The equivalence between the statements Statement 1: Suppose for sake of contradiction that there is no non-negative rational number $x$ for which $x^2 < 2 < (x+ \epsilon)^2 $ .
Statement 2: This means that whenever $x$ is non-negative and $x^2 < 2$, we must also have $(x + \epsilon)^2 < 2$.
How does statement 2 logically follows from statement 1? I am not able to understand that statement 1 and statement 2 are same.
(Reference: http://www.math.ucla.edu/~tao/resource/general/131ah.1.03w/week2.pdf, proposition 7, page no. 9)
 A: To prove logical entailment, you assume the first statement and then show the second. So assume that there is no $x$ so that $x^2 < 2 < (x+ \epsilon)^2 $ (1).
Now, let's say $x$ is non-negative and $x^2 < 2$, we want to show that $(x + \epsilon)^2 < 2$. Well consider $(x + \epsilon)^2$. By total ordering of numbers we have that $(x + \epsilon)^2 \leq 2$ or $2 < (x + \epsilon)^2$. The second case is immediately a contradiction with (1), so now we know that $(x + \epsilon)^2 \leq 2$. But this itself has two cases: $(x + \epsilon)^2 = 2$ and $(x + \epsilon)^2 < 2$. The first case can't be true, because the sum of two rationals is rational, and $\sqrt{2}$ is irrational. So we must have that $(x + \epsilon)^2 < 2$, as required.
A: 
Statement 1: Suppose for sake of contradiction that there is no non-negative rational number $x$  for which $x^2 <2<(x+ϵ)^2$
Statement 2: This means that whenever $x$  is non-negative and $x^2 <2$ , we must also have $(x+ϵ)^2 <2$ .
How does statement 2 logically follows from statement 1? I am not able to understand that statement 1 and statement 2 are same.

They are not (quite) the same, the implication includes the fact that $\sqrt 2$ is irrational, and that $\epsilon$ is rational.   By reason that if $x$ and $\epsilon$ are both rational, then their sum cannot equal a square root of $2$.
Statement 1 is: $\neg \exists x \Big(x\in\Bbb Q^+\cap\{0\} \wedge x^2< 2< (x+\epsilon)^2\Big)
\\ \Updownarrow \quad (\text{dual negation})
\\ \forall x \neg \Big(x\in\Bbb Q^+\cap\{0\} \wedge (x^2< 2) \wedge (2<(x+\epsilon)^2)\Big)
\\ \Updownarrow \quad (\text{DeMorgan's Law})
\\ \forall x \Big(x\notin\Bbb Q^+\cap\{0\} \vee \neg (x^2< 2) \vee \neg(2 < (x+\epsilon)^2)\Big)
\\ \Updownarrow\quad(\text{Implication Equivalence})
\\ \forall x \Big(\neg\big(x\notin\Bbb Q^+\cap\{0\} \vee \neg (x^2< 2)\big) \to \neg(2 < (x+\epsilon)^2)\Big)
\\ \Updownarrow\quad(\text{DeMorgan's Law})
\\ \forall x \Big(\big(x\in\Bbb Q^+\cap\{0\} \wedge (x^2< 2)\big) \to ((x+\epsilon)^2\leq 2)\Big)
\\ \Downarrow \quad (\epsilon\in\Bbb Q, \sqrt 2\notin\Bbb Q)
\\ \forall x \Big(\big(x\in\Bbb Q^+\cap\{0\} \wedge (x^2< 2)\big) \to ((x+\epsilon)^2 < 2)\Big)
$
Which is statement 2.
