Answer
See explanation
Work Step by Step
Below is a typical way to present the argument using the three functions \(H\), \(I\), and \(J\). These functions illustrate how two sets (\(2\mathbb{Z}\) and \(\mathbb{Z}\)) can have a bijection between them (same cardinality) yet also admit other functions that fail to be onto or fail to be one-to-one.
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## 1. A One-to-One Correspondence \(H\colon 2\mathbb{Z} \to \mathbb{Z}\)
A simple bijection is
\[
H(n) \;=\; \frac{n}{2}.
\]
- **Domain**: \(2\mathbb{Z} = \{\, \dots, -4, -2, 0, 2, 4, \dots\}\).
If \(n \in 2\mathbb{Z}\), then \(n = 2k\) for some \(k \in \mathbb{Z}\).
- **Codomain**: \(\mathbb{Z} = \{\, \dots, -2, -1, 0, 1, 2, \dots\}\).
**Why \(H\) is a bijection**
1. **One-to-one**: If \(H(2k_1) = H(2k_2)\), then \(\frac{2k_1}{2} = \frac{2k_2}{2}\), so \(k_1 = k_2\). Hence \(2k_1 = 2k_2\).
2. **Onto**: Given any \(m \in \mathbb{Z}\), pick \(n = 2m \in 2\mathbb{Z}\). Then \(H(n) = H(2m) = \tfrac{2m}{2} = m\).
Thus, \(H\) is a **one-to-one and onto** map from \(2\mathbb{Z}\) to \(\mathbb{Z}\), proving these sets have the same cardinality.
---
## 2. A One-to-One but Not Onto Function \(I\colon 2\mathbb{Z} \to \mathbb{Z}\)
Define
\[
I(n) \;=\; n \quad \text{(the “inclusion” function)},
\]
where the domain is still \(2\mathbb{Z}\) but the codomain is \(\mathbb{Z}\). In other words, it just “treats” each even integer as an integer.
- **One-to-one**: If \(I(n_1) = I(n_2)\), then \(n_1 = n_2\).
- **Not onto**: The image of \(I\) contains **only even integers** in \(\mathbb{Z}\). No odd integer can ever be an output of \(I\). For instance, there is no \(n \in 2\mathbb{Z}\) such that \(I(n) = 1\).
Hence \(I\) is injective but **not** surjective.
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## 3. An Onto but Not One-to-One Function \(J\colon \mathbb{Z} \to 2\mathbb{Z}\)
Define
\[
J(n) \;=\; 2 \Bigl\lfloor \tfrac{n}{2} \Bigr\rfloor,
\]
where the domain is now \(\mathbb{Z}\) and the codomain is \(2\mathbb{Z}\).
- **Onto**: For any even integer \(2m \in 2\mathbb{Z}\), you can choose \(n = 2m\) (or \(n = 2m + 1\)). Then
\[
\Bigl\lfloor \tfrac{n}{2} \Bigr\rfloor
= \Bigl\lfloor \tfrac{2m}{2} \Bigr\rfloor \text{ or } \Bigl\lfloor \tfrac{2m+1}{2} \Bigr\rfloor
= m,
\]
which gives \(J(n) = 2m\). Thus every even integer in \(2\mathbb{Z}\) is hit by some integer \(n\), so \(J\) is onto.
- **Not one-to-one**: Notice that \(J(2m) = 2m\) and \(J(2m+1) = 2m\). Distinct inputs \(2m\) and \(2m+1\) yield the same output, so \(J\) fails to be injective.
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## Key Takeaway
- **\(H\)** (from \(2\mathbb{Z}\) to \(\mathbb{Z}\)) is both **one-to-one** and **onto** (a bijection).
- **\(I\)** (from \(2\mathbb{Z}\) to \(\mathbb{Z}\)) is **one-to-one** but **not** onto.
- **\(J\)** (from \(\mathbb{Z}\) to \(2\mathbb{Z}\)) is **onto** but **not** one-to-one.
This highlights that two sets can have the **same cardinality** (there exists a bijection \(H\)) but still admit other natural functions (like \(I\) or \(J\)) that fail one or the other property (injectivity or surjectivity).
