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How would I show that the algebraic numbers are dense in R. I know that the rationals and irrationals are dense in R.

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HINT: Every rational number is also what kind of number?

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    Every rational number is an algebraic number...2012-10-01
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    @Alti: And you know that the rationals are dense, so ... ?2012-10-01
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    The algebraic numbers are dense? Is this because using the rationals, I can construct an irrational algebraic number? (Ex. multiplying w in Q by (2)^(1/2))2012-10-01
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    @Alti: You’re making this **much** too hard. A set $D$ is dense in $\Bbb R$ if every open interval $(a,b)$ contains a member of $D$. The rationals are dense, so every open interval contains a rational, and therefore automatically contains an algebraic number. Therefore?2012-10-01
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    Therefore the algebraic numbers are dense in R. Wow, I didn't know it could be so simple. Thank you!2012-10-01
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    @Alti: You’re welcome! (Sometimes it really is simple, amazingly enough.)2012-10-01
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    You (Alti) mean the real algebraic numbers of course. Not all algebraic numbers are real, for example, $i$ is an algebraic number.2012-10-01