# Another Analysis statement

• Jan 28th 2009, 05:58 PM
Skinner
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?
• Jan 28th 2009, 06:03 PM
chabmgph
Quote:

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.
• Jan 28th 2009, 06:26 PM
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.

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

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.