Proof by contradiction. Statement negation This should be an easy question. Yet, the provided solution confuses me.
The question comes from "Understanding analysis" by S. Abbot, 2nd edition (Exercise 1.2.11). 

Negate the statement. Make an intuitive guess as to whether the claim or its negation is the true statement.

(b) There exists a real number $x > 0$ such that $x < 1/n\;\;\forall n \in \mathbb{N}$.


The provided solution says:


The solution seems correct, apart from: shouldn't the negation be with $\exists n \in \mathbb{N}$, i.e.:
$$\forall x >0 \;\; \exists n \in \mathbb{N}: x \geq 1/n$$
?
 A: Regarding Ex.1.2.11 (b) :

Form the logical negation of [...] there exists a real number $x > 0$ such that  $x < \dfrac 1 n$ for all $n \in \mathbb N$,

the formula to be negated is :

$\exists x > 0 \ \forall n \in \mathbb N \ (x < \dfrac 1 n)$.

Thus, tou are right. The correct negation will be :


$\forall x > 0 \ \exists n \in \mathbb N \ (x \ge \dfrac 1 n)$.



But in the solution provided, the author exhibits a proof of the statement; from this, we have to assume that the formula above is not what the author alludes to.
We have instead to transalte the semi-formal statement with :

$\forall n \in \mathbb N  \ \exists x > 0 \ (x < \dfrac 1 n)$

which is true.
Its negation will then be : $\exists n \in \mathbb N  \ \forall x > 0 \ (x \ge \dfrac 1 n)$.
A: This is why putting quantifiers at the end of a formula is a bad practice. It creates ambiguity. The statements  


*

*$(\forall n \in \mathbb{N})(\exists x > 0)(x < \frac{1}{n})$

*$(\exists x > 0)(\forall n \in \mathbb{N})(x < \frac{1}{n})$
are not equivalent. The second one is obviously false, however it's more likely to interpret your formulation as the second statement. Undoubtedly, the first statement is what's actually meant. For proving by contradiction, we need its negation which goes as follows:
$$(\exists n \in \mathbb{N})(\forall x > 0)\left(x \ge \frac{1}{n}\right)$$
A: 

There exists a real number $x > 0$ such that $x < 1/n\;\;\forall n \in \mathbb{N}$.


Is this statement really valid? Let's check.
$nx<1 \;\;\forall n\in\mathbb{N}$ is valid only for $x\leq 0$.
(Because $n$ becomes very large, and if $x\gt 0$ then $nx$ diverges to infinity)
So there is no such $x$ exist. 
