Answer
See explanation
Work Step by Step
We want to show, by **structural induction**, that **every string in \(S\)** (when interpreted as a decimal number) **represents an odd integer**.
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## 1. Definition of the Set \(S\)
The set \(S\) is defined recursively as follows:
1. **Base**: The single‐digit strings \(1\), \(3\), \(5\), \(7\), \(9\) are all in \(S\).
(These are precisely the one‐digit odd numbers.)
2. **Recursion**: If \(s \in S\), then each of the following is also in \(S\) (formed by **prefixing** one digit to \(s\)):
- \(4s \in S\)
- \(2s \in S\)
- \(8s \in S\)
- \(6s \in S\)
3. **Restriction**: Nothing is in \(S\) other than what is obtained from (1) and (2).
When each string is viewed as a decimal numeral, we must prove that **every** such numeral is an **odd** integer.
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## 2. Key Observation
In decimal notation, **the parity (odd/even) of an integer depends only on its last digit**. Because our rules place new digits to the **left** (prefix) and never alter the **rightmost** digit, the rightmost digit remains one of \(\{1,3,5,7,9\}\) for every string in \(S\). Consequently, each string in \(S\) ends in an odd digit, so it represents an odd integer.
We formalize this via structural induction.
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## 3. Proof by Structural Induction
### Inductive Hypothesis
Assume for an arbitrary string \(s \in S\), when \(s\) is interpreted as a decimal number, it is odd (i.e.\ its last digit is one of \(\{1,3,5,7,9\}\)).
### Base Case
By rule (1), the strings in the base are the single‐digit strings \(1,3,5,7,9\). Each of these is obviously an odd integer. So the property (being an odd integer) holds for all base strings.
### Inductive Step
Suppose \(s \in S\) is already known (by the inductive hypothesis) to represent an odd integer. We must check each new string formed from \(s\) by rule (2):
- **\(4s\)**: If we prepend ‘4’ to \(s\), the resulting string \(4s\) has the **same last digit** as \(s\). Since \(s\) was odd, \(4s\) remains odd.
- **\(2s\)**: Prepending ‘2’ to \(s\) also leaves the last digit unchanged. Hence \(2s\) is odd.
- **\(8s\)**: Same reasoning; \(8s\) is odd.
- **\(6s\)**: Same reasoning; \(6s\) is odd.
In each case, the newly formed string still ends in the same last digit as \(s\), which is one of \(\{1,3,5,7,9\}\). Thus the new string also represents an odd integer.
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## 4. Conclusion
By structural induction on the rules defining \(S\), every string in \(S\) ends with an odd digit and hence represents an odd integer. Formally:
\[
\boxed{\text{All strings in }S\text{ represent odd integers.}}
\]
