I figured it out.
If x is any real number? I'd choose a number more interesting than [tex]x=\pi[/tex]. Remember that the claim employs "either/or", so if both [tex]\pi - x[/tex] and [tex]x+\pi[/tex] are irrational the claim is false.
So the proof is incorrect because the statement is false? How do you know its an exclusive or?
Yes. Give a counterexample to disprove the claim. It's an exclusive or because that's the language you use when it's an exclusive or. EITHER x-pi is irrational OR x+pi is irrational is what the claim is for any real x, so if they're both irrational then that isn't "either", that's "both".
If you changed it to either A... or...B, or both, then the statement and proof would be true?
well strictly speaking just dropping "either" would work, but nobody ever uses "or" to mean logical or. But yes what you say is right.
Edit: Yeah Cristo is right I had totally forgotten about any "proof" by the time I wrote this.
The proof is incorrect on a more fundamental level: namely, it is not a proof. It gives one example, but it is claimed that this holds for any x, hence a proof must show this. Remember, you cannot prove a general case by one example.
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