What's wrong with the following proof of the theorem?

Question

I am reading How to prove it.

In Section 3.2, it asked us what's wrong with the following proof of the theorem?

Suppose $x$ and $y$ are real numbers and $x+y = 10$. Then $x \neq 3$ and $y \neq 8$.

Proof:

Suppose the conclusion of the theorem is false. Then $x=3$ and $y = 8$. But then $x+y = 11$, which contradicts the given information that $x+y = 10$. Therefore the conclusion must be true.

I can feel that the mistake is in $x=3$ and $y=8$. Should it be $x=3$ or $y=8$ since it negates $x=3$ and $y=8$?

But I intuitively think the theorem is correct because $x$ cannot be 3 and $y$ cannot be 8 at the same time if $x + y = 10$ is true. Can anyone explain?

Answer 1

The theorem is false. 3 + 7 = 10, so it can in fact be the case that "x + y = 10" is true while "x is not 3 AND y is not 8" is false. So "x + y = 10" does not imply "x is not 3 AND y is not 8".

"x + y = 10" does imply "x is not 3 OR y is not 8". And this is what the proof in the text is a proof for: the author assumes that "x + y = 10", then assumes that "x is 3 AND y is 8" (which is the negation of "x is not 3 OR y is not 8") and (correctly!) derives a contradiction. So what's wrong with the proof isn't really wrong with the proof as such; it's just not a proof for that theorem. (Nothing is a proof for that theorem, of course, since it is, as noted, false.)

Answer 2

It should be $x=3$ or $x=8,$ which does not give a contradiction.

The theorem is wrong.

Try $x=3$ and $y=7$.

Also, try $y=8$ and $x=2$.