the points themselves. This is a closed interval because it includes its endpoints. Understanding this concept is super useful when you're dealing with inequalities, especially in calculus and analysis.

Now, let's break this down further to make sure we've really got it. The key to solving these absolute value inequalities is remembering that |x| ≤ a is equivalent to -a ≤ x ≤ a. This is because any number within a distance 'a' of zero must fall between -a and a. This simple rule is super handy and will save you a lot of time and effort. Also, remember that sqrt(2) is approximately 1.414, so we're talking about numbers between roughly -1.414 and 1.414. This can give you a better sense of the scale and the range of values we're dealing with.

Example 2: C = {x ∈ R | -1 ≤ (2x + 11)/5 < 3}

Next up, we've got a slightly more complex set. We need to find all real numbers x that satisfy the inequality -1 ≤ (2x + 11)/5 < 3. To solve this, we need to isolate x in the middle. Let's multiply all parts of the inequality by 5:

-5 ≤ 2x + 11 < 15

Now, subtract 11 from all parts:

-16 ≤ 2x < 4

Finally, divide all parts by 2:

-8 ≤ x < 2

So, we can write this set as the interval:

C = [-8, 2)

Notice that we use a square bracket on the left side because x can be equal to -8, but we use a parenthesis on the right side because x must be strictly less than 2. This is a half-open interval. Half-open intervals are pretty common when you're dealing with inequalities that have both 'less than or equal to' and 'strictly less than' conditions. When working with these types of inequalities, always remember to perform the same operation on all parts of the inequality to keep it balanced. It's like a mathematical seesaw – you gotta keep it even!

Let's recap the steps we took to solve this inequality. First, we multiplied by 5 to get rid of the fraction. Then, we subtracted 11 to isolate the term with x. Finally, we divided by 2 to solve for x. By doing this step-by-step, we can avoid making mistakes and keep everything organized. Also, it's a good idea to check your answer by plugging in a value from within the interval into the original inequality to make sure it holds true. For example, you could try plugging in x = 0, which is between -8 and 2, and see if it satisfies the original inequality. If it does, that's a good sign that you've solved it correctly.

Example 3: A = {x ∈ R | |x| > 2}

Alright, let's move on to another example. This time, we want to find all real numbers x such that the absolute value of x is greater than 2. This means that x is more than 2 units away from zero. This gives us two separate cases:

x > 2 or x < -2

In interval notation, this is:

A = (-∞, -2) ∪ (2, ∞)

Here, we use the union symbol (∪) because x can be in either of these intervals. The parentheses indicate that -2 and 2 are not included in the set. This is an open interval because it doesn't include its endpoints. When dealing with inequalities involving 'greater than' or 'less than,' you'll often end up with open intervals that extend to infinity. Visualizing this on a number line can be super helpful. Imagine a number line with zero in the middle. Then, mark -2 and 2 on either side. The interval represents everything to the left of -2 and everything to the right of 2, not including the points -2 and 2 themselves. This type of interval is super common in calculus when you're dealing with limits and infinity.

Now, let's talk about why we have two separate cases. The absolute value of a number is its distance from zero. So, if the absolute value of x is greater than 2, that means x is either more than 2 units to the right of zero (x > 2) or more than 2 units to the left of zero (x < -2). That's why we end up with two separate intervals. Remember, when you're dealing with absolute value inequalities, it's super important to consider both positive and negative cases to make sure you're capturing all possible solutions.

Example 4: C = {x ∈ R | |2x - 3| ≥ 3}

This one's a bit more involved, but we can handle it! We want to find all real numbers x such that the absolute value of 2x - 3 is greater than or equal to 3. This means:

2x - 3 ≥ 3 or 2x - 3 ≤ -3

Let's solve each inequality separately.

For 2x - 3 ≥ 3:

2x ≥ 6 x ≥ 3

For 2x - 3 ≤ -3:

2x ≤ 0 x ≤ 0

So, the interval notation is:

C = (-∞, 0] ∪ [3, ∞)

We use square brackets here because x can be equal to 0 or 3. Again, we use the union symbol because x can be in either of these intervals. This is another example of open intervals combined with closed intervals. When solving absolute value inequalities like this, always remember to split it into two cases, one where the expression inside the absolute value is greater than or equal to the value, and one where it's less than or equal to the negative of the value. This will ensure that you capture all possible solutions. Also, it's a good idea to check your answer by plugging in values from within each interval into the original inequality to make sure they hold true.

Let's recap the steps we took to solve this inequality. First, we split it into two cases based on the definition of absolute value. Then, we solved each inequality separately to find the values of x that satisfy each case. Finally, we expressed the solution as a union of two intervals, using square brackets to indicate that the endpoints are included. By breaking it down into smaller steps, we can make the process more manageable and avoid making mistakes.

Example 5: B = {x ∈ R | |x - sqrt(3)| ≤ sqrt(3)}

Here we have another absolute value inequality. We're looking for all real numbers x such that the absolute value of x - sqrt(3) is less than or equal to sqrt(3). This translates to:

-sqrt(3) ≤ x - sqrt(3) ≤ sqrt(3)

To isolate x, we add sqrt(3) to all parts of the inequality:

0 ≤ x ≤ 2*sqrt(3)

Therefore, the set as an interval is:

B = [0, 2*sqrt(3)]

This interval includes all numbers from 0 to 2sqrt(3), including 0 and 2sqrt(3) themselves. This is a closed interval because it includes its endpoints. This type of inequality is common when you're dealing with transformations of functions, such as translations and scaling. Understanding how to solve these inequalities is super important for understanding how these transformations affect the domain and range of a function.

Let's break down why this works. The key to solving this absolute value inequality is remembering that |x - a| ≤ b is equivalent to -b ≤ x - a ≤ b. This is because we're looking for all numbers x such that their distance from 'a' is less than or equal to 'b'. This simple rule is super handy and will save you a lot of time and effort. Also, remember that sqrt(3) is approximately 1.732, so we're talking about numbers between roughly 0 and 3.464. This can give you a better sense of the scale and the range of values we're dealing with.

Example 6: D = {x ∈ R | |x + 2| < 1}

Last but not least, let's tackle this final example. We're looking for all real numbers x such that the absolute value of x + 2 is less than 1. This means:

-1 < x + 2 < 1

Subtract 2 from all parts of the inequality to isolate x:

-3 < x < -1

So, we can express this set as the interval:

D = (-3, -1)

This interval includes all numbers between -3 and -1, but not -3 and -1 themselves. The parentheses indicate that the endpoints are not included in the set. This is an open interval. This type of inequality is super common when you're dealing with limits and continuity in calculus. Understanding how to solve these inequalities is super important for understanding the behavior of functions near a particular point.

Let's recap the steps we took to solve this inequality. First, we recognized that |x + 2| < 1 is equivalent to -1 < x + 2 < 1. Then, we subtracted 2 from all parts of the inequality to isolate x. Finally, we expressed the solution as an open interval, using parentheses to indicate that the endpoints are not included. By doing this step-by-step, we can avoid making mistakes and keep everything organized. Also, it's a good idea to check your answer by plugging in a value from within the interval into the original inequality to make sure it holds true. For example, you could try plugging in x = -2, which is between -3 and -1, and see if it satisfies the original inequality. If it does, that's a good sign that you've solved it correctly.

Conclusion

So, there you have it! We've worked through several examples of expressing sets defined by absolute value inequalities as intervals. Remember the key steps:

1. Understand the definition of absolute value.
2. Split the inequality into cases if necessary.
3. Isolate x by performing the same operations on all parts of the inequality.
4. Express the solution in interval notation, using brackets for included endpoints and parentheses for excluded endpoints.
5. Use the union symbol (∪) when x can be in multiple intervals.

With practice, you'll become a pro at converting sets to intervals. Keep up the great work, and you'll master these concepts in no time! Remember, math is all about practice, so keep solving problems, and you'll eventually get the hang of it. Good luck, guys!