# What is the cardinality of $A=\{(a,b)\in \mathbb{R}\times \mathbb{R}\mid 2a+b\in \mathbb{N}\text{ and }a-2b\in \mathbb{N}\}$

Let $A=\{(a,b)\in \mathbb{R}\times \mathbb{R}\mid 2a+b\in \mathbb{N}\text{ and }a-2b\in \mathbb{N}\}$ What is $|A|$?

I think $|A|={{\aleph }_{0}}$ ,but i am not sure how to prove it.

• Can you work backwards? Suppose that $2a+b=m$ and $a-2b=n$. Can you try writing $a$ and $b$ in terms of $m$ and $n$? – Steven Stadnicki Dec 13 '14 at 16:41

Since

$$\det \begin{pmatrix} 2 & 1\\ 1 & -2 \end{pmatrix} =-5\neq 0$$

for each pair $(x, y) \in \mathbb{N} \times \mathbb{N}$ there is a unique solution of

$$2a+b = x\\ a-2b = y$$ So you have at most ${\aleph }_{0}$ solutions and you can find directly ${\aleph }_{0}$ solutions.

We get that $$2\cdot(2a+b) + (a-2b) = 5a \in \mathbb{N}$$ and thus $$5\cdot(2a+b)-2\cdot 5a = 5b \in \mathbb{N}$$ So $A$ is a subset of $B\times B$ where $$B = \{\frac{a}{5} \mid a \in \mathbb{N} \}$$ Now clearly $|B| = |\mathbb{N}|$ so $$A \leq |B\times B| = |\mathbb{N} \times \mathbb{N} | = |\mathbb{N}| = {\aleph }_{0}$$ And it's not hard to see that $A$ must be infinite. Therefore, $|A| = {\aleph }_{0}$

$\square$

If $(a,b) \in A$, then we know that $2a + b \in \mathbb{N}$ and $a - 2b \in \mathbb{N}$.

Since sums of natural numbers are still in $\mathbb{N}$, we then find that $4a + 2b \in \mathbb{N}$ and hence that $5a \in \mathbb{N}$. This means there are $\aleph_0$ possible values for $a$.

For a fixed $a$, $2a + b \in \mathbb{N}$ gives $\aleph_0$ possible values for $b$.

Hence $\lvert A \rvert \leq \aleph_0 \cdot \aleph_0 = \aleph_0$.

But $\lvert A \rvert \geq \aleph_0$ since $\{(n,0)|n \in \mathbb{N}\} \subseteq A$.

Therefore $\lvert A \rvert = \aleph_0$.

If $(a,b)$ is in that set, then both $5a$ and $5b$ are in $\mathbb{Z}$, are they not? Then can you find the cardinality of $F\times F$ where $F = \{ \frac15 x \mid x\in\mathbb{Z} \}$?