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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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  • $\begingroup$ Every rational number is an algebraic number... $\endgroup$ – Alti Oct 1 '12 at 6:32
  • $\begingroup$ @Alti: And you know that the rationals are dense, so ... ? $\endgroup$ – Brian M. Scott Oct 1 '12 at 6:33
  • $\begingroup$ 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)) $\endgroup$ – Alti Oct 1 '12 at 6:37
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    $\begingroup$ @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? $\endgroup$ – Brian M. Scott Oct 1 '12 at 6:38
  • $\begingroup$ Therefore the algebraic numbers are dense in R. Wow, I didn't know it could be so simple. Thank you! $\endgroup$ – Alti Oct 1 '12 at 6:42

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