Let A = {x ∈ Z : x = 9a−2 for some integer a}, and let B = {y ∈ Z : y = 9b+16 for some integer b}. Prove A = B.

Let x ∈ A. Substitute x = 9a - 2 into the equation for A. A = {x ∈Z: x=9a2 for some integer a} A = \{9a - 2 / a \in Z\} Rewrite the set using the additive inverse property. A = \{9a + (- 2) / a \in Z\}

Use the commutative property of addition.

A = \{- 2 + 9a / a \in Z\} Add the numbers. A = \{9b + 16 / b \in Z\}

Step 2

Find the integers that belong to B.

Let y ∈ B. Substitute y = 9b + 16 into the equation for B. B=\ y \in Z / y = 9b + 16 for some integer b B = \{9b + 16 / b \in Z\}

Step 3

Notice that the sets A and B contain the same integers. Lambda = \{9b + 16 / b \in Z\} B = \{9b + 16 / b \in Z\} Therefore, A = B.

Solution

The sets A and B are equal.

To prove A = B, we need to show that every element of A is also an element of B, and every element of B is also an element of A.

First, let's take an arbitrary element x ∈ A. Then, x = 9a - 2 for some integer a. We can rewrite this as x = 9(a - 1) + 7. Since a - 1 is also an integer (call it b), we have x = 9b + 7, which means x ∈ B.

Now, let's take an arbitrary element y ∈ B. Then, y = 9b + 16 for some integer b. We can rewrite this as y = 9(b + 1) - 2. Since b + 1 is also an integer (call it a), we have y = 9a - 2, which means y ∈ A.

Since we've shown that every element of A is also in B, and every element of B is also in A, we can conclude that A = B.

We can prove that A = B by showing that any element in A can also be expressed as an element in B, and vice versa.

Proof:

* Let's take an arbitrary element x from set A. So, x = 9a - 2 for some integer a.

* We can rewrite x as x = (9a - 2) + 18.

* Factoring out 9, we get x = 9(a + 2).

* Here, a + 2 is another integer (since a is an integer).

* Therefore, x can be expressed as y = 9b + 16, where b = a + 2, which shows that x belongs to set B.

Similarly, we can prove the other direction:

* Let's take an arbitrary element y from set B. So, y = 9b + 16 for some integer b.

* We can rewrite y as y = 9b + 16 - 18.

* Factoring out 9, we get y = 9(b - 2).

* Here, b - 2 is another integer (since b is an integer).

* Therefore, y can be expressed as x = 9a - 2, where a = b - 2, which shows that y belongs to set A.

In conclusion, we have shown that every element in A can also be expressed as an element in B, and vice versa. Therefore, the sets A and B are equal.

To prove that $ A = B $, we need to show that every element in set A is also in set B, and vice versa.

Let's start by proving that every element in set A is in set B:

1. Let $ x $ be an arbitrary element in set A. This means $ x = 9a - 2 $ for some integer $ a $.

2. We need to express $ x $ in the form $ 9b + 16 $ for some integer $ b $.

3. So, $ 9a - 2 = 9b + 16 $.

4. Solving for $ b $, we get $ b = a - 2 $.

5. Since $ a $ is an integer, $ b = a - 2 $ is also an integer.

6. Thus, $ x = 9b + 16 $, which means $ x $ is in set B.

Now, let's prove that every element in set B is in set A:

1. Let $ y $ be an arbitrary element in set B. This means $ y = 9b + 16 $ for some integer $ b $.

2. We need to express $ y $ in the form $ 9a - 2 $ for some integer $ a $.

3. So, $ 9b + 16 = 9a - 2 $.

4. Solving for $ a $, we get $ a = b + 2 $.

5. Since $ b $ is an integer, $ a = b + 2 $ is also an integer.

6. Thus, $ y = 9a - 2 $, which means $ y $ is in set A.

Since we've shown that every element in set A is in set B and every element in set B is in set A, we conclude that $ A = B $.