Why Does Assuming x Equals 7 Not Prove the Theorem?

[johnnyies]

Homework Statement

Theorem. Suppose x is a real number and x \neq 4. If \frac{2x - 5}{x - 4} = 3 then x = 7

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

Proof. Suppose x = 7. Then \frac{2x - 5}{x - 4} = \frac{2(7) - 5}{7 - 4} = \frac{9}{3} = 3. Therefore if \frac{2x - 5}{x - 4} = 3 then x = 7.

Homework Equations

The Attempt at a Solution

 

[johnnyies]

Crap. I accidentally clicked submit instead of preview post.

Attempt at solution:

Our two givens for this proof is that x is a real numberand x is not 4, and the statement is in an if P, then Q: P \rightarrow Q. However, the proof is in the form of the converse, if Q, then P: Q \rightarrow P, which is not equal to the statement. The proof should be in form of P \rightarrow Q or its contrapositive \negQ \rightarrow \neg P

I'm not sure if that's the reason why the proof is incorrect. The theorem seems fine.

 

[Mark44]

In your proof in your first post, you are assuming that x = 7, and then concluding that x = 7. This is not very interesting.

The second thing to point out is that you say

Therefore if \frac{2x - 5}{x - 4} = 3
then x = 7.

That (the part after if) is precisely what you need to prove, and you have completely glossed over it.

This is a very simple proof, so you don't need to (and shouldn't) work with the contrapositive. Just start from the three given conditions, a very important one of which is (2x - 5)/(x - 4) = 3, and conclude that x = 7.

 

[vela]

Crap. I accidentally clicked submit instead of preview post.

Attempt at solution:

Our two givens for this proof is that x is a real numberand x is not 4, and the statement is in an if P, then Q: P \rightarrow Q. However, the proof is in the form of the converse, if Q, then P: Q \rightarrow P, which is not equal to the statement. The proof should be in form of P \rightarrow Q or its contrapositive \negQ \rightarrow \neg P

I'm not sure if that's the reason why the proof is incorrect. The theorem seems fine.

Yeah, that's the problem with the proof. Assuming what you want to prove and showing the assumption leads to a true statement doesn't tell you anything about whether the initial assumption was correct because (F→T)=T.

 

[johnnyies]

In your proof in your first post, you are assuming that x = 7, and then concluding that x = 7. This is not very interesting.

The second thing to point out is that you say


That (the part after if) is precisely what you need to prove, and you have completely glossed over it.

This is a very simple proof, so you don't need to (and shouldn't) work with the contrapositive. Just start from the three given conditions, a very important one of which is (2x - 5)/(x - 4) = 3, and conclude that x = 7.

Ya, you're right; that's the question though.

The book told me to refer to a proof strategy in which you have givens, and the statement you want to prove is P \rightarrow Q in which you assume P as a given, and prove Q. Which is similar to what you're saying?

 

[Mark44]

That's not only similar to what I'm saying, that's what I'm saying.

 

[johnnyies]

I don't still understand how assuming P, then proving Q shows that <br /> P \rightarrow Q <br /> is true?

 

[vela]

How does it not?

 

[Mark44]

johnnyies, IMO, you're getting wrapped around the axle fretting about P --> Q and contrapositives and such, when you should just jump in and prove the thing.

You know that x is a real number other than 4, and you know that (2x - 5)/(x - 4) = 3, so what can you conclude from these conditions? By "conclude" I mean what statement can you arrive at by using ordinary mathematical operations.

 

[vela]

I don't still understand how assuming P, then proving Q shows that <br /> P \rightarrow Q <br /> is true?

When you assume P, you mean statement P is true. When you prove Q, you mean statement Q is true. From the truth table for P→Q, you can see P→Q is true.

From a more intuitive viewpoint, for the theorem P→Q to be true, the truth table says if the conditions are met, i.e. P=true, then the conclusion must also hold, i.e. Q=true. If the conditions aren't met, i.e. P=false, then the theorem tells you nothing because the conclusion could hold or not hold, i.e. Q either true or false, and the theorem P→Q would still be true.

On the other hand, if the conditions are met, i.e. P=true and the conclusion doesn't hold, i.e. Q=false, the theorem P→Q must be false.

 

[Char. Limit]

Don't think logically, think mathematically. Take your rational equation and solve for x, using only reversibles.

 

[vela]

I don't think the OP's question is about how to prove this specific theorem. It's about the logical structure of proofs, what's acceptable and what's not. He or she was given an example proof and asked to determine whether and why it was valid or not. I'm guessing this is from a "how to write proofs" course.

 

[Char. Limit]

In that case, the proof begs the question.

 

[Mark44]

I don't think the OP's question is about how to prove this specific theorem. It's about the logical structure of proofs, what's acceptable and what's not. He or she was given an example proof and asked to determine whether and why it was valid or not. I'm guessing this is from a "how to write proofs" course.

I hadn't considered that possibility, but it could very well be the case here.

Looking at the OP's question in that light, the problem was to find the flaw in the proof that P --> Q, where P is the statement, (2x - 5)/(x - 4) = 3, and Q is the statement, x = 7. I am omitting the assumption that x is a real number not equal to 4.

The proof was actually for Q --> P, the converse of the original implication.

 

[HallsofIvy]

I don't still understand how assuming P, then proving Q shows that <br /> P \rightarrow Q <br /> is true?

Because that's exactly what P\rightarrow Q means! "If P is true then Q is true". Proving that any time P is true (assuming P), Q is also true is exactly what proving P\rightarrow Q means.