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If t is, how can we prove?
Thank you

2007-08-13 08:09:41 · 1 answers · asked by Edson 1 in Science & Mathematics Mathematics

1 answers

Yes it is. Let (a, b), with a>= 0, be any non empty open interval of R. sqrt(2) is irrational but algebraic, for it's root of the polynomial P(x) = x^2 - 2, whose coefficients are integer. There exists a rational r such that r* sqrt(2) is in (a, b). r * sqrt(2) is a root of the polynomial Q(x) = x^2 - 2r^2. Since the coefficients of Q are rational, r sqrt(2) is algebraic. And since r is rational and sqrt(2) is irrational, r sqrt(2) is irrational.

So, every nonempty open interval (a, b), with a>=0, contains an irrational algebraic. By a similar reasoning, we see the same is true for non empty open intervals contained in (-oo, 0). So, the set of the irrational algebraics is dense in R.

2007-08-13 08:23:57 · answer #1 · answered by Steiner 7 · 2 0

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