0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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 $n$? $\endgroup$ – Steven Stadnicki Dec 13 '14 at 16:41
1
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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, $|A| = {\aleph }_{0}$

$\square$

$\endgroup$
0
$\begingroup$

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 $\lvert A \rvert = \aleph_0$.

$\endgroup$
0
$\begingroup$

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} \} $?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.