2 ms·

> a + b√3 I assume there are no values a and b (including irrationals) which would make that sum an integer?

by l0b0 3y ago

kadoban 3y ago
b = sqrt(3) gets you back into integers, sqrt(3)^2 is 3. Any factor of that with other integers also works. b=0 also works but that one seems rather cheap.
jameshart 3y ago
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.
jlokier 3y ago
Here are some positive a and b pairs. a = 2-√3, b = 1 a = 1, b = 1/√3 a = 1, b = √3 At least one √3 always appears somewhere in a or b (if they are non-zero), so they can't both be (non-zero) rationals. In some ways the √3 behaves like the imaginary unit i in complex numbers.
- quickthrower2 3y ago
  There is a bijection between a + bi numbers and a + b√3 numbers, where a, b are integers*. In addition, with the operation of addition this is an isomorphism. Simply because from a + bi you can extract a and b, and you can also do this for a + b√3. Proof: Given a + b√3 = c + d√3, I will show that a = c and b = d: a + b√3 = c + d√3 => a - c = (d - b)√3 For purpose of contradiction, assume a != c, Then b != d otherwise it is clearly not equal. => √3 = (a - c) / (d - b) => √3 = rational number Which is a contradiction You can do the same to show b != d is false too. *conjecture: works with rational numbers too.

> a + b√3 I assume there are no values a and b (including irrationals) which would make that sum an integer?

by l0b0 3y ago

> a + b√3 I assume there are no values a and b (including irrationals) which would make that sum an integer?

b = sqrt(3) gets you back into integers, sqrt(3)^2 is 3. Any factor of that with other integers also works. b=0 also works but that one seems rather cheap.

jameshart 3y ago

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.

jlokier 3y ago

Here are some positive a and b pairs. a = 2-√3, b = 1 a = 1, b = 1/√3 a = 1, b = √3 At least one √3 always appears somewhere in a or b (if they are non-zero), so they can't both be (non-zero) rationals. In some ways the √3 behaves like the imaginary unit i in complex numbers.

quickthrower2 3y ago
There is a bijection between a + bi numbers and a + b√3 numbers, where a, b are integers*. In addition, with the operation of addition this is an isomorphism. Simply because from a + bi you can extract a and b, and you can also do this for a + b√3. Proof: Given a + b√3 = c + d√3, I will show that a = c and b = d: a + b√3 = c + d√3 => a - c = (d - b)√3 For purpose of contradiction, assume a != c, Then b != d otherwise it is clearly not equal. => √3 = (a - c) / (d - b) => √3 = rational number Which is a contradiction You can do the same to show b != d is false too. *conjecture: works with rational numbers too.