Disproving an incorrect theorem?

[YamiBustamante]

Incorrect Theorem:
Suppose x and y are real numbers and x + y = 10, then x != 3 and y != 8.

(a) What’s wrong with the following proof of the theorem?

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.

(b) Show that the theorem is incorrect by finding a counterexample.So according to the answer it's false because x != 3 can't be proven with x = 3 because that's not the negation, but even so, isn't the theorem true because 3 + 8 = 11 which does contradict the premise... I'm very confused by this...

For b, the counter example was x = 3 and x = 7 but how does that disprove it? I'm still very confused by counter examples.
So would it be written as "Suppose x and y are real numbers and x + y = 10, then x = 3 and y = 7" Is that how the counter example would be written?

Please explain!

 

[micromass]

Proof. Suppose the conclusion of the theorem is false. Then x = 3 and y = 8.

That is not the negation of the theorem.

 

[YamiBustamante]

That is not the negation of the theorem.

So the proof is false so therefore the theorem is false or is the theorem already false to begin with, so proof would also be false...

 

[YamiBustamante]

That is not the negation of the theorem.

Never mind. I figured it out. Thank you.

 

[Mark44]

In the statement, the conclusion is "then $x \ne 3 \text{ and } y \ne 8$"

The negation of the conclusion is $x = 3 \text{ or } y = 8$. I believe this is what micromass was alluding to.