# Thread: Another Analysis statement

1. ## Another Analysis statement

Here's a new one, all values are assumed to be real:

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

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

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

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

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

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

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

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

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 $\displaystyle xz=y$ is true. We do not need to consider what z is. This fails when $\displaystyle x=0$ and $\displaystyle y\neq 0$. i.e. if $\displaystyle y=5$ there no z such that $\displaystyle 0*z = 5$.

And second statement is false like you have stated.

3. Okay, Let me make sure that I understand.

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

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.

Is that sound about right?

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

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

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.

Is that sound about right?
Correct