# Proof By Contradiction With Rational and Irrational Numbers

The question I am working on is:

Use a proof by contradiction to prove that the sum of an irrational number and a rational number is irrational.

I am looking exclusively at Charles' answer. I understand the proof, and its steps, I just don't understand how it proves the original statement is true; I only see it as proving that, if you add two rational numbers, you get another rational number. Could someone help me?

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Do you understand proof by contradiction? – Thomas Andrews Feb 14 '13 at 13:31
@ThomasAndrews Yes, I do. Essentially, the idea is that you proof that something, say $q$, implies something that is, say $p$, and something that isn't, say $\neg p$. Hence, $q \implies (p \wedge \neg p)$ Which is absurd: something can't both be and not be at the same time--unless, of course, we are discussing quantum mechanics. – Mack Feb 14 '13 at 13:37
Okay, now define your $p$ and $q$. – Joe Z. Feb 14 '13 at 13:39
Well, I know that $p$ would definitely be that "$i$ is irrational"; however, I am not sure exactly what $q$ would be. Would $q$ be "the sum of a rational and irrational number?" – Mack Feb 14 '13 at 13:42

He starts by assuming you can find rational $r$ and irrational $i$ that have rational sum $s$. Using a direct proof that the difference of two rationals is rational, he shows that this assumption leads to a contradiction. Therefore you cannot find a rational and irrational that sum to a rational, so the sum of a rational and irrational is always irrational.
Actually he proves more than that: he shows that if $r$ is rational and $i\in\mathbb R$, then the sum $r+i$ is rational if and only if the number $i$ is rational: $$r\in\mathbb Q ~~\text{and}~~ r+i\in\mathbb Q \quad\Rightarrow\quad i\in\mathbb Q$$ hence $$r\in\mathbb Q ~~\text{and}~~ i\notin\mathbb Q \quad\Rightarrow\quad r+i\notin\mathbb Q$$