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I've just been reading this question about the existence (or lack thereof) of contradictions in maths.

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!

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    $\begingroup$ Relevant keywords: "Intuitionistic logic" and "constructive mathematics". $\endgroup$ – Arthur Aug 1 '14 at 21:38
  • $\begingroup$ @Arthur Thanks! From your point of view, does the question make sense as is? $\endgroup$ – beep-boop Aug 1 '14 at 21:42
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    $\begingroup$ 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. $\endgroup$ – Dustan Levenstein Aug 1 '14 at 21:47
  • $\begingroup$ 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. $\endgroup$ – Jack M Aug 1 '14 at 21:51
  • $\begingroup$ There's a difference between a theory proving $\phi \Rightarrow \psi\wedge\neg\psi$ and proving $\psi\wedge\neg\psi$. Thank goodness... $\endgroup$ – Malice Vidrine Aug 1 '14 at 22:01
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The reality (existence) of the particular contradiction that you mention is excluded by the definition of rational or irrational numbers. The irrational numbers are defined to be precisely those real numbers that are not rational. You ask:

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

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.

Proof by contradiction is therefore a valid method in this instance precisely because we have not assumed that a contradiction cannot occur. On the contrary (ha ha), it is the strict definition of terms that prevents a contradiction. We are only authorised to name something a "contradiction" when we know that it is impossible. We are only authorised to claim to know that something is impossible because we have definitions.

Consider the claim "all red pencils are pencils". If you find a "red pencil" that is not a "pencil", I would claim that is not a "red pencil". This only works if we both agree at the outset that a "red pencil" is a "pencil" that happens to be "red". If we are in agreement with this definition, then the claim "all red pencils are pencils" is not an empirical claim. Similarly, the denial of contradictions in mathematics is not a question of evidence or lack thereof.

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Well if the sum of a rational and an irrational number is rational, then you could have something like your example, x=p/q-a/b, and we already know that irrational numbers can't equal rational numbers.

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