Answer
See explanation
Work Step by Step
To prove that \(25\mathbb{Z} = \{\,25k : k \in \mathbb{Z}\}\) has the same cardinality as \(2\mathbb{Z} = \{\,2k : k \in \mathbb{Z}\}\), it suffices to exhibit a bijection between them. One straightforward function is
\[
f : 25\mathbb{Z} \;\to\; 2\mathbb{Z},
\quad f(25n) \;=\; 2n.
\]
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### 1. Injectivity
Suppose \(f(25n_1) = f(25n_2)\). Then
\[
2n_1 \;=\; 2n_2
\quad\Longrightarrow\quad
n_1 \;=\; n_2
\quad\Longrightarrow\quad
25n_1 \;=\; 25n_2.
\]
Thus, if two elements in \(25\mathbb{Z}\) map to the same element in \(2\mathbb{Z}\), they must be the same element. Hence, \(f\) is injective.
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### 2. Surjectivity
Every element of \(2\mathbb{Z}\) can be written as \(2m\) for some \(m \in \mathbb{Z}\). We see that
\[
f(25m) = 2m.
\]
Hence, for any \(2m \in 2\mathbb{Z}\), there is a \(25m \in 25\mathbb{Z}\) that maps to it. Therefore, \(f\) is surjective.
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Since \(f\) is both injective and surjective, it is a **bijection**. Consequently, \(25\mathbb{Z}\) and \(2\mathbb{Z}\) have the same cardinality. This is another illustration of the fact that any nonzero integer multiple of \(\mathbb{Z}\) has the same cardinality as \(\mathbb{Z}\) (and hence any two such sets also share the same cardinality).
