order of multiple quantifiers Problem: For a, b, c, d restricted to the universe of positive integers, explain why
∀a ∃b ∀c ∃d   a/b = c/d   is true, but ∀a ∃d ∀c ∃b a/b = c/d    is false. 
I understand that the order of quantifiers is significant when there are 2 quantifiers. Here there are 4 and I'm unable to explain why the latter is false. 
Is it perhaps that in the former, we are able to choose the proportion (a/b) then duplicate it on the RHS with (c/d)? 
 A: 
$\forall a\exists b\forall c\exists d : a/b=c/d$

Proof: Let $a$ any integer. Consider $b=a$. Then, for any $c$, there exists $d$ (namely, $d=c$), such that $a/b=c/d$.

$\forall a\exists d\forall c\exists b : a/b=c/d$ (false)

Counterexample: Let $a=1$. Then for any $d$ there exists some $c$ (for example, $c=2d$) such that the equality $a/b=c/d$ is impossible, since it is actually $1/b=2$.
A: Let's just take it slow and re-write them in English. The first statement:

For all positive integers $a$, there exists a positive integer $b$, such that for all positive integers $c$, there exists a positive integer $d$, such that $$\frac{a}{b}=\frac{c}{d} \implies ad = bc$$

First we are given an integer $a$. Given $a$, we want to see if we can choose a $b$ such that the condition holds, because if we can choose such a $b$, such a $b$ must exist.
Take $b=a$. Then for any positive integer $c$, there also exists an integer $d$, namely $c$ itself, such that equation is satisfied. $$ac=ac$$ Thus the first statement is true.
The second statement:

For all positive integers $a$, there exists a positive integer $d$, such that for all positive integers $c$, there exists a positive integer $b$, such that $$\frac{a}{b}=\frac{c}{d} \implies ad = bc$$

First we are given an $a$. In order to show that the second statement is false, we need to show that no choice of $d$ will work to make the statement true. I.e. the statement must be false for every $d$. So let's show this. At this point, we are now assuming that $a$ and $d$ are given. Now the next part of the statement gives us a $c$ (i.e. we are not free to choose), so now the statement we need to show is always false is: 

Given positive integers $a,d,$ and $c$, there (always) exists a positive integers $b$ such that $$ad=bc \implies b=\frac{ad}{c}$$

Now we can clearly see why this is false; not every fraction of the form $\frac{ad}{c}$ reduces to a positive integer, hence cannot equal any choice of positive integer $b$. Therefore there does not always exist a positive integer $b$ such that the condition holds, and thus the statement is false (since it says that there should always exist such a $b$).
I hope this helps.
