Proving by contradiction that if $a\in\mathbb Q,b\in \mathbb R\setminus \mathbb Q$ then $a+b\in \mathbb R \setminus \mathbb Q$

Question

I'm trying to prove by contradiction that if $a\in\mathbb Q,b\in \mathbb R\setminus \mathbb Q$ then $a+b\in \mathbb R \setminus \mathbb Q$, I already proved it with contra position and a direct proof seems impossible.

Suppose $a\in\mathbb Q,b\in \mathbb R\setminus \mathbb Q$ show that $a+b\in \mathbb Q$.

So $a=\frac p q, p,q\in \mathbb Z,q\neq 0$, so we have $\frac p q+b=\frac {p+bq}q$ and now I'm basically back to square one since I can't assume that a sum and product of integers and irrationals isn't an integer.

Answer 1

To put it simply, you only need to use that the sum of rationals is again a rational. If $a+b \in \Bbb Q$, you would get that $b = (a+b) - a$ would be a rational, contradiction.

Answer 2

Or if this isn't correct (by contradiction) then let $f_a:\mathbb R \setminus \mathbb Q\to \mathbb Q$ with $f_a(b)=a+b$. Then $f_a$ is $1-1$ which is wrong because $\mathbb Q$ is countable.

Answer 3

Your problem is a misunderstanding of the technique of proof by contradiction.

When performing a proof by contradiction, you make your usual assumptions (e.g., $a\in\mathbb Q, b\in\mathbb R\setminus \mathbb Q$) and you assume the logical negation of what you're trying to prove (e.g., $a+b\in\mathbb Q$).

Per your description, it looks like you made your usual assumptions, and then tried to prove the logical negation of what you were trying to prove; that is impossible to do.