If there is some continuous function in $\mathbb{R}$ that satisfies $f(x)\notin \mathbb{Q}$ for every x. Is f then a constant function? How would I show this?
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To think about this problem, you'll want to ponder two facts: (1) continuous functions satisfy the intermediate value theorem, and (2) rational numbers are "dense" in $ \mathbb{R} $. Can you go from there? – Jake Mirra Nov 12 '19 at 01:55
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Suppose $f$ is continuous and takes on two distinct values $a,b\in\mathbb R$ with $a<b$. By the intermediate value theorem, it also takes on all values in the interval $(a,b)$. Since $\mathbb Q$ is dense in $\mathbb R$, $(a,b)$ contains some rationals, and so $f$ must take on some rational values.
Hence, if $f$ only takes on irrational values, then it must be constant.
Franklin Pezzuti Dyer
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Yes. Equivalently to franklin's answer, the Reals are connected and a continuous function preserves connectedness, while the Irrationals ( the intended range/codomain of your function) are (totally) disconnected in that maximal connected components are singletons. For your image to be connected, the image must be a singleton, i.e., it must be constant.
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