# What if 'proof by contradiction' is not a valid method of proof?

I've been wondering:

What if 'proof by contradiction' is not a valid method to (dis)prove a statement? What if the 'absurdity' is actually a contradiction?

A proof by contradiction says that to disprove a statement $P$, assume $P$ is true and show that it leads to some contradiction, therefore $P$ is false. But what if the contradiction in this proof was actually a valid contradiction and $P$ is true (and we are just dismissing the contradiction)?

I'll demonstrate my question with an example.

$\underline{\text{Proof that the sum of a rational number and an irrational number is irrational}}$:

Let $\frac{a}{b}$ be the rational number and let $x$ be the irrational number Assume for a contradiction that $\frac{a}{b}+x$ is rational. i.e. assume that $\frac{a}{b}+x=\frac{p}{q},$ for $p, q \in \mathbb{Z}$.

Then $x=\frac{p}{q}-\frac{a}{b}=\frac{pb-aq}{qb}$ which is rational, a contradiction. Therefore, $\text{rational+irrational = irrational}. \square$

But what if the sum of a rational and an irrational number is in fact rational (and this is a contradiction in mathematics)?

If anyone can trim this question to make it more concise and/or articulate, feel free!

• Relevant keywords: "Intuitionistic logic" and "constructive mathematics". – Arthur Aug 1 '14 at 21:38
• @Arthur Thanks! From your point of view, does the question make sense as is? – beep-boop Aug 1 '14 at 21:42
• In order to lead to this conclusion, you must first prove (for example) that the sum of a rational number and an irrational number might be rational. As your linked question alludes to, we can't prove that it's impossible to do this; we can only guess that it's impossible based on the massive amount of circumstantial evidence that no one has discovered a contradiction in mathematics yet. – Dustan Levenstein Aug 1 '14 at 21:47
• I mean, the simplest answer is that it's obviously valid. You may as well call into question modus ponens. There is a point where you just have to accept certain rules of logic as correct. – Jack M Aug 1 '14 at 21:51
• There's a difference between a theory proving $\phi \Rightarrow \psi\wedge\neg\psi$ and proving $\psi\wedge\neg\psi$. Thank goodness... – Malice Vidrine Aug 1 '14 at 22:01

You have shown, in your proof, that the sum cannot be rational. The only way it could be is if $x$ (the irrational number) were a rational number. If $x$ were a rational number, it would, by definition, not be irrational and thus would not satisfy the hypothesis that $x$ is irrational.