Rational + Irrational = Irrational

A rational number + an irrational number results in an irrational number $\forall a\in Q,b\notin Q,a+b\notin Q$

Proof By Contradiction

Let $a\in Q,b\notin Q$ be arbitrary. Assume $a+b\in Q$ It follows that $a+b=\frac{p}{q}$ where $p,q\in Z$ and $q\ne 0$ (by definition of Rational Number) Furthermore it can be said that, $a=\frac{p_1}{q_1}$ where $p_1,q_1\in Z$ and $q_1\ne 0$ (by definition of Rational Number) $(a+b)-a=\frac{p}{q}-\frac{p_1}{q_1}$ $b=\frac{p{q}_1-p_{1}q}{q{q}_1}$ Note that $p{q}_1-p_{1}q\in Z$ and $q{q}_1\in Z$ (by laws of Integers) This means that $b\in Q$ (by definition of Rational Number) This contradicts our statement $b\notin Q$ Therefore our assumption $a+b\in Q$ is false. Hence, $a+b\notin Q$ $\square$