Prove: every irrational number q , given e > 0 , there exists natural numbers



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Prove: every irrational number q, given e > 0, there exists natural numbers N and M such that | N q M | < e

Answer & Explanation



Beginner2022-07-16Added 21 answers

Consider the quotient R / Z which can be identified to the circle S 1 . The irrationality of q is equivalent to the injectivity of the quotient map π : R  S 1 restricted to Z q. The claim serves to demonstrate that for each neighborhood that is open V  0 in S 1 there is always a n q ¯  V with n  0, i.e. π ( Z q )  ( V  ( 0 ) )  
Think about the translations V n = n q ¯ + V  S 1 for n  Z (still open, of course). IIf you can locate an m  Z such that there exists n q ¯  V m  ( m q ¯ ) then ( n  m ) q ¯  V  ( 0 ) and you are done.
If not, it means that for all m  Z π ( Z q )  V m = { m q ¯ }. But this says that π ( Z q ) is discrete in S 1 and since S 1 is compact, this is in contradiction with its infinitness.

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