Bridge to abstract math: what is wrong with following proof

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Homework Help Overview

The discussion revolves around a proof related to logical expressions and their truth values, specifically examining the statement involving implications and disjunctions. Participants are analyzing the validity of a proof that claims to demonstrate a falsehood in a given logical statement.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the truth table for the expression (P or Q) => R and its equivalence to (P=>R) ^ (Q=>R). Questions arise regarding the correctness of the original proof and where the misunderstanding may lie.

Discussion Status

The discussion includes various interpretations of the logical expressions involved. Some participants provide hints about using truth tables for verification, while others question the initial setup of the problem and its assumptions. There is acknowledgment of a potential error in the problem statement itself.

Contextual Notes

There is mention of a specific case where the statement may not hold true, particularly for certain values of x. Additionally, a participant notes that the original problem may have been miswritten, which could affect the validity of the proof being discussed.

Aziza
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See attached picture.

The question asks to prove that the statement which I have written on the first line is true. But I somehow proceeded to proving it is false. Basically what I did was simplify the given expression into the form (P or Q) => R and said this is equivalent to (P=>R) ^ (Q=>R). Then just looking at P=>R I arrived at a contradiction, so because F^T if F, the entire initial proposition is false. Where am I going wrong..?
 

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Interesting... [itex](P \lor Q) \rightarrow R[/itex]
What is the truth table for this? That is one way of showing it. Namely, what is the value of R for various values of P and Q.

Similarly, what are those values in the case of [itex](P \rightarrow R) \wedge (Q \rightarrow R)[/itex]

Hint: Wolfram Alpha is amazing for checking your truth tables...
 
UNChaneul said:
Interesting... [itex](P \lor Q) \rightarrow R[/itex]
What is the truth table for this? That is one way of showing it. Namely, what is the value of R for various values of P and Q.

Similarly, what are those values in the case of [itex](P \rightarrow R) \wedge (Q \rightarrow R)[/itex]

Hint: Wolfram Alpha is amazing for checking your truth tables...

No that's not my question...I know that this part of my proof is right. It is page 35 of my book lol. My question is that something else must be wrong with my proof
 
Your result is correct. The statement you are trying to prove is obviously false for x = 3. I should also add that it does not seem to make much sense in the first place: it starts for "all x", but there are exactly two values of x for which it could possibly be non-trivial. Are you sure you wrote it down properly?
 
It is very hard to read what you have written- it surely would have been less trouble just to type the problem here! It appears to ask you to prove "if x^2= 12- x then either x= -1 or (x+ 3)/(x+1) is greater than or equal to 12".

The only thing wrong is just what you say- this is NOT true. If x^2= 12- x then x is either 3 or -4. Neither of those is -1 so the "x= -1" part is false. If x= 3 then (x+3)/(x+ 1) is (3+3)(3+ 1)= 6/4= 3/2 which is NOT "greater than or equal to 12". If x= -4 then (x+ 3)/(x+ 1) is (-4+ 3)/(-4+1)= -1/(-3)= 1/3 which also is OT "greater than or equal to 12".

Check to see if you haven't copied the problem incorrectly.
 
the professor wrote it wrong i found out today, but thanks anyways!
 

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