Prove that a language is in co-NP
Languages are defines with respect to a finite alphabet. Thus, integers cannot be represents just by giving a triplet as you suggest, you need to encode it.

For convenience, let's assume a decimal encoding, with the "$-$" sign for negation, and the character $\#$ separating the numbers. Thus, for example, the word $8\#12\#-3$ represents the triplet $(8,12,-3)$ (note that usually we would use a binary encoding, but it doesn't really matter). Call our alphabet $\Sigma$.

Now, a word $w$ is in the language you describe iff the following hold:


$w=a\#b\#c$ where $a,b,c$ are either words in {0,...,9}^$\setminus {0}^. That is, they represent positive numbers (we allow leading 0's here, but we could also formulate without them).
There does not exist $x\in \mathbb{Z}_+$ such that $x&lt;a$.


Note that we did not do anything yet, just wrote down the language. In particular, we purposely ignore the fact that condition 2 simply means that $a\le 1$.

The formal complement of your language is the set of words that either don't satisfy condition 1, or don't satisfy condition 2 (or both). 

You could write this more succinctly or more explicitly, as words that are either not of the form $a\#b\#c$ with the condition 1 above, or words that are of that form, but where $a$ is not of the form $0^*1$.

It should be mentioned that once you are familiar with the notion of languages, we usually don't dig down to this level of formality, and in fact we would ignore the encoding somewhat. Then, you might see things such as 
$\overline{L}=\{(a,b,c)\in \mathbb{Z}: \exists x\in \mathbb{Z}_+ \wedge x&lt;a\}$.

Also note that throughout this answer we do not distinguish between a word and its numeric value, but if you really want to nitpick, then such a distinction should be made.