Sets A, B, C, D: Divisibility In Natural Numbers

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Hey guys! Today, we're diving into a fun little math problem involving sets and divisibility within the set of natural numbers. We've got four sets, cleverly named A, B, C, and D, each defined by a specific divisibility rule. Our mission is to figure out exactly what elements belong in each of these sets. So, grab your thinking caps, and let's get started!

Set A: Multiples of 3

Set A is defined as A = {x | x ∈ ℕ and 3 | x}. In simpler terms, this set contains all natural numbers (ℕ) that are divisible by 3. This one's pretty straightforward, guys. Natural numbers are the positive whole numbers (1, 2, 3, ...), and we're looking for those that, when divided by 3, leave no remainder.

To find the elements of set A, we simply list the multiples of 3 that are also natural numbers. The first few elements are 3, 6, 9, 12, and so on. We can represent set A as:

A = {3, 6, 9, 12, 15, 18, 21, ...}

Key Characteristics of Set A:

  • Infinite Set: Set A is an infinite set because there's no largest multiple of 3.
  • Arithmetic Progression: The elements of set A form an arithmetic progression with a common difference of 3.
  • Divisibility by 3: Every element in set A is, by definition, divisible by 3.

Understanding Divisibility. It's important to remember what divisibility actually means. When we say "3 divides x," we mean that x can be written as 3 times some other natural number. For instance, 6 is in set A because 6 = 3 * 2. Similarly, 15 is in set A because 15 = 3 * 5. The concept of divisibility is fundamental to number theory, and it pops up all over the place in mathematics. Understanding it deeply helps to solve problems like this one easily. So, set A is all about those nice, clean multiples of 3 that fit perfectly within the natural numbers. Think of it like counting by threes forever! This set is a foundational example of how we can define sets using divisibility rules, and it serves as a great warm-up for tackling the more complex sets B, C, and D that we'll explore next. Remember, the key is to always go back to the fundamental definition of the set and work your way through the possibilities. And most importantly, have fun with it!

Set B: 4 Divides (2x - 1)

Set B is defined as B = {x | x ∈ ℕ and 4 | (2x - 1)}. This means we are looking for natural numbers x such that when 2x - 1 is divided by 4, the remainder is zero. In other words, 2x - 1 must be a multiple of 4. This already gives us a clue – 2x - 1 will always be an odd number. Let's explore this further.

To figure out the elements of set B, we need to test natural numbers to see if they satisfy the condition 4 | (2x - 1). Let's try a few values:

  • If x = 1, then 2x - 1 = 2(1) - 1 = 1. Is 4 | 1? No.
  • If x = 2, then 2x - 1 = 2(2) - 1 = 3. Is 4 | 3? No.
  • If x = 3, then 2x - 1 = 2(3) - 1 = 5. Is 4 | 5? No.
  • If x = 4, then 2x - 1 = 2(4) - 1 = 7. Is 4 | 7? No.
  • If x = 5, then 2x - 1 = 2(5) - 1 = 9. Is 4 | 9? No.

Wait a minute, guys! Notice something? 2x is always even for any natural number x, which means 2x - 1 will always be odd. But multiples of 4 are always even. An odd number can never be a multiple of 4! Therefore, there are no natural numbers x that satisfy the condition 4 | (2x - 1). Consequently, set B is an empty set.

B = {}

Key Takeaway for Set B:

  • Empty Set: Set B is an empty set, meaning it contains no elements.
  • Contradiction: The condition 4 | (2x - 1) leads to a contradiction because 2x - 1 is always odd, and multiples of 4 are always even.

Why is this important? This example shows that not all sets defined by mathematical conditions will have elements. Sometimes, the conditions themselves are contradictory, leading to an empty set. Recognizing such contradictions is a crucial skill in mathematics. It helps us avoid wasting time searching for elements that simply cannot exist. Understanding the properties of even and odd numbers, combined with the definition of divisibility, allowed us to quickly determine that set B is empty. So, while set B might seem like a bit of a downer since it's empty, it actually teaches us a valuable lesson about the importance of carefully analyzing the conditions that define a set.

Set C: 5 Divides (3x - 1)

Set C is defined as C = {x | x ∈ ℕ and 5 | (3x - 1)}. This means we're looking for natural numbers x such that 3x - 1 is divisible by 5. In other words, 3x - 1 must be a multiple of 5.

To find the elements of set C, we need to test natural numbers to see if they satisfy the condition 5 | (3x - 1). Let's try a few values:

  • If x = 1, then 3x - 1 = 3(1) - 1 = 2. Is 5 | 2? No.
  • If x = 2, then 3x - 1 = 3(2) - 1 = 5. Is 5 | 5? Yes!
  • If x = 3, then 3x - 1 = 3(3) - 1 = 8. Is 5 | 8? No.
  • If x = 4, then 3x - 1 = 3(4) - 1 = 11. Is 5 | 11? No.
  • If x = 5, then 3x - 1 = 3(5) - 1 = 14. Is 5 | 14? No.
  • If x = 6, then 3x - 1 = 3(6) - 1 = 17. Is 5 | 17? No.
  • If x = 7, then 3x - 1 = 3(7) - 1 = 20. Is 5 | 20? Yes!

It seems like we are getting some values. We found that x = 2 and x = 7 satisfy the condition. Let's try to find a pattern. We want to find x such that 3x - 1 = 5k for some integer k. We can rewrite this as 3x = 5k + 1. We need to find values of k that make 5k + 1 divisible by 3.

Let's test a few values of k:

  • If k = 1, 5k + 1 = 6, which is divisible by 3 (x = 2).
  • If k = 2, 5k + 1 = 11, which is not divisible by 3.
  • If k = 3, 5k + 1 = 16, which is not divisible by 3.
  • If k = 4, 5k + 1 = 21, which is divisible by 3 (x = 7).
  • If k = 5, 5k + 1 = 26, which is not divisible by 3.
  • If k = 6, 5k + 1 = 31, which is not divisible by 3.
  • If k = 7, 5k + 1 = 36, which is divisible by 3 (x = 12).

So, x = (5k + 1) / 3. We need to find the values of k that result in a natural number for x. We see that k must be of the form 3n + 1, where n is a non-negative integer. Substituting k = 3n + 1 into the equation for x, we get:

x = (5(3n + 1) + 1) / 3 = (15n + 5 + 1) / 3 = (15n + 6) / 3 = 5n + 2

Therefore, the elements of set C can be represented as x = 5n + 2, where n is a non-negative integer.

C = {2, 7, 12, 17, 22, 27, 32, ...}

Key Characteristics of Set C:

  • Infinite Set: Set C is an infinite set.
  • Arithmetic Progression: The elements of set C form an arithmetic progression with a common difference of 5.
  • Form: The elements can be represented as 5n + 2, where n is a non-negative integer.

Think of it like this. We are looking for numbers that, when multiplied by 3 and then decreased by 1, become multiples of 5. It requires a little trial and error initially, but once you spot the pattern, you can express the entire set in a general form. So, set C is all about finding those specific natural numbers that satisfy this particular divisibility rule.

Set D: 4 Divides (x + 2)

Set D is defined as D = {x | x ∈ ℕ and 4 | (x + 2)}. This means we are looking for all natural numbers x such that x + 2 is divisible by 4. In other words, x + 2 must be a multiple of 4.

To find the elements of set D, we need to find natural numbers x such that when we add 2 to them, the result is divisible by 4. Let's test a few values:

  • If x = 1, then x + 2 = 1 + 2 = 3. Is 4 | 3? No.
  • If x = 2, then x + 2 = 2 + 2 = 4. Is 4 | 4? Yes!
  • If x = 3, then x + 2 = 3 + 2 = 5. Is 4 | 5? No.
  • If x = 4, then x + 2 = 4 + 2 = 6. Is 4 | 6? No.
  • If x = 5, then x + 2 = 5 + 2 = 7. Is 4 | 7? No.
  • If x = 6, then x + 2 = 6 + 2 = 8. Is 4 | 8? Yes!

We are starting to see a pattern. We found that x = 2 and x = 6 satisfy the condition. We want to find x such that x + 2 = 4k for some integer k. We can rewrite this as x = 4k - 2.

Let's test a few values of k:

  • If k = 1, 4k - 2 = 2.
  • If k = 2, 4k - 2 = 6.
  • If k = 3, 4k - 2 = 10.
  • If k = 4, 4k - 2 = 14.

We can see that the elements of set D can be represented as x = 4k - 2, where k is a natural number.

D = {2, 6, 10, 14, 18, 22, 26, ...}

Key Characteristics of Set D:

  • Infinite Set: Set D is an infinite set.
  • Arithmetic Progression: The elements of set D form an arithmetic progression with a common difference of 4.
  • Form: The elements can be represented as 4k - 2, where k is a natural number.

In essence, Set D is defined by natural numbers that, when you add 2 to them, give you a multiple of 4. It's another example of how a simple divisibility rule can define an infinite set of numbers. So, understanding the pattern and how to express the general form is essential to completely define set D.

Conclusion

Alright, guys, we've successfully determined the elements of sets A, B, C, and D. Here’s a quick recap:

  • Set A: A = {3, 6, 9, 12, 15, 18, 21, ...} (Multiples of 3)
  • Set B: B = {} (Empty set)
  • Set C: C = {2, 7, 12, 17, 22, 27, 32, ...} (Numbers of the form 5n + 2)
  • Set D: D = {2, 6, 10, 14, 18, 22, 26, ...} (Numbers of the form 4k - 2)

We saw how divisibility rules can define sets, sometimes leading to infinite sets, and sometimes even to the empty set. Understanding these concepts is fundamental in mathematics, especially in areas like number theory and set theory. Keep practicing, and you'll become a pro at solving these kinds of problems! Keep exploring, and have fun with math!