Is This Proof that 1=2 Valid or Fallacious?

[Enjoicube]

Homework Statement

Alright here it is:

Theorem: if there exists an x belonging to reals such that (x^2)-x-2=(x^2)-4 then 1=2.

Remark: note that there is such an x belonging to reals.

Proof:

1) by hypothesis assume there exists an X belonging to reals such that (x^2)-x-2=(x^2)-4

2)factor each side,

3)resulting in (x-2)(x+1)=(x-2)(x+2)

4)divide each side by (x-2),

5)resulting in x+1=x+2

6)subtract x from each side, resulting in 1=2

1) What terminology (quantifiers, predicates) can be used to express the entire statements 1,3,5

2)Why is this proof fallacious, refer to statements by their numbers. Hint: What are the domains for each statement?

Homework Equations

The Attempt at a Solution

Ok, my attempt at part one (is
Theorem: (\existsX \in\Re) \right arrow ((x^2)-x-2=(x^2)-4))

Statement 1: \existsX \in\Re ((x^2)-x-2=(x^2)-4))

Statement 3:\existsX \in\Re(x-2)(x+1)=(x-2)(x+2)

Statement 5:\existsX \in\Rex+1=x+2

For part 2, i am quite lost, the problem is the use of existential quantifier. My guess is a division by zero somewhere, but otherwise, I need a bigger hint.

Homework Statement

 

[HallsofIvy]

Homework Statement

Alright here it is:

Theorem: if there exists an x belonging to reals such that (x^2)-x-2=(x^2)-4 then 1=2.

Remark: note that there is such an x belonging to reals.

Proof:

1) by hypothesis assume there exists an X belonging to reals such that (x^2)-x-2=(x^2)-4

2)factor each side,

3)resulting in (x-2)(x+1)=(x-2)(x+2)

4)divide each side by (x-2),

5)resulting in x+1=x+2

6)subtract x from each side, resulting in 1=2

1) What terminology (quantifiers, predicates) can be used to express the entire statements 1,3,5

2)Why is this proof fallacious, refer to statements by their numbers. Hint: What are the domains for each statement?

Homework Equations

The Attempt at a Solution

Ok, my attempt at part one (is
Theorem: (\existsX \in\Re) \right arrow ((x^2)-x-2=(x^2)-4))

Statement 1: \existsX \in\Re ((x^2)-x-2=(x^2)-4))

Statement 3:\existsX \in\Re(x-2)(x+1)=(x-2)(x+2)

Statement 5:\existsX \in\Rex+1=x+2

For part 2, i am quite lost, the problem is the use of existential quantifier. My guess is a division by zero somewhere, but otherwise, I need a bigger hint.

Homework Statement

The theorem starts "there exists an X belonging to reals such that (x^2)-x-2=(x^2)-4". Okay, is that true? If not then the conclusion is false because the hypothesis is false! If it is true, what is that x? In other words, solve x^2- x- 2= x^2- 4. I will tell you right now that the hypothesis is true but once you have determined what that x is, you will see why dividing both sides of the equation by x- 2 is an error.

 

[Enjoicube]

Aha! got it. Thank you so much for that. In retrospect I really should have noticed this x value.

 

[HallsofIvy]

In retrospect, I should have been a genius!

 

[statdad]

Actually, in logic, when there are several hypotheses, if any single hypothesis is false the implication (hypotheses imply conclusion) is always true, simply by the nature of the formal definition of a valid argument.

the problem with this "proof" is simply that dividing by x-2 is division by zero, since 2 is a solution to x^2 - x -2 = x^4 - 4