Answer
No inverse
Work Step by Step
Given the function
\[
D : S \;\to\; \mathbb{Z}
\quad\text{by}\quad
D(s) = \bigl(\text{number of 0’s in }s\bigr)\;-\;\bigl(\text{number of 1’s in }s\bigr),
\]
where \(S\) is the set of all finite strings of 0’s and 1’s, and \(\mathbb{Z}\) is the set of all integers.
We want to determine whether \(D\) is one-to-one (injective) and whether it is onto (surjective).
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## (a) Is \(D\) one-to-one (injective)?
A function \(D\) is injective if whenever \(D(s_1) = D(s_2)\), it follows that \(s_1 = s_2\).
- **Counterexample**:
- Consider the string \(s_1 = "0"\). It has 1 zero and 0 ones, so \(D("0") = 1 - 0 = 1.\)
- Consider another string \(s_2 = "001"\). It has 2 zeros and 1 one, so \(D("001") = 2 - 1 = 1.\)
Here \(D(s_1) = D(s_2) = 1\), but clearly \(s_1 = "0"\) and \(s_2 = "001"\) are different strings. Hence \(D\) is **not** injective.
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## (b) Is \(D\) onto (surjective)?
A function \(D\) is surjective if for **every** integer \(n \in \mathbb{Z}\), there is **some** string \(s \in S\) with \(D(s) = n\).
- **Constructing a preimage for an arbitrary integer \(n\)**:
- If \(n = 0\), we can take \(s\) to be the empty string (which has 0 zeros and 0 ones, difference \(0\)).
- If \(n > 0\), pick a string of \(n\) zeros and no ones (e.g., \("00\cdots0"\) with \(n\) zeros). Then \(D(s) = n - 0 = n.\)
- If \(n < 0\), say \(n = -k\) for \(k>0\). Pick a string of \(k\) ones and no zeros (e.g., \("11\cdots1"\) with \(k\) ones). Then \(D(s) = 0 - k = -k = n.\)
In all cases, we can construct a string whose 0/1 difference is \(n\). Thus \(D\) **is** onto \(\mathbb{Z}\).
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### Conclusion
- \(D\) is **not** one-to-one.
- \(D\) **is** onto (surjective).
