Quantified statement For universal quantified , we just need to find a counter example to prove the statement.Isn't it ?So how about existential quantified?Isn't we just have to find at least one is true and proof it in general? Im quite confused with both of them on how to proof it.
 A: If we have a universally quantified statement such as $\forall x.P(x)$, then we only need to find a single counterexample (i.e. some $x$ where $P(x)$ does not hold) to disprove the statement. However, in order to prove the statement, we have to show that $P(x)$ holds for any choice of $x$.
Conversely, for a statement with an existential quantifier such as $\exists x.Q(x)$, we only have to find a single witness $x$ where $Q(x)$ holds in order to prove the statement. However, to disprove the statement, we would need to show that $Q(x)$ does not hold for any $x$.

In fact, this relationship between proving and disproving is nothing more than the basic laws
$$\neg \exists x.P(x) \equiv \forall x.\neg P(x)$$
$$\neg \forall x.Q(x) \equiv \exists x.\neg P(x)$$

As an example, let $P(a,b)$ be $a/b < 1$, $A = \{2,3,5\}$, and $B = \{2,4,6\}$. Now, to prove the statement
$$\forall a \in A .\exists b \in B . P(a,b)$$
we would have to go through every $a \in A$ and for each of them pick a $b \in B$ such that $a/b < 1$. In order to disprove it, we would have to pick an $a \in A$ and go through every $b \in B$ and show that $a/b < 1$ does not hold.
In your comment you say let $a = 2$ and $b = 2$, then $2/2 \not < 1$. Sure, this is true, but in order to disprove the statement, we would have to have $a/b \not < 1$ for every choice of $b\in B$, you have merely shown it for a single $b \in B$. Indeed, if we take $b = 4$, then we get $2/4 < 1$, so we have not actually disproved it. Let's try to prove it instead. We go through the $a$'s one by one.
If $a = 2$, then we can pick $b = 4$ and we get $2/4 < 1$.
If $a = 3$, then we can pick $b = 6$ and we get $3/6 < 1$.
If $a = 5$, then we can pick $b = 6$ and we get $5/6 < 1$.
So we have gone through every $a \in A$ and shown that there exists for each of these $a$ a $b \in B$ such that $P(a,b)$. Hence we have proved that $\forall a \in A.\exists b \in B. P(a,b)$.
