Question

Prove or disprove: The set \(A=\left\\{\frac{\sqrt{2}}{n}: n \in \mathbb{N}\right\\}\) countably infinite.

Short answer

Yes, the set \(A=\left\{\frac{\sqrt{2}}{n}: n \in\mathbb{N}\right\}\) is countably infinite.

Step-by-step solution

Define the function that will be used for the correspondence

Let's denote by \(f(n)\) the function whose image gives the elements of the set. In this case, we have \(f(n) = \frac{\sqrt{2}}{n}, \forall n \in\mathbb{N}\). That is, every natural number \(n\) is associated with a unique number from the set \(A\).

Prove that the function is injective

A function is injective or one-to-one if every element of the domain corresponds to a unique element in the codomain. In other words, if \(f(n_1) = f(n_2)\), then \(n_1 = n_2\). In this case, if \(\frac{\sqrt{2}}{n_1} = \frac{\sqrt{2}}{n_2}\), then by simplifying it would imply that \(n_1 = n_2\). Therefore, the function \(f(n)\) is injective.

Prove that the function is surjective

A function is surjective or onto if every element of the codomain corresponds to an element in the domain. In this case, every element of the set \(A\) is the image of some natural number \(n\) under the function \(f(n)\). This is evident from the way the set is defined. Therefore, the function \(f(n)\) is surjective.

Conclude that the set is countably infinite

Since the function \(f(n)\) is both injective and surjective, it means that there is a one-to-one correspondence between the set of natural numbers and the set \(A\). Therefore, the set \(A\) is countably infinite.