3 ms·

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

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.