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antennense 2022-07-11 Answered
Proof Irrational number x, given N N exists ϵ > 0 so all rationals V ϵ ( x ) have denominator larger than N.
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Jenna Farmer
Answered 2022-07-12 Author has 17 answers
Since there are only finitely many rational points in [ 0 , 1 ] whose denominator is not larger than N there has to be a ε so small that ( x ε , x + ε ) contains none of them; just take
ε = min { | x y | : y  is rational and its denominator is N } .

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