Irrational Rectangles

Please comment your solutions, questions and remarks..

This riddle is not easy. I solved it only after I got a hint. I got the hint without wanting it though, and I was already thinking in the right direction; so with hindsight I say that I probably was able to solve it without the hint. Given time. And passion.

Suppose a rectangle R is divided into smaller rectangles in a way similar to that pictured below, i.e. the small rectangles cover R and they do not mutually intersect except at their boundaries. Suppose that each of the smaller rectangles has the following property: its height or length (or both) is an integer. Show that then R has the same property: the height or length (or both) of R is an integer.



I can offer hints in the comments by request. Good luck!

About Vadim Kulikov

For details see this
This entry was posted in Geometry, Mathematics, Wednesday Problem. Bookmark the permalink.

9 Responses to Irrational Rectangles

  1. Anonymous says:

    Either I’m missing something or this is not too hard (probably the former but please point out what I’m missing). Assume that the heights of the smaller rectangles are integers (the other case is similar). Consider the set S of those small rectangles that intersect the left side of R. Since the heights of the rectangles of S are bounded from below by a positive constant (namely 1), you can show that S is finite. Since the rectangles of S intersect each other only at their boundaries, the length of the left side of R (i.e. height of R) is the sum of the heights of the rectangles of S. These heights are integers by assumption so the sum must be an integer too.

    • What you did is that you solved just a special case of the riddle where all the rectangles are integer in the same direction, i.e. for instance the height of all the rectangles is integer. But the riddle is more general: there might be rectangles with just integer height and rectangles with integer width but not height.

    • See image
      In these pictures there are both: integer-width and integer-height small rectacgles.

  2. Jaakko Seppälä says:

    At least the second proof by me in that wikibooks page has a mistake. The second integral is missing dx dy. Then everything follows from the Fubini’s theorem.

  3. Mindworm says:

    Suppose A is a rectangle with irrational height, length and consisting of smaller rectangles with at least 1 integer side. Let A be minimal in the number of rectangles it consists of (there must be such an A, since this is a natural number. Also it is greater than 1). Then there is some rectangle B at the left boundary of A which has integer length (if not, all the heights would be rational and thus the side of A). If we cut A along the inner side of B (but all the way to the sides) we get a smaller rectangle C to the left of B which has irrational length and height and consists of at least one less rectangle (B). All cut rectangles (at the boundary) were cut by an integer amount, which leaves integer sides integer and irrational sides irrational. So the rectangles C consists of have at least one integer side, if they had one before. Contradiction, since A was already minimal.

    This didn’t seem too hard. Is there a flaw in my reasoning?

    • You are making the same kind of a mistake that many many others have done. The part which is not true is this claim:

      All cut rectangles (at the boundary) were cut by an integer amount, which leaves integer sides integer and irrational sides irrational.

      Namely there might be a rectangle D whose distance from the left boundary is 1/2 and whose length equals that of B and whose height is not an integer. After your cut, this D becomes non-integer length and height.

Leave a Reply

Your email address will not be published. Required fields are marked *