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String
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String asked in Science & MathematicsMathematics · 1 decade ago

Number theory: can any irrational number be approximated as relatively good as desired?

Suppose z is irrational and |z-a/b| < 1/b^n for some rational a/b. By this I define the rational approximation a/b of the irrational z to be "level n good" when n is the maximal integer satisfying the inequality.

For any irrational it is always possible to find an approximation that is "level 1 good" whatever b may be. But does there always exist approximations that are good at higher levels than 1?

Update:

Further I want b to be unlimited so that a/b is not only good, but b is as big as one wishes too, ie. one may always find another fraction a'/b' with b'>b so that a'/b' is just as good as a/b.

Update 2:

@Nemo Captain: I follow you - great work!

So this answers my question negatively for n>2. Thus it states conclusively that the set of all intervals of the form ( a/b-b⁻ⁿ, a/b+b⁻ⁿ ) is not a covering of the real axis for n>2. We know it (clearly) is a covering for n=1.

Next thing is to consider whether the real axis is covered for n=2.

I therefore close this thread and put forward this new question for further investigation.

1 Answer

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  • Nemo Captain
    Lv 5
    1 decade ago
    Favorite Answer

    A very interesting question.

    I could say that: if n>2 then there exists irrationals z that could not be a "level n good".

    Let me take an example:

    z=√2.

    Is there any positive intergers a,b such that:

    |z-a/b| < 1/bⁿ

    (a,b) = 1.

    Let see if b>4, n>2.

    Then:

    -1/bⁿ < z-a/b < 1/bⁿ

    <=> a/b - 1/bⁿ < z < a/b + 1/bⁿ

    <=> a.bⁿ⁻¹ - 1 < z.bⁿ < a.bⁿ⁻¹ + 1

    <=> a.bⁿ⁻¹ - 1 < √2.bⁿ < a.bⁿ⁻¹ + 1

    <=> (a.bⁿ⁻¹ - 1)² < 2.b²ⁿ < (a.bⁿ⁻¹ + 1)²

    <=> a².b²ⁿ⁻² - 2.abⁿ⁻¹ + 1 < 2.b²ⁿ < a².b²ⁿ⁻² + 2.abⁿ⁻¹ + 1. (1)

    Let prove this inequality:

    bⁿ⁻¹ > 2a. (2)

    Of course, hénce: |√2-a/b| < 1/bⁿ < 1/2² = 1/4 , then 2>a/b>1 then 2b>a.

    So that:

    bⁿ⁻¹ ≥ b² ≥ 4b > 2a.

    That means (2) is ok.

    (2)

    <=> b²ⁿ⁻² > 2a.bⁿ⁻¹

    => (a²-1)b²ⁿ⁻² + 1 < a².b²ⁿ⁻² - 2.abⁿ⁻¹ + 1;

    and, a².b²ⁿ⁻² + 2.abⁿ⁻¹ + 1 < (a²+1)b²ⁿ⁻²

    Then, by (1):

    (a²-1)b²ⁿ⁻² + 1 < 2.b²ⁿ < (a²+1)b²ⁿ⁻²;

    Because,

    b²ⁿ⁻² | 2.b²ⁿ

    And there's only an integer on interval ( (a²-1)b²ⁿ⁻² + 1, (a²+1)b²ⁿ⁻² ), that could divides by b²ⁿ⁻², that is a²b²ⁿ⁻², then:

    2.b²ⁿ = a²b²ⁿ⁻²

    <=> 2b² = a² (impossible).

    So that, there is no positive integers a,b such that:

    |√2-a/b| < 1/bⁿ

    (a,b) = 1.

    and b>4, n>2.

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