No two integers are equidistant from the square root of 2.This is a rather simple consequence of the fact that √2 is irrational. Those who understood the proof or came up with one of their own won't be surprised at the following generalization: No two rational numbers are equidistant from the square root of 2. We may try to generalize even further. First we claimed that no two integers are equidistant from √2. In other words, the set of all integers does not contain a pair of points equidistant from √2. Next, we established that another, bigger set - the set Q of all rational numbers - has the same property, i.e., no two of its elements lie at the same distance from √2. Can we augment the latter set still preserving the property in question? So another question arises naturally: What is the biggest set such that no two of its points are equidistant from √2. The first of the two question is simple, the second is less so (I have thoughts about the latter, but not a complete answer.) We can augment the set Q by, say, √3 and even consider the extension field Q[√3]. On top of this, extension field Let's now switch the view. From the stand point of √2, all distances from it to the set Q of rational numbers are different. Are there any other points with the same property? Let S denote the set of all such points. What can be said about it? First, S contains no rational points. Second, if a real number is equidistant from two distinct rational points, the number is necessarily the average of the two points, and is, therefore, rational. From here, S consists of all irrational points. The latter result has been generalized by Schoenberg [Cofman, p 187] to higher dimensions. In two dimensions, consider a grid Q2 of pairs of rational numbers. Schoenberg gave a characterization of the set S of points in the plane each of which lies at different distances from the grid points. According to Schoenberg, S consists of those points a1x1 + a2x2 + a3 = 0, where a1, a2, a3 are rational numbers not all equal to 0. As an example, point (√2, √2) lies on the line x1 - x2 = 0 and, therefore, does not belong to S. On the other hand, point (√2, √3) does belong to S. Stepping back from the grid Q2 to pairs of integer points, we may now answer a question posed by Hugo Steinhaus in 1957: Is it possible to construct a circle such that its interior contains exactly n grid points, for any (non-negative) integer n? Not only such circles exist for any n, the circles can be selected with the same center. Let, for example, point References
|Contact| |Front page| |Contents| |Up| |Store| Copyright © 1996-2012 Alexander Bogomolny Simple consequenceAssume that for two different integers m and n |m - √2| = |n - √2|. Then, after squaring both sides of the equality and some simplifications, we get Why?Field is a set closed under arithmetic operations. A field always contains a zero (0) and a unit (1) elements. Since Now, if two numbers a and b are equidistant from √2, then they lie on the opposite sides from √2 with the latter serving as their average. S contains no rational pointsA rational number r is equidistant from two rational numbers: 0 and 2r. Therefore, no rational number may belong to S. According to SchoenbergPoints (x1, x2) in the plane equidistant from the grid points (p1, p2) and (q1, q2) satisfy (x1 - p1)2 + (x2 - p2)2 = (x1 - q1)2 + (x2 - q2)2, or (2q1 - 2p1)x1 + (2q2 - 2p2)x2 + p12 - q12 + p22 - q22 = 0 which is a linear equation with rational coefficients:
Conversely, let line L be given by equation (*). Note that both a1 and a2 may not be 0; for, otherwise, a3 will be 0 as well in contradiction with the requirement that not all three coefficients are 0. Assume, e.g., that a1 is not 0. Point M(-a3/a1, 0) then lie on L. By direct verification, points |Contact| |Front page| |Contents| |Up| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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