Stuck in determining truth value using proof I need your advice/help regarding these questions.
Q. Let p(x,y) denotes the predicate x divides y. Determine truth value of each statement and give an example or a counter-example for the statements. 
1) ∀x ∃y p(x,y)
A. I guess it is true, but I'm basically having difficulty in understanding how is it true?   
My logic says that for example if we suppose x = 3 and y = 15 then y =15 can be divided by this x =3 and also x = 5 so it is true. But confusion is how this covers every x (How every x can divide y?) 
2) ∀y ∃x p(x,y)
This looks false, but how? 
3) ∃x ∀y p(x,y)
A. ?  
4) ∃y ∀x p(x,y)
A.
?  
I would appreciate if someone helps me in these questions.  I'm basically struggling the in the concept of For all x and for some y here in such questions.   
EDIT: For Q.4 Answer:
(0 | 0 and 1 | 0 and 2 | 0 and 3 | 0 and 4 | 0 .....)
     ----------------OR----------------------------
(0 | 1 and 1 | 1 and 2 | 1 and 3 | 1 and 4 | 1 .....)
--------------------OR---------------------------
(0 | 2 and 1 | 2 and 2 | 2 and 3 | 2 and 4 | 2 .....)  
So, First Row is False as 0 | 0 is false and similarly second and third row is also false so it makes the whole statement false.
 A: I think what you need most is a translation (from math into English). 
1) For each given $x$ there exists a $y$ such that $x|y$ (try to find one). 
2) For each given $y$ there exists an $x$ such that $x|y$ (try to find one). 
3) There exists a value of $x$ that satisfies $x|y$ for each given $y$. 
4) There exists a value of $y$ that satisfies $x|y$ for each given $x$. 
A: $$ \newcommand{\and} {~\text{and}~}
 \newcommand{\or} {~~\text{or}~~}
 \newcommand{\undef} {~~\text{undefined}~~}$$
Write the statements without quantifiers.  I'll be using $x|y$ to indicate that $x$ divides $y$, that is, that $y$ is divisible by $x$:

1) $$\forall x \exists y ~ p(x,y)  $$

$$\begin{array} {c}
(0|0 \or 0|1 \or 0|2 \or 0|3 \dots )\\
\and \\
(1|0 \or 1|1 \or 1|2 \or 1|3 \dots )\\
\and \\
(2|0 \or 2|1 \or 2|2 \or 2|3 \dots )\\
\and \\
\vdots \\
\end{array}$$
Division by zero is undefined.  Otherwise, each horizontal line has an expression of the form $n|n$ (the second line has $1|1$, the third has $2|2$, etc).  So the expression is:
$$\begin{array} {c}
(\undef \or \undef \or \undef \or \undef \dots )\\
\and \\
(1|0 \or \top \or 1|2 \or 1|3 \dots )\\
\and \\
(2|0 \or 2|1 \or \top \or 2|3 \dots )\\
\and \\
\vdots \\
\end{array}$$
which is
$$\undef \and \top \and \top \dots$$
So the entire expression is undefined.

2) $$\forall y \exists x ~ p(x,y)  $$

$$\begin{array} {c}
(0|0 \or 1|0 \or 2|0 \or 3|0 \dots )\\
\and \\
(0|1 \or 1|1 \or 2|1 \or 3|1 \dots )\\
\and \\
(0|2 \or 1|2 \or 2|2 \or 3|2 \dots )\\
\and \\
\vdots \\
\end{array}$$
Although division by zero is undefined, zero is divisible by everything (as $0/x$ is always an integer, $0$).  Otherwise everything is divisible by itself:
$$\begin{array} {c}
(\undef \or \top \or \top \or \top \dots )\\
\and \\
(\undef \or \top \or 2|1 \or 3|1 \dots )\\
\and \\
(\undef \or 1|2 \or \top \or 3|2 \dots )\\
\and \\
\vdots \\
\end{array}$$
which is:
$$\top \land \top \land \top \dots$$
which is true.

3) $$\exists x \forall y ~ p(x,y)  $$

$$\begin{array} {c}
(0|0 \and 0|1 \and 0|2 \and 0|3 \and 0|4 \dots )\\
\or \\
(1|0 \and 1|1 \and 1|2 \and 1|3 \and 1|4 \dots )\\
\or \\
(2|0 \and 2|1 \and 2|2 \and 2|3 \and 2|4 \dots )\\
\or \\
(3|0 \and 3|1 \and 3|2 \and 3|3 \and 3|4 \dots )\\
\or \\
\vdots \\
\end{array}$$
This time we look at how $n$ never divides $n + 1$, except when $n=1$, because $1$ divides everything:
$$\begin{array} {c}
(\undef \and \undef \and \undef \and \undef \and \undef \dots )\\
\or \\
(\top \and \top \and \top \and \top \and \top \dots )\\
\or \\
(2|0 \and 2|1 \and 2|2 \and \bot \and 2|4 \dots )\\
\or \\
(3|0 \and 3|1 \and 3|2 \and 3|3 \and \bot \dots )\\
\or \\
\vdots \\
\end{array}$$
Which is:
$$\undef \or \top \or \bot \or \bot \dots$$
which is true.  That one is pretty complicated.

4) $$\exists y \forall x ~ p(x,y)  $$

Can you do this one?
