Question

Suppose \(A=\\{(m, n) \in \mathbb{N} \times \mathbb{R}: n=\pi m\\} .\) Is it true that \(|\mathbb{N}|=|A| ?\)

Short answer

Yes, it is true that \(|\mathbb{N}| = |A|\) because there is a one-to-one mapping from \(\mathbb{N}\) to \(A\).

Step-by-step solution

Understanding the sets

The first step is to understand what the two sets are. The set \(\mathbb{N}\) is the set of all natural numbers. The set \(A\) is a little more complicated. It's a set that contains pairs \((m, n)\) such that \(m\) comes from the natural numbers and \(n\) equals \(\pi m\). In other words, \(n\) is \(\pi\) multiplied by a natural number.

Identify the mapping

What we can see here is that for every natural number \(m\), there is a corresponding real number \(n\) in the set \(A\). This real number is simply \(\pi\) times the natural number.

Comparing the cardinalities

Since this is a one-to-one correspondence from \(\mathbb{N}\) to \(A\), it implies that the size or the cardinality of both sets is the same. Every element in \(\mathbb{N}\) corresponds uniquely to an element in \(A\) and vice versa.