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I'm trying to find the cardinality of this set:

$$\{ m + \sqrt 5 n + \sqrt3 p : m, n, p \in\mathbb Q\}.$$

However, this is somewhat a tougher set to calculate a cardinality for. How do we go about calculating the cardinality for this set?

user915841
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    Hint: $\mathbb{Q}^3$ is countable. – Matthew H. Apr 16 '21 at 23:01
  • It doesn’t really matter much for the cardinality, but was the term $m+\sqrt5n+\sqrt3p$ or $m+\sqrt{5n}+\sqrt{3p}?$ – Thomas Andrews Apr 16 '21 at 23:01
  • @ThomasAndrews the square root is just over the 3 and 5, not n and p! – user915841 Apr 16 '21 at 23:05
  • @MatthewPilling sorry, could you please elaborate on how that helps me solve this, my cardinality skills are very rusty right now unfortunately, thank you! – user915841 Apr 16 '21 at 23:06
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    You can write your set as the untion $$\bigcup_{(m,n,p)\in \mathbb{Q}^3}A(m,n,p)$$ where $$A(m,n,p)={m+n\sqrt{3}+p\sqrt{5}}$$ $A(m,n,p)$ is finite ($\implies$ countable) and $\mathbb{Q}^3$ is countable. Now use the fact that a countable union of countable sets is also countable. – Matthew H. Apr 16 '21 at 23:12

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Let $S$ be the set in question ($S = \{m + n \sqrt{3} + p \sqrt{5} : m, n, p \in \mathbb{Q}\}$).

Clearly, there is a surjection $s : \mathbb{Q}^3 \to S$ defined by $s(m, n, p) = m + n \sqrt{3} + p \sqrt{5}$. Since $S$ is infinite and $\mathbb{Q}^3$ is countable, it follows that $S$ is countable.

Mark Saving
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