There is an important distinction to be made here. Examples in this thread show cases of irrational numbers multiplied by or added to other irrational numbers producing real numbers, but in the special case of a rational number added to or multiplied by an irrational number, the result is always irrational.
Otherwise, supposing for instance that (n/m)x is rational for integers n, m, both non-zero, and irrational x, we can express (n/m)x as a ratio of two integers p, q, q non-zero: (n/m)x = p/q if and only if x = (mp)/(qn). Since integers are closed under multiplication, x is rational, against supposition; thus by contradiction (n/m)x is irrational for any rational r = (n/m), with integers n, m both non-zero. Similarly for the case of addition.
Otherwise, supposing for instance that (n/m)x is rational for integers n, m, both non-zero, and irrational x, we can express (n/m)x as a ratio of two integers p, q, q non-zero: (n/m)x = p/q if and only if x = (mp)/(qn). Since integers are closed under multiplication, x is rational, against supposition; thus by contradiction (n/m)x is irrational for any rational r = (n/m), with integers n, m both non-zero. Similarly for the case of addition.