Prove a real number must be either positive, negative, or zero I came up with a few cases about the components of the corresponding Cauchy sequence of a real x, like having all positive rationals, but got stuck to prove with a sequence having mixing positive and negative rationals, but all the negative rationals are in first n positions, and all positive rationals are in the rest. 
How do you prove that? and how do you prove if using my approach?
Update: Limit should not be used here, because in the analysis book I am reading, the section having such question does not introduce about the definition of limit yet, but rather a placeholder notion LIM, such that a real is equal to LIM of a Cauchy sequence, without further say how LIM works. The books continue to introduce basic operations of reals like sum and product using such notion. 
The book I am reading is Analysis I of Tao, FYI.
 A: You can prove that such a sequence with a mix of positive and negative numbers has an equivalent Cauchy Sequence which is positively(or negatively) bounded away from 0.

but all the negative rationals are in first n positions, and all positive rationals are in the rest. 

Assuming the positive portion is a Cauchy sequence, you can set all the negative portion to some $\delta \gt 0$ since we have some value of $\delta$ such that $|a_i|>\delta$ for all $i>=1$ (This stems from x being a non-zero real number, which is hence not equivalent to the zero-sequence). You now have a positive Cauchy sequence bounded away from zero, which is equivalent to the example described in the question. And hence a positive real number.
You can use equivalent sequences to prove the whole statement. An outline of the whole proof would go like this: 
You can prove that utmost one of the 3 statements is true for a real number $x = LIM_{n \to \inf} a_n$. No two sequences (let alone Cauchy sequences) can be both positively and negatively bound from zero. So we cannot simultaneously have $x \gt 0$ and $x \lt 0$. A positively bound sequence is bounded away from zero. So $x \gt 0$ and $x = 0$ doesn't hold simultaneously. The same holds for $x \lt 0$ and $x = 0$. 
We now prove that at least one of the three statements is true. Since real numbers are well defined, x's sequence $a_n$ is either equivalent to 0-sequence or not. In the former case, we have $x = 0$ (Case 1). In the latter case, since $a_n$ is not equivalent to 0-sequence, there exists a positive rational $\delta$ such that: $|a_i - 0| = |a_i| > \delta$ for all $i \geq 1$. Since $x$ is a real number, the sequence $a_n$ is Cauchy. So it is $\frac{\delta}{2}$-steady for some $N$. But since there is some $i$ such that $|a_i|>\delta$, we have: 
$$ \delta \lt |a_i| \\ = |a_i - 0| \\ \leq |a_i - a_j| + |a_j - 0| \\ \leq \frac{\delta}{2} + |a_j| $$Thus, for a fixed $a_i \gt \delta$, we have $a_j \gt \frac{\delta}{2}$ for all $i, j \geq N$. For a fixed $i$ and any $j$, we are now left with four cases:


*

*$a_i \gt \delta \gt 0$.


*

*If $a_j \gt \frac{\delta}{2} \gt 0$: In this case, for all $n \geq N$,
$a_n \gt 0$. A sequence where $b_n := \delta$ for all $n \lt N$ and
$b_n := a_n$ for all $n \geq N$ is positively bounded away from
zero, and is equivalent to the original sequence $a_n$, making $x =
   LIM_{n \to \inf} a_n$ a positive value. (Case 2)

*If $a_j \leq -\frac{\delta}{2} \lt 0$: Then $|a_i - a_j| \gt \frac{\delta}{2}$ which is a contradiction. 


*$a_i \lt -\delta \lt 0$ and $a_j \lt -\frac{\delta}{2} \lt 0$. We can generate an equivalent sequence negatively bounded away from zero, thus making x a negative value (Case 3). When $a_i \lt 0$, $a_j$ cannot be positive, as it would contradict $\frac{\delta}{2}$-steadiness.


None of $a_n$ can be zero since $x$ is a non-zero real number.  
