Problem 2 - Discrete Mathematics with Applications

Question

A set \(S\) is defined recursively. Find four elements in each case. i) \(1 \in S\) ii) \(x \in S \rightarrow 2^{x} \in S\)

Step-by-Step Solution

Answer

In conclusion, for both cases, using the given recursive rules, we can find four elements in the set S: \(1 \in S\), \(2 \in S\), \(4 \in S\), and \(16 \in S\).

Step 1: Case 1: 1 ∈ S

For the first case, we know that the number 1 is an element of the set S. 1 ∈ S

Step 2: Case 1: Applying the second rule for the first time

Using rule (ii), if x ∈ S, then 2^x ∈ S. Since 1 ∈ S, let x = 1: 2^1 ∈ S 2 ∈ S

Step 3: Case 1: Applying the second rule for the second time

Now we have another element in the set S, which is 2. Using rule (ii) again with x = 2: 2^2 ∈ S 4 ∈ S

Step 4: Case 1: Applying the second rule for the third time

Now, having the number 4 in the set S, we can apply rule (ii) once more with x = 4: 2^4 ∈ S 16 ∈ S Using these application of the second rule, we have found four elements in the set S for case 1: 1, 2, 4, 16

Step 5: Case 2: 1 ∈ S (different approach)

For the second case, we could use the same first element (1 ∈ S) while still applying the second rule. 1 ∈ S

Step 6: Case 2: Applying the second rule for the first time

Using rule (ii), if x ∈ S, then 2^x ∈ S. Since 1 ∈ S, let x = 1: 2^1 ∈ S 2 ∈ S

Step 7: Case 2: Applying the second rule for the second time

Now we have another element in the set S, which is 2. Using rule (ii) again with x = 2: 2^2 ∈ S 4 ∈ S

Step 8: Case 2: Applying the second rule for the third time

Instead of applying the second rule to 4 (as we did in the first case), let's apply the second rule to 2 itself (x = 2) once again: 2^2 ∈ S 4 ∈ S We end up in the same situation as case 1, where the four elements in set S are: 1, 2, 4, 16 In conclusion, using the given recursive rules, we can find four elements in the set S, which are 1, 2, 4, and 16.

Key Concepts

Recursive Definition

A recursive definition is a unique way to define a set or a sequence of numbers using its preceding elements. It's like building a staircase where each step relies on the one before it. In our exercise, the set \(S\) is defined recursively with two rules:

• First, the element \(1\) belongs to the set \(S\).
• Second, if any number \(x\) is part of \(S\), then \(2^x\) must also be part of \(S\).

So, starting with the number 1, which we know is in the set, we use the second rule to generate new numbers like \(2^1 = 2\), \(2^2 = 4\), \(2^4 = 16\), and so on. This recursive process continues indefinitely, creating the set in a methodical, step-by-step manner.

Set Theory

Set theory is an essential mathematical discipline that studies collections of objects, known as sets. In the context of our problem, we're working with a set \(S\) that is defined by specific rules.
A set, in general, can be anything from numbers to cars to letters of the alphabet. For our specific problem:

• The set \(S\) started with the known element \(1\).
• Through the recursive definition, subsequent elements like \(2, 4, 16\), and any future elements derived will also be part of this set.
• We learned that using these recursive methods allows us to efficiently generate massive and complex sets.

It's like having a magic formula where you plug in parts and get new parts without defining each one individually from scratch.

Exponential Growth

Exponential growth in mathematics appears when numbers increase very quickly, growing in powers or multiples. In our exercise, you may notice how the elements in the set \(S\) grow exponentially.
Starting with 1, the next element is \(2^1 = 2\), which is double the previous. As you continue, \(2^2 = 4\), quadruples from the starting point.

• With each application of the recursive rule, the resulting number becomes a power of two.
• This means every subsequent number is exponentially larger than it originally was. For instance, \(2^4 = 16\), which is significantly larger than \(4\).

This showcases exponential growth, where small initial numbers quickly develop into very large values, demonstrating how powerful these recursive processes can be in generating rapid expansions.