Bad assumption. It's an integer for all b=0, for a start. Also, among the multiples of √3, as you keep picking higher and higher values of b, you'll keep findi

Bad assumption. It's an integer for all b=0, for a start. Also, among the multiples of √3, as you keep picking higher and higher values of b, you'll keep finding results that are closer and closer to (but never equal to) integers. And from them, you can choose any value of a to get a result equally close to any other integer. So you can also find an infinite number of combinations of an and nonzero b that get you arbitrarily close to any integer. The formulation for the Einstein tile grid points uses integer a and b, so those are the only cases of interest ('irrational' a or b doesn't matter) In the context of the Einstein grid, go look at the tiles and it makes sense. They have sides that are integer multiples of a unit length. They have 90 degree angles in them. So you can place them up against each other so you can walk along the edges taking only integer steps and 90 degree turns - staying on the integer cartesian grid. But once the tiling forces you to take one of the 120° angles you veer off at an angle and you're going to be an irrational distance from the grid. You might then find yourself turning back onto the grid and moving integer steps again - but those will not bring you back to the gridlines. To get back to the grid you have to take exactly the same number of irrational steps in the opposite direction.