# Homework Help: Logic question

1. Sep 4, 2013

### Miike012

I dont understant how the statement is false. Because I gave a reason why it is true. Can someone explain please? thank u

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2. Sep 4, 2013

### Staff: Mentor

As long as y ≠ 0, it has a multiplicative inverse 1/y. Then if x = 1/y, xy = (1/y)(y) = 1.

3. Sep 4, 2013

### Miike012

Yes, F is the "correct" answer
But I believe the correct answer is T

4. Sep 4, 2013

### Dick

Well, it's not true. Pay attention the quantifiers, it says there exists a fixed x such that for all y. x can't depend on y.

5. Sep 5, 2013

### Miike012

So if an existential quantification of x is before the univeral quantification of y then that means that the value of x must be fixed?

what if it was the universal quant of y then the existential quan of x? Then would x still have to be fixed?

6. Sep 5, 2013

### Miike012

The way I am reading it is...
There exists a real number x such that for all real numbers y not equal to zero , the expression xy = 1.

Or basically how I am interpreting it is, Let y = a where a is a real number and not equal to zero, then we can find a value of x such that xa = 1. The value of x that we are looking for is x = 1/a. Then choose a number y = b such that b =/= a and not equal to zero, then we can find a value of x such that xb = 1.
We would repeat this process for all values y = a where a is all real numbers.
So as you can see in my understanding of the sentence, the value of x is not "fixed" as u say it is.

7. Sep 5, 2013

### CompuChip

$\exists x \forall y (y \neq 0 \implies xy = 1)$
is false.

$\forall y \exists x (y \neq 0 \implies xy = 1)$
is true.