# Another Analysis statement

• January 28th 2009, 05:58 PM
Skinner
Another Analysis statement
Here's a new one, all values are assumed to be real:

$(\forall x) (\forall y), (\exists z) \ni xz=y$

To me, this says: For all x and y values, there's some z where x times z equals y.

This is false, but I answered true. The reason why is because if z=1, any x would be equal to any possible y. I thought this meant it was true, but apparently it's not.

Wait, just now I was thinking:

The statement

$(\forall x)(\forall y), x=y$

is false, isn't it? Does this mean that any possible values of x and y have to be equal to each other?
• January 28th 2009, 06:03 PM
chabmgph
Quote:

Originally Posted by Skinner
Here's a new one, all values are assumed to be real:

$(\forall x) (\forall y), (\exists z) \ni xz=y$

To me, this says: For all x and y values, there's some z where x times z equals y.

This is false, but I answered true. The reason why is because if z=1, any x would be equal to any possible y. I thought this meant it was true, but apparently it's not.

Wait, just now I was thinking:

The statement

$(\forall x)(\forall y), x=y$

is false, isn't it? Does this mean that any possible values of x and y have to be equal to each other?

The answer for the first part is false, however, your reasoning is not quite correct. The statement is that given any x and y, there exist z such that $xz=y$ is true. We do not need to consider what z is. This fails when $x=0$ and $y\neq 0$. i.e. if $y=5$ there no z such that $0*z = 5$.

And second statement is false like you have stated.
• January 28th 2009, 06:26 PM
Skinner
Okay, Let me make sure that I understand.

This is false because if $x=0$, Then no $z$ would exist to make $xz=y$ true for all real values $y$.

We know that this is false because it must work for every possible x and y value, and since it doesn't for this one case, we can be sure that it is false.

• January 28th 2009, 06:37 PM
chabmgph
Quote:

Originally Posted by Skinner
Okay, Let me make sure that I understand.

This is false because if $x=0$, Then no $z$ would exist to make $xz=y$ true for all real values $y$.

We know that this is false because it must work for every possible x and y value, and since it doesn't for this one case, we can be sure that it is false.