# Is this correct(proof)?

1. Oct 17, 2008

### tronter

I figured it out.

Last edited: Oct 17, 2008
2. Oct 17, 2008

### jhicks

If x is any real number? I'd choose a number more interesting than $$x=\pi$$. Remember that the claim employs "either/or", so if both $$\pi - x$$ and $$x+\pi$$ are irrational the claim is false.

3. Oct 17, 2008

### tronter

So the proof is incorrect because the statement is false? How do you know its an exclusive or?

4. Oct 17, 2008

### jhicks

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".

5. Oct 17, 2008

### tronter

If you changed it to either A... or...B, or both, then the statement and proof would be true?

6. Oct 17, 2008

### jhicks

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.

Last edited: Oct 17, 2008
7. Oct 17, 2008

### cristo

Staff Emeritus
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.