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

Patatiniuh 2022-07-15 Answered
Prove: every irrational number q, given e > 0, there exists natural numbers N and M such that | N q M | < e
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SweallySnicles3
Answered 2022-07-16 Author has 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 assertion amounts to prove that for any open neighborhood V 0 in S 1 there is always a n q ¯ V with n 0, i.e. π ( Z q ) ( V ( 0 ) )
Consider the translates V n = n q ¯ + V S 1 for n Z (still open, of course). If you can find 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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