$\mathbb{C}$ is isomorphic to $\bar{\mathbb{Q}}$?
I know this is false but I can not give a valid argument to justify this, why can not this be given? Thank you very much.
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12$\begingroup$ cardinality${}$? $\endgroup$– Angina SengCommented Sep 21, 2017 at 17:44
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2 Answers
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The set of algebraic numbers is countable, the set of complex numbers is uncountable. Therefore, there cannot be a bijection between them.
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2$\begingroup$ To be more explicit, $\bar Q$ contains precisely all the algebraic numbers, and there are plenty of transcendental real (and complex) numbers. $\endgroup$ Commented Sep 21, 2017 at 17:51
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$\begingroup$ Also, the algebraic numbers are countable because you have a countable number of polynomials (you write them with finite combinations of symbols from a finite alphabet) and each polynomial has a finite number of roots. $\endgroup$ Commented Sep 21, 2017 at 19:31
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$\begingroup$ @TedShifrin Won't we use the cardinality of $\overline{\mathbb{Q}}$ and $\mathbb{C}$ to show existence of transcendental (non-algebraic) complex numbers ? $\endgroup$– reunsCommented Sep 22, 2017 at 7:52
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$\begingroup$ By all means, @reuns, uncountability of $\Bbb R$ shows the existence of transcendentals. And plenty of’em. But with a number sitting in your palm, you need a proof to show that it’s transcendental. $\endgroup$– LubinCommented Sep 24, 2017 at 2:21
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No:
$\mathbb C$ contains $\pi$, which is transcendental over $\mathbb Q$.
$\overline {\mathbb Q}$ contains only elements that are algebraic over $\mathbb Q$.
