|
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 Q[3, 5] also has the desired property. Similarly to the theory of constructible numbers, we may even form an expanding sequence of extensions Q[m1, m2, ..., mn] each of which will suite our purpose as long as it does not contain 2 itself. (If one of these fields does contain 2, then, being a field, it of necessity also contains 0 and 22 that are equidistant from 2. The converse is also true: if a field contains two numbers equidistant from 2 then it contains 2. Why?) We can construct such a sequence in many ways, e.g., by selecting distinct mi's from the set of prime numbers.
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 (x1, x2) that do not belong to any line that admits an equation
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 (2, 3) be chosen as the center of the circles in question. If the radius is sufficiently small, the circle contains no (integer) grid points. Let the radius grow. At some point, the circle will pass through a grid point for the first time. No other grid point may lie on the same circle. From that moment on and until the circle passes through a grid point for the second time, the circle will contain exactly 1 grid point. Afterwards and until it passes through another grid point for the third time, it will contain only 2 grid points, and so on. If f(r) is the number of grid points inside the circle with radius r, then f(r) changes in jumps of 1, never 2 or more. And clearly it grows without bound as r grows. Therefore, for any N there is radius r (not uniquely determined) such that the circle with radius r contains exactly N grid points.
References
- J. Cofman, What To Solve?, Oxford Science Publications, 1996.
Copyright © 1996-2008 Alexander Bogomolny
Simple consequence
Assume 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 22(n - m) = n2 - m2. Or 22 = n + m. Which implies rationality of 2. Contradiction.
Why?
Field is a set closed under arithmetic operations. A field always contains a zero (0) and a unit (1) elements. Since 1 + 1 = 2 one may be prompted to assert that every field contains 2 as well. But this is not true in general. E.g., the finite field Z2 that contains no other elements but 0 and 1, does not, in particular, contain 2. All extension fields of Q naturally contain 2. They, therefore, contain the average (a + b)/2 of any two elements.
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 points
A rational number r is equidistant from two rational numbers: 0 and 2r. Therefore, no rational number may belong to S.
According to Schoenberg
Points (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 A(-a3/a1 + a1, a2) and B(-a3/a1 - a1, -a2) lie on a line through M perpendicular to L. Therefore, all points of L are equidistant from the grid points A and B.
Copyright © 1996-2008 Alexander Bogomolny
|
