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.
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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
irrational number + irrational number could be rational or irrational.
5 - sqrt(2) is irrational
sqrt(2) is irrational
Add them up you get 5, which is rational
[1] If it were rational, you will be able to construct a rational representation of the irrational number using this equation.