Suppose that a sequence of (rational) fractions p

kokoszzm

kokoszzm

Answered question

2022-06-26

Suppose that a sequence of (rational) fractions p / q converge to an irrational number r. Show that q converges to infinity.

Answer & Explanation

mallol3i

mallol3i

Beginner2022-06-27Added 20 answers

Let r n = p n / q n . r n x. x Q .
Suppose q n y. Then for 1 / 2 > ϵ > 0 there exists and N such that | q n q m | | q n y | + | y q m | < ϵ + ϵ < 1 for all m , n > N. But q n , q m are integers so | q n q m | < 1 q n = q m . So there exists an N where all q n ; n > N are an equal constant integer, y.
If r n x then { r n y } x y. For n > N, r n y = p n y / q n = p n y / y = p n Z . So, by the exact same argument as above: r n y becomes a sequence of integers for n > N. As this converges there is an M where r n y is an equal constant integer, p, for all n > M.
So for n > M, r r n y = p r n = p / y and r n p / y = x Q .
This A contradiction.
Lydia Carey

Lydia Carey

Beginner2022-06-28Added 9 answers

Let { ( p q ) k } be a sequence of of rational numbers such that { ( p q ) k } = { p n } { q n } and so that the sequence converges to r R Q .
Suppose that lim n q n = Q.
If p n converges, we have a contradiction to the irrationality or r, since this would imply that it converges to some z Z .
If p n diverges, p n Q also diverges to infinity, contradicting the fact that the sequence is convergent.

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