Sets A, B, C: Union, Intersection, And Difference Calculations

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Hey guys! Let's dive into a cool math problem involving sets. We've got three sets here – A, B, and C – defined in terms of natural numbers and some conditions. Our mission, should we choose to accept it (and we totally do!), is to figure out their unions, intersections, and differences. Sounds like fun, right? So, grab your thinking caps, and let's get started!

Understanding the Sets

Before we jump into calculations, we need to understand what each set contains. This is super important because if we misinterpret the sets, all our hard work will lead us down the wrong path. Let's break it down:

Set A: The Foundation

Set A is defined as {x ∈ N | 2 ≤ x < 7}. This means Set A contains all natural numbers (N) that are greater than or equal to 2 and strictly less than 7. Natural numbers are the positive whole numbers (1, 2, 3, ...), and sometimes include 0 depending on the convention. In this case, since we're dealing with inequalities, we'll stick to the positive whole numbers. So, Set A includes 2, 3, 4, 5, and 6. We can write it out explicitly as A = {2, 3, 4, 5, 6}.

To be crystal clear, let's reiterate what this definition implies. The symbol '∈' means "is an element of," and 'N' denotes the set of natural numbers. The vertical bar '|' is read as "such that." So, the entire expression reads: Set A is the set of all x that are elements of the natural numbers such that x is greater than or equal to 2 and less than 7. Understanding this notation is crucial for tackling set theory problems.

Set B: A Subset of A

Next up, we have Set B, defined as {x ∈ A | x ≠ 4, x ≠ 5}. This tells us that Set B contains elements that are also in Set A, but with a twist! Set B excludes the numbers 4 and 5. The symbol '≠' means "is not equal to." So, we start with Set A (which we already know is {2, 3, 4, 5, 6}) and remove 4 and 5. This leaves us with Set B = {2, 3, 6}.

The key here is recognizing that Set B is a subset of Set A. A subset is a set whose elements are all contained within another set. In our case, every element in Set B is also an element in Set A. This relationship between sets is fundamental in set theory and will influence how we perform operations like union and intersection.

Set C: A Little Inequality Puzzle

Now, for Set C, which is defined as {x ∈ N* | 2 < 2x - 4 ≤ 8}. This one looks a bit more complex because it involves an inequality. N* usually denotes the set of positive natural numbers (excluding 0). We need to find the positive natural numbers x that satisfy the condition 2 < 2x - 4 ≤ 8. To do this, we'll solve the inequality.

Let's break down the inequality step-by-step. First, we have 2 < 2x - 4. Adding 4 to both sides gives us 6 < 2x. Dividing both sides by 2, we get 3 < x. This means x must be greater than 3.

Next, we have 2x - 4 ≤ 8. Adding 4 to both sides gives us 2x ≤ 12. Dividing both sides by 2, we get x ≤ 6. This means x must be less than or equal to 6.

Combining these two results, we have 3 < x ≤ 6. So, x must be a positive natural number that is greater than 3 and less than or equal to 6. This gives us the numbers 4, 5, and 6. Therefore, Set C = {4, 5, 6}.

Calculating Set Operations

Now that we know exactly what's in Sets A, B, and C, we can start calculating the operations requested. This is where the fun really begins! We'll tackle each operation one by one, explaining the logic behind each step.

1. A ∪ B (Union of A and B)

The union of two sets is a new set that contains all the elements that are in either set, or in both. Think of it as merging the two sets together. We have Set A = {2, 3, 4, 5, 6} and Set B = {2, 3, 6}. To find A ∪ B, we combine all the elements from both sets, without repeating any elements. So, A ∪ B = {2, 3, 4, 5, 6}.

Notice that since Set B is a subset of Set A (all elements of B are also in A), the union A ∪ B simply results in Set A itself. This is a common occurrence when dealing with subsets and unions.

2. B ∪ C (Union of B and C)

Next, we need to find the union of Sets B and C. We have Set B = {2, 3, 6} and Set C = {4, 5, 6}. Again, we combine all the elements from both sets, avoiding duplicates. This gives us B ∪ C = {2, 3, 4, 5, 6}.

Here, we see that the element 6 is present in both Sets B and C, but it only appears once in the union. Remember, the union operation only includes each unique element once.

3. A \ B (Difference of A and B)

The difference of two sets, denoted by A \ B, is the set of elements that are in Set A but not in Set B. In other words, we take Set A and remove any elements that are also in Set B. We have Set A = {2, 3, 4, 5, 6} and Set B = {2, 3, 6}. So, we remove 2, 3, and 6 from Set A, leaving us with A \ B = {4, 5}.

This operation highlights the concept of set subtraction. We're essentially taking away the elements of Set B from Set A.

4. B \ C (Difference of B and C)

Now, let's find the difference between Sets B and C. We have Set B = {2, 3, 6} and Set C = {4, 5, 6}. We need to remove any elements from Set B that are also in Set C. The only element that appears in both sets is 6. Removing it from Set B gives us B \ C = {2, 3}.

Notice that the order matters in the difference operation. B \ C is not the same as C \ B. We'll calculate C \ A in the next step to illustrate this further.

5. C \ A (Difference of C and A)

To find C \ A, we need to remove any elements from Set C that are also in Set A. We have Set C = {4, 5, 6} and Set A = {2, 3, 4, 5, 6}. All the elements in Set C (4, 5, and 6) are also in Set A. Therefore, removing them leaves us with an empty set, which is denoted by {} or ∅. So, C \ A = {}.

This example clearly shows that the set difference is not commutative, meaning A \ B ≠ B \ A and C \ A ≠ A \ C. In this case, C \ A resulted in an empty set, while A \ C would be different.

6. A ∪ B ∪ C (Union of A, B, and C)

To find the union of three sets, we combine all the elements from all three sets, without repeating any elements. We have Set A = {2, 3, 4, 5, 6}, Set B = {2, 3, 6}, and Set C = {4, 5, 6}. Combining all the elements gives us A ∪ B ∪ C = {2, 3, 4, 5, 6}.

In this case, the union of all three sets is the same as Set A. This is because both Set B and Set C are subsets of Set A, meaning all their elements are already present in Set A.

7. A ∩ C (Intersection of A and C)

The intersection of two sets is a new set that contains only the elements that are common to both sets. Think of it as finding the overlap between the sets. We have Set A = {2, 3, 4, 5, 6} and Set C = {4, 5, 6}. The elements that are present in both sets are 4, 5, and 6. Therefore, A ∩ C = {4, 5, 6}.

Here, the intersection of Sets A and C is equal to Set C. This indicates that Set C is a subset of Set A, as all elements of C are also present in A.

8. A ∩ B (Intersection of A and B)

Next, let's find the intersection of Sets A and B. We have Set A = {2, 3, 4, 5, 6} and Set B = {2, 3, 6}. The elements common to both sets are 2, 3, and 6. So, A ∩ B = {2, 3, 6}.

Again, the intersection of Sets A and B results in Set B. This is because Set B is a subset of Set A, so the overlap between the two sets is precisely the elements in Set B.

9. B ∩ C (Intersection of B and C)

Now, we find the intersection of Sets B and C. We have Set B = {2, 3, 6} and Set C = {4, 5, 6}. The only element that is present in both sets is 6. Therefore, B ∩ C = {6}.

This intersection shows that only one element is shared between Sets B and C. This tells us something about the relationship between these sets – they are not entirely disjoint (they share an element), but neither is a subset of the other.

10. (A ∩ B) ∪ C (Union of the Intersection of A and B with C)

Finally, we need to calculate (A ∩ B) ∪ C. This operation involves two steps. First, we find the intersection of Sets A and B, which we already calculated as A ∩ B = {2, 3, 6}. Then, we find the union of this result with Set C. We have (A ∩ B) = {2, 3, 6} and Set C = {4, 5, 6}. Taking the union of these sets gives us (A ∩ B) ∪ C = {2, 3, 4, 5, 6}.

This final calculation combines the concepts of intersection and union. We first found the common elements between Sets A and B, and then we merged that result with Set C. The order of operations is important here; we had to perform the intersection (A ∩ B) before taking the union with Set C.

Conclusion

Phew! We made it through all the set operations. By carefully defining the sets and applying the rules of union, intersection, and difference, we were able to calculate each result. Remember, the key to success with set theory problems is to break them down step-by-step and to understand the meaning of each operation.

So, to recap, we found:

  • A ∪ B = {2, 3, 4, 5, 6}
  • B ∪ C = {2, 3, 4, 5, 6}
  • A \ B = {4, 5}
  • B \ C = {2, 3}
  • C \ A = {}
  • A ∪ B ∪ C = {2, 3, 4, 5, 6}
  • A ∩ C = {4, 5, 6}
  • A ∩ B = {2, 3, 6}
  • B ∩ C = {6}
  • (A ∩ B) ∪ C = {2, 3, 4, 5, 6}

Great job, guys! You've tackled a complex set theory problem and come out on top. Keep practicing, and you'll be a set operation pro in no time!