Approximation by default, with significant digits I have a statement that says:

In relation with the number $-\frac{23}{7}$, what of the following
  statements are correct ?
$I)$ When writing it with three significant digits we obtain an
  approximation by default.
$II)$ When truncating to the thousandth we obtain an approximation by
  excess.
$III)$ The difference between, truncate it to the unit and write it with
  a one significant digit It is zero.

My development was:
$I)$ first, the fraction in decimal is equal to $-3.28571429$, and with 3 significant digits is $-3.28$ Also, when we talk about significant figures, there are 2 types of approximation, by excess and by default. It is by excess when the approximate number is greater than the original. It is by default when the approximate number is less than the original.
$ -3.28 $ is greater than $ -3.28571429 $, therefore it is false.
$II)$ When truncating to the thousandth, I get:$ -3.285$  that is greather than the original value, therefore is true
$III)$ If i truncate it to the unit: $-3$ and if i write it with one significant digit is $-3$, so $-3 + 3 = 0$, therefore is true
so, i answered $II)$ and $III)$ are correct, but here the problem, since all were correct. What am I wrong about? 
 A: There is a difference between making a number significant and truncating a number. Significant means a rounded number (instead of just dropping the $n$-th decimal you turn the previous decimal to the "integer" closed to the previous decimal part: with "integer" I mean considering it as an integer within decimals), truncated means dropped digits (just throwing them away without rounding them).
What you did in $II)$ is correct: you truncated $−3.28571429...$ by dropping $...571429$.
In $I)$ you should round $-3.28571429...$ at the number $8$, which means that $...8571429$ should become $...9$. In "integer" style: $8,571429...$ lies closest to $9$ (minus sign doesn't change anything due to symmetry around $0$ in the real numbers). This means that $-3.28571429...$ becomes $-3.29$ and we know that $-3.29<-3.28571429...$, so it is indeed an approximation by default.
In $III)$ you happen to have a correct answer, since both truncating and making the number significant yield the same result.

In conclusion: truncating is different from making a number significant.
