Sets A And B: Finding Elements And Proving Disjoint

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Hey guys! Let's dive into a fun math problem involving sets. We're given two sets, A and B, defined by specific rules, and we need to find some elements and then prove something interesting about their relationship. This is a classic example of how number theory and set theory can come together to create cool challenges. So, let's break it down step by step and make sure we understand each part.

Understanding the Sets

Before we jump into finding elements or proving anything, let's make sure we really understand what these sets A and B are all about. This is super important because if we don't get the definitions right, everything else we do will be based on shaky ground. So, what do these equations x = 5k + 7 and x = p^2 actually tell us?

Set A: The 5k + 7 Club

The definition of set A, {x | x = 5k + 7, k ∈ N}, might look a bit intimidating at first, but it's really not that scary. Let's break it down piece by piece. The x represents the elements of the set – these are the numbers that belong to A. The equation x = 5k + 7 tells us how to generate these numbers. The k ∈ N part is crucial; it means that k can be any natural number (0, 1, 2, 3, and so on). So, to get an element of A, we pick a natural number for k, multiply it by 5, and then add 7. Let's try a few examples to make this crystal clear.

  • If k = 0, then x = 5(0) + 7 = 7. So, 7 is an element of A.
  • If k = 1, then x = 5(1) + 7 = 12. So, 12 is also in A.
  • If k = 2, then x = 5(2) + 7 = 17. Yep, 17 is part of the club too.

See? It's just a matter of plugging in different values for k. The set A is essentially all the numbers you can get by following this 5k + 7 rule. You might notice that these numbers leave a remainder of 2 when divided by 5 (since 5k is always divisible by 5, and adding 7 leaves a remainder of 2). This is a key characteristic of set A that will become important later when we think about its relationship with set B.

Set B: The Perfect Squares Squad

Now let's tackle set B, defined as {x | x = p^2, p ∈ N}. This one might be a little more familiar. Again, x represents the elements of the set. The equation x = p^2 tells us that the elements of B are perfect squares. And just like before, p ∈ N means that p is a natural number (0, 1, 2, 3, and so on). So, to get an element of B, we simply square a natural number. Easy peasy!

  • If p = 0, then x = 0^2 = 0. So, 0 is in B.
  • If p = 1, then x = 1^2 = 1. So, 1 is also a member of B.
  • If p = 2, then x = 2^2 = 4. You guessed it, 4 is in B too.
  • If p = 3, then x = 3^2 = 9. 9 joins the squad!

Set B is the collection of all numbers that can be obtained by squaring a natural number. These are the perfect squares: 0, 1, 4, 9, 16, 25, and so on. Understanding that set B consists of perfect squares is crucial for the next part of our problem, where we'll be looking at the intersection of A and B.

Finding Elements: Part (a)

Now that we have a solid understanding of what sets A and B represent, let's tackle the first part of the problem: finding two elements from each set. We've already done some of this work in our explanation above, so this should be a breeze. Remember, we just need to plug in different natural numbers for k (for set A) and p (for set B) to generate elements.

Elements of Set A

We already found a few elements when we were explaining set A, but let's list them out explicitly to answer the question. We can choose any two values for k and calculate the corresponding x values.

  • Let's use k = 0: x = 5(0) + 7 = 7. So, 7 is an element of A.
  • Let's use k = 1: x = 5(1) + 7 = 12. So, 12 is another element of A.

So, two elements of set A are 7 and 12. We could easily find more by using other values of k, but two is all we need for this part of the problem.

Elements of Set B

We did the same thing for set B earlier, so let's just pick two of the perfect squares we found.

  • Let's use p = 0: x = 0^2 = 0. So, 0 is an element of B.
  • Let's use p = 1: x = 1^2 = 1. So, 1 is also an element of B.

Therefore, two elements of set B are 0 and 1. Again, we could find many more, but we only need two for now.

So, to recap, we've found the following elements:

  • Set A: 7 and 12
  • Set B: 0 and 1

We've successfully completed part (a) of the problem! Now, let's move on to the more interesting challenge of proving that the intersection of A and B is the empty set.

Proving A ∩ B = ∅: Part (b)

This is where things get a bit more interesting! We need to show that the intersection of sets A and B is the empty set, denoted by ∅. What does this mean? It means that there are no elements that belong to both set A and set B. In other words, there's no number that can be written in the form 5k + 7 (where k is a natural number) and also be a perfect square (p^2, where p is a natural number). This might seem obvious intuitively, but we need to provide a solid mathematical argument to prove it.

The Proof by Contradiction Approach

One common and powerful technique for proving statements like this is called proof by contradiction. The idea is to assume the opposite of what we want to prove, and then show that this assumption leads to a logical absurdity or contradiction. If our assumption leads to a contradiction, then it must be false, which means the original statement we wanted to prove must be true. Clear as mud, right? Let's see how this works in practice.

So, let's assume the opposite of what we want to prove. We want to prove that A ∩ B = ∅, so let's assume that A ∩ B ≠ ∅. This means that there is at least one element that belongs to both A and B. Let's call this element x. If x belongs to both A and B, then it must satisfy the definitions of both sets. This is the crucial step – we're connecting the properties of the sets to a single element.

Since x belongs to A, we know that it can be written in the form x = 5k + 7 for some natural number k. And since x belongs to B, we know that it can be written in the form x = p^2 for some natural number p. So, we have two expressions for x:

  1. x = 5k + 7
  2. x = p^2

Since both expressions are equal to x, we can set them equal to each other:

5k + 7 = p^2

This equation is the key to our proof. We need to show that this equation leads to a contradiction. Let's think about this equation modulo 5. What does that mean? It means we're looking at the remainders when we divide both sides of the equation by 5. This is a clever trick that often helps in number theory problems.

Analyzing the Equation Modulo 5

Let's take the equation 5k + 7 = p^2 and consider it modulo 5.

  • The left side, 5k + 7, modulo 5 is simply 7 modulo 5, which is 2 (because 7 divided by 5 leaves a remainder of 2). The 5k term disappears because it's divisible by 5.
  • So, we have p^2 ≡ 2 (mod 5). This means that p^2 leaves a remainder of 2 when divided by 5.

Now, let's think about the possible remainders when we square a natural number and divide by 5. We only need to consider the remainders of p when divided by 5, which can be 0, 1, 2, 3, or 4. Let's square each of these and see what remainders we get:

  • If p ≡ 0 (mod 5), then p^2 ≡ 0^2 ≡ 0 (mod 5)
  • If p ≡ 1 (mod 5), then p^2 ≡ 1^2 ≡ 1 (mod 5)
  • If p ≡ 2 (mod 5), then p^2 ≡ 2^2 ≡ 4 (mod 5)
  • If p ≡ 3 (mod 5), then p^2 ≡ 3^2 ≡ 9 ≡ 4 (mod 5)
  • If p ≡ 4 (mod 5), then p^2 ≡ 4^2 ≡ 16 ≡ 1 (mod 5)

Notice anything interesting? The possible remainders for p^2 when divided by 5 are 0, 1, and 4. We never get a remainder of 2! This is a crucial observation.

The Contradiction!

We've shown that if 5k + 7 = p^2, then p^2 must leave a remainder of 2 when divided by 5. But we've also shown that a perfect square can never leave a remainder of 2 when divided by 5. This is our contradiction! Our assumption that there exists an element x in both A and B has led us to a logical impossibility.

Since our assumption has led to a contradiction, it must be false. Therefore, the opposite must be true: A ∩ B = ∅. We have successfully proven that the intersection of sets A and B is the empty set.

Conclusion

Wow, we've tackled a really cool problem today! We started by carefully understanding the definitions of sets A and B, then we found some elements of each set, and finally, we used a proof by contradiction to show that their intersection is empty. This problem beautifully illustrates how different areas of math, like set theory and number theory, can come together to create interesting challenges. And remember, the key is to break down the problem into smaller, manageable steps and to really understand the definitions and concepts involved. Keep practicing, guys, and you'll become math superstars in no time!