Does ( a/b-1/b², a/b+1/b² ) cover the real axis?

Question

Let a real number, z, be given. Does there always exist an infinite sequence, {bₓ}, of natural numbers such that some multiple of 1/bₓ is closer than 1/(bₓ)² to z.

Put otherwise: is there a infinite sequnce of naturals, bₓ, such that |z-a/bₓ|<1/(bₓ)² for some integer, a.

Cập nhật:

This question is a follow up upon the question here:

http://answers.yahoo.com/question/index;_ylt=AhU67...

Cập nhật 2:

The following PDF contains more info:

http://www.ziddu.com/download/8907149/Doesfraction...

Cập nhật 3:

Thanks to both of you (gianlino, Nemo Captain).

I will look the answer from Nemo Captain through before we reach the deadline for this thread (in 16 hours). If you are right (I suppose you are), I believe it shows an interesting property for lines with irrational slopes:

Let z be irrational and consider lines of the form y=z*x+b. If we consider the set of all such lines with the same irrational number, z, through lattice points, then b is in a dense subset of the real axis, ie. you cannot find any interval (however small it may be) on any side of one line y=z*x+b1 through some lattice point without infinitely many other lines y=z*x+b each through some lattice point somewhere in the plane.

Answer 1

Câu trả lời yêu thích

The other example of this, if z=√n, where n - integer but in form not a², with a-positive integer.

Then the Pell's equation:

x²-ny²=1

have infinitely positive solutions.

And the number x/y does satisfy:

|x/y-√(n)| = 1/y² . 1/(x/y+√n) < 1/y².

More details about this:

http://en.wikipedia.org/wiki/Pell's_equation.

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As gianlino's suggestion, now I'm training my self with continued fraction with hoping that I'd solve this question.

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Finally I've solved it (by reading :

http://en.wikipedia.org/wiki/Continued_fraction

somethings I will type here is only the repeating of wikipedia.

For any irrational z, then there is always undefinite continued fraction

z = (a(0); a(1), a(2), ..., a(n), ....)

That a(i) are positive integers.

And we call:

h(n)/k(n) = (a(0); a(1), a(2), ..., a(n))

Then if we assume

h(-2) = 0, k(-2) = 1,

h(-1) = 1, k(-1) = 0,

by inducing, we can prove the theorem 1,and then we have these:

h(n+1) = a(n+1) . h(n) + h(n-1)

k(n+1) = a(n+1) . k(n) + k(n-1)

h(n)/k(n) - h(n-1)/k(n-1) = (-1)^(n-1) / (k(n).k(n-1)),

then we have the formula:

z = a(0) + ∑(n=0->+∞) (-1)^n / [k(n).k(n+1)].

Now let prove the theorem 5 (or exactly a half of it):

|z-h(n)/k(n)| < 1/[k(n).k(n+1)] (*)

With

h(n)/k(n) = a(0) + ∑(i=0->(n-1)) (-1)^i/[k(i).k(i+1)]

|z-h(n)/k(n)| = | ∑(i=n->+∞) (-1)^i / [k(i).k(i+1)] |.

By the formula:

k(i+2) = a(i+2).k(i+1) + k(i) > k(i+1), then

1/[k(i).k(i+1)] > 1/[k(i+1).k(i+2)] for every i ≥ 0.

then

| ∑(i=n->+∞) (-1)^i / [k(i).k(i+1)] |

= {1/[k(n).k(n+1)] - 1/[k(n+1).k(n+2)]} + {1/[k(n+2).k(n+3)] - 1/[k(n+3).k(n+4)]} + ... + {1/[k(n+2i)k(n+2i+1)] - 1/[k(n+2i+1).k(n+2i+2)]} + ...

< {1/[k(n).k(n+1)] - 1/[k(n+1).k(n+2)]} + {1/[k(n+1)k(n+2)] - 1/[k(n+2)k(n+3)]} + {1/[k(n+2).k(n+3)] - 1/[k(n+3).k(n+4)]} + {1/[(k(n+3)k(n+4)] - 1/[k(n+4)k(n+5)]} + ... + {1/[k(n+2i)k(n+2i+1)] - 1/[k(n+2i+1).k(n+2i+2)]} + {1/[k(n+2i+1)k(n+2i+2)] - 1/[k(n+2i+2)k(n+2i+3)]} +...

= 1/[k(n).k(n+1)] - 1/[k(n+1).k(n+2)] + 1/[k(n+1)k(n+2)] - 1/[k(n+2)k(n+3)] + 1/[k(n+2).k(n+3)] - 1/[k(n+3).k(n+4)] + 1/[(k(n+3)k(n+4)] - 1/[k(n+4)k(n+5)] + ... + 1/[k(n+2i)k(n+2i+1)] - 1/[k(n+2i+1).k(n+2i+2)] + 1/[k(n+2i+1)k(n+2i+2)] - 1/[k(n+2i+2)k(n+2i+3)] +...

= 1/[k(n).k(n+1)] + {- 1/[k(n+1).k(n+2)] + 1/[k(n+1)k(n+2)]} + { - 1/[k(n+2)k(n+3)] + 1/[k(n+2).k(n+3)]} + { - 1/[k(n+3).k(n+4)] + 1/[(k(n+3)k(n+4)]} + { - 1/[k(n+4)k(n+5)] + ... + 1/[k(n+2i)k(n+2i+1)]} + { - 1/[k(n+2i+1).k(n+2i+2)] + 1/[k(n+2i+1)k(n+2i+2)]} + { - 1/[k(n+2i+2)k(n+2i+3)] +...

= 1/[k(n).k(n+1)] + 0 + 0 + 0 + 0 + .... + 0 + 0+ ...

= 1/[k(n).k(n+1)].

that means:

(*) exactly.

And we have:

|z - h(n)/k(n)| < 1/[k(n).k(n+1)] < 1/k(n)².

|z-h(n)/k(n)| < 1/k(n)².

With z- any given irrational.

That menas there always exists an infinite sequence, {bₓ}, of natural numbers such that some multiple of 1/bₓ is closer than 1/(bₓ)² to z. (In number theory, it's called the Dirichlet's theorem).

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Even more, the Hurwitz's theorem:

http://en.wikipedia.org/wiki/Hurwitz's_theorem_(nu...

It told that:

( a/b-1/(√5.b²), a/b+1/(√5.b²) ) cover all the real axis.

And more about Langrange numbers:

http://en.wikipedia.org/wiki/Hurwitz's_theorem_(nu...

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:)

Maths is so beautiful!

Answer 2

This seems to be related to continued fractions. But your constant 1 may be too ambitious; at least I think I remember that there is an explicit constant K so that

( a/b-K/b², a/b+K/b² ) cover the real axis. Anyway this is a very well studied topic. So either you solve it for fun and for yourself, or you look up the litterature on continued fractions.

(Các) Nguồn: http://en.wikipedia.org/wiki/Continued_fraction

Answer 3

Multiply by potential of a + b + c: (a + b + c)(a million/a) + (a + b + c)(a million/b) + (a + b + c)(a million/c) = a million a/a + b/a + c/a + a/b + b/b + c/b + a/c + b/c + c/c = a million a million + b/a + c/a + a/b + a million + c/b + a/c + b/c + a million = a million (b + c)/a + (a + c)/b + (a + b)/c = -2 ... no longer likely particular what you're after...? And what do you mean "2 of those numbers are opposite"?