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Take for example $Q ∩ [ - \sqrt(2) , \sqrt (2)]$? Would it be $[ - \sqrt(2) , \sqrt (2)]$ or is this untrue since they are not in $Q$?

Patrick Stevens
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user310160
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  • It wouldn't be the $[-\sqrt2, \sqrt2]$ since that contains (uncountably many) irrational numbers. Usually we just write the intersection of $Q$ with an interval as exactly that. – David Kleiman Feb 03 '16 at 06:56
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    Find it? When was it lost? It's thoroughly and completely defined and described as it is "the rational numbers between negative sqrt 2 and sqrt 2". That's utterly clear and nothing more needs to be said. The only thing to discuss is that the end points aren't in it and if you a presuming that Q is you universal set you don't have to state the intersect. But then you should call the end points by sqrt 2. – fleablood Feb 03 '16 at 07:13

2 Answers2

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Of course, it cannot be $[-\sqrt{2}, \sqrt{2}]$ as the endpoints are not rationals. Your question is not exactly clear, the set you have mentioned is closed in the rationals, that's all I can say for now.

aalo
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It would usually be written $$\mathbb{Q} \cap [-\sqrt{2}, \sqrt{2}]$$ That is equal to $$\mathbb{Q} \cap (-\sqrt{2}, \sqrt{2})$$ because $\sqrt{2} \not \in \mathbb{Q}$.

Note that this is not $[-\sqrt{2}, \sqrt{2}]$ because (for example) $\frac{\sqrt{2}}{2}$ is in the interval $(-\sqrt{2}, \sqrt{2})$ but is not rational.

The set can also be written $$\{ x \in \mathbb{Q} \mid x^2 < 2\}$$ if you want to stay entirely within the rationals.

Patrick Stevens
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