To put the first equation more formally, we know that ℚ is closed under addition¹, so given k∊ℝ\ℚ, l∊ℚ then if k+l=m∊ℚ, then m-l=m+(-l)∊ℚ, but m-l=k which is no

To put the first equation more formally, we know that ℚ is closed under addition¹, so given k∊ℝ\ℚ, l∊ℚ then if k+l=m∊ℚ, then m-l=m+(-l)∊ℚ, but m-l=k which is not in ℚ so k+l∉ℚ. ⸻ 1. For p,q∊ℚ, let p=a/b, q=c/d, a,b,c,d∊ℤ, then p+q=(ad+bc)/bd, but the products and sums of integers are integers, so p+q∊ℚ