On rational approximations to an irrational number

1977 ◽  
Vol 47 (1) ◽  
pp. 241-256 ◽  
Author(s):  
A. M. Ostrowski
Keyword(s):  
Irrational Number ◽  
Rational Approximations

scholarly journals A Geometrical Proof of a Theorem of Hurwitz

1916 ◽  
Vol 35 ◽  
pp. 59-65 ◽  
Author(s):  
Lester R. Ford
Keyword(s):  
Infinite Number ◽  
Irrational Number ◽  
Rational Approximations ◽  
Irrational Numbers ◽  
Geometrical Proof ◽  
Positive Quantity ◽  
Image Position

In the study of rational approximations to irrational numbers the following problem presents itself: Let ω be a real irrational number, and let us consider the rational fractions satisfying the inequalityhow small can the positive quantity k be chosen with the certainty that there will always be an infinite number of fractions satisfying the inequality whatever the value (irrational) of ω?

scholarly journals Partitioning the positive integers with higher order recurrences

1991 ◽  
Vol 14 (3) ◽  
pp. 457-462 ◽  
Author(s):  
Clark Kimberling
Keyword(s):  
Positive Integer ◽  
Positive Root ◽  
Irrational Number ◽  
Higher Order ◽  
Typical Result ◽  
Algebraic Integers ◽  
Positive Integers

Associated with any irrational numberα>1and the functiong(n)=[αn+12]is an array{s(i,j)}of positive integers defined inductively as follows:s(1,1)=1,s(1,j)=g(s(1,j−1))for allj≥2,s(i,1)=the least positive integer not amongs(h,j)forh≤i−1fori≥2, ands(i,j)=g(s(i,j−1))forj≥2. This work considers algebraic integersαof degree≥3for which the rows of the arrays(i,j)partition the set of positive integers. Such an array is called a Stolarsky array. A typical result is the following (Corollary 2): ifαis the positive root ofxk−xk−1−…−x−1fork≥3, thens(i,j)is a Stolarsky array.

Macromodeling of I/O Buffers via Compressed Tensor Representations and Rational Approximations

2016 ◽  
Vol 6 (10) ◽  
pp. 1522-1534 ◽  
Author(s):  
Gianni Signorini ◽  
Claudio Siviero ◽  
Stefano Grivet-Talocia ◽  
Igor S. Stievano
Keyword(s):  
Rational Approximations ◽  
Tensor Representations

On Rational Approximations of Groups of Operators

1980 ◽  
Vol 17 (1) ◽  
pp. 119-125 ◽  
Author(s):  
Philip Brenner ◽  
Vidar Thomée
Keyword(s):  
Rational Approximations

Surprisingly Accurate Rational Approximations

2002 ◽  
Vol 75 (4) ◽  
pp. 307-310 ◽  
Author(s):  
Tom M. Apostol ◽  
Mamikon A. Mnatsakanian
Keyword(s):  
Rational Approximations

Paper-folding and cutting activities to demonstrate five-fold symmetry

2018 ◽  
Vol 102 (555) ◽  
pp. 413-421
Author(s):  
King-Shun Leung
Keyword(s):  
Irrational Number ◽  
Simple Fraction ◽  
Regular Pentagon ◽  
Paper Folding ◽  
Pointed Star

We can obtain a two-fold symmetric figure by folding a square sheet of paper in the middle and then cutting along some curves drawn on the paper. By making two perpendicular folds through the centre of the paper and then cutting, we can obtain a four-fold symmetric figure. We can also get an eight-fold symmetric figure by making a fold bisecting an angle made by the two perpendicular folds before cutting. But it is not possible to obtain a three-fold, five-fold or six-fold symmetric figure in this way; we need to make more folds before cutting. Making a three-fold (respectively five-fold and six-fold) figure involves the division of the angle at the centre (360°) of a square sheet of a paper into six (respectively ten and twelve) equal parts. In other words, we need to construct the angles 60°, 36° and 30°. But these angles cannot be obtained by repeated bisections of 180° by simple folding as in the making of two-fold, four-fold and eight-fold figures. In [1], we see that each of the constructions of 60° and 30° applies the fact that sin 30° = ½ and takes only a few simple folding steps. The construction of 36° is more tedious (see, for example, [2] and [3]) as sin 36° is not a simple fraction but an irrational number. In this Article, we show how to make, by paper-folding and cutting a regular pentagon, a five-pointed star and create any five-fold figure as we want. The construction obtained by dividing the angle at the centre of a square paper into ten equal parts is called apentagon base. We gained much insight from [2] and [3] when developing the method for making the pentagon base to be presented below.

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