What is the cardinality of $A=\{(x,y)\in \mathbb R\times \mathbb R|(x-y=\sqrt 2) \land (x+y\in \mathbb N)\}$?

Question

What is the cardinality of $A=\{(x,y)\in \mathbb R\times \mathbb R|(x-y=\sqrt 2) \land (x+y\in \mathbb N)\}$?

I wonder if this is possible that $x-y$ is an irrational number while $x+y$ is a natural number. Could it be that $A$ is an empty set and of cardinality $0$?

Answer 1

You have $x+y=n\in\Bbb N$ and $x-y=\sqrt2$. Adding gives $2x=n+\sqrt2$: so $x=\frac12(n+\sqrt2)$. Likewise $y=\frac12(n-\sqrt2)$. So, there's a solution for each $n\in\Bbb N$.

Answer 2

Define $f: \mathbb N \to A$ by

$$f(n):=\left(\frac12(n+\sqrt2), \frac12(n-\sqrt2\right).$$

Now it is easy to see that $f$ is bijective, Hence $A$ is countable.