In statement 1 it's true as it is a rational no. and set of rational no. is countable hence set S is countable. But in the second statement taking the primes p= 3 and Q= 2 we get b=√(3/2) = 1.224745.... Which is non repetitive and non-repititive and non-terminating and hence it is a irrational no. and the set of irrational numbers is uncountable . Hence set T should be uncountable. But answer says both the sets are countable . I hope my explanation is correct and that the answer may be wrong. Please correct me if iam wrong.
query on countability of a set
-3
$\begingroup$
set-theory
-
0You have some serious formatting issues that you should work on. (T is countable because it is a set of square roots of rational numbers, so the cardinality is not greater than that of $\mathbb{Q}$) – 2017-02-09
-
0A subset of a countable set will always be countable (or finite). A subset of an uncountable set could be uncountable, or it could be countable, or it could be finite (for example, the set containing $\pi, 2\pi, 3\pi, \ldots$ is clearly countable). You need to find some way to prove that the set is "equal" in size to another set whose countability you know. – 2017-02-09
2 Answers
3
The set $T$ is countable, because you can put it in one-to-one correspondence with the set of order pairs of distinct primes, $\{(p,q): p \neq q\}$ where $p,q$ are prime. Since the primes are a subset of $\mathbb{N}$, they are countable. The set of pairs of primes is a subset of $\mathbb{N} \times \mathbb{N}$. If you also know that the cartesion product of countable sets is countable, then you're done.
0
T is countable. See T consists those numbers $b$ which are of the form $\sqrt{\frac{p}{q}}$ where p and q are primes and $\{\frac{p}{q}:p,q \text{ primes}\} $ $\subset \mathbb{Q}$ i.e you are taking square root of certain rational numbers and hence it is countable.