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

Question

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, $\text{rational+irrational = irrational}. \square$

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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!

Answer 1

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.

Answer 2

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.