Dividing a square with different parallel lines and minimize distance between.

A square of side 1 is divided into "a" strips by "a-1" equally spaced

red lines parallel to a side, and into "b" strips by "b-1" equally spaced

blue lines parallel to the red lines. Suppose that a does not divide b

and that b does not divide a. What is the smallest possible distance

between a red line and a blue line?

Edit: Actually, I think case one may be that assume (a,b) =/= 1. In that case since a does not divide b and b does not divide a, there must be at least one c such that (a,b) = c. In this case the red and blue lines most coincide, and so the minimum distance is 0. What about the case when (a,b) = 1?

Re: Dividing a square with different parallel lines and minimize distance between.

Quote:

Originally Posted by

**libzdolce** A square of side 1 is divided into "a" strips by "a-1" equally spaced

red lines parallel to a side, and into "b" strips by "b-1" equally spaced

blue lines parallel to the red lines. Suppose that a does not divide b

and that b does not divide a. What is the smallest possible distance

between a red line and a blue line?

Edit: Actually, I think case one may be that assume (a,b) =/= 1. In that case since a does not divide b and b does not divide a, there must be at least one c such that (a,b) = c. In this case the red and blue lines most coincide, and so the minimum distance is 0. What about the case when (a,b) = 1?

I assume you mean that the strips must be of equal width, so that the red lines are at distances $\displaystyle k/a\ (1\leqslant k\leqslant a-1)$ from one side of the square, and similarly for the blue lines. It then looks as though the minimum distance from a red line to a blue line should be $\displaystyle 1/(ab).$

Reason: given that (a,b) = 1, there exist integers p, q such that $\displaystyle |pa-qb| = 1.$ Then $\displaystyle \left|\tfrac pb - \tfrac qa\right| = \tfrac1{ab}.$