Why does switching the quantifiers make the statement false?

Question

According to my teacher,

• for every $y$ there exists an $x$, such as $y=0$ OR $xy>0$

is a true statement, while

• there exists an $x$ for each $y$, such as $y=0$ OR $xy>0$

is a false statement.

Why is the second statement false?

Answer 1

The order in which things are written may make a BIG difference.

The statement: for every $y$ there exists an $x$, such that $y=0$ OR $xy>0$. $$(\forall y\in \mathbb R)(\exists x\in \mathbb R)\quad(y=0)\vee(xy>0)$$ is true.

The statement: there exists an $x$ for each $y$, such that $y=0$ OR $xy>0$. $$(\exists x\in \mathbb R)(\forall y\in \mathbb R)\quad(y=0)\vee(xy>0)$$ is false. It says that there exists a fixed real number $x$ which simultaneously makes $xy>0$ for every $y\ne 0$.

Answer 2

"For all x in a Set A, x have property P" is same as saying,

Given any x in A, x has property p

The famous analogue to understand the difference between change of quantifiers is knowing the difference between "Given any man x, there exists a woman, such that x,y are dating"

"There exists a woman such that given any man x, x and y are dating"

The first one means for a man there is a woman... While the second one means every man dating a single woman...

This analogue or example is made for understanding purpose only and for the OP to appreciate the difference happens while changing the quantifiers...

So for you doubt after you changed the quantifiers the statment is... There exists a x such that for each y, either y=0 or xy>0...

This is false suppose such an x exists then x can be positive or negative, and x cannot be zero... So for y, nonzero xy>0... But what about -y... It will be -xy<0 so we found that there exists a z=-y such that xz<0 and so your statement is false...