# Disproving an incorrect theorem?

1. Feb 16, 2016

### 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?

2. Feb 16, 2016

### micromass

Staff Emeritus
That is not the negation of the theorem.

3. Feb 16, 2016

### YamiBustamante

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....

4. Feb 16, 2016

### YamiBustamante

Never mind. I figured it out. Thank you.

5. Feb 18, 2016

### Staff: Mentor

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.