How is it possible to have a transitive yet not complete relation? How can we say we cannot compare between pairs (no completeness) and yet we can have transitivity?
Let's assume the relation is a preference relation.
For instance if my set has: $(a, b, c)$ How can I say that a is preferred to b, b is preferred to c and hence a is preferred to c (transitivity) if the relation is not complete and thus I cannot compare a to b and b to c?
 A: Transitivity says that if $\langle a,b\rangle\in R$ and $\langle b,c\rangle\in R$, then $\langle a,c\rangle$ must also be in $R$. That’s a conditional statement; when the condition isn’t met, it doesn’t apply and therefore imposes no requirement on $R$.
Suppose that $R$ is a relation on a set $A$. You can think of the elements of $A$ as stepping stones in a river; $\langle a,b\rangle\in R$ means that you can step from $a$ to $b$, while $\langle a,b\rangle\notin R$ means that you cannot. $R$ is transitive if it has the following property: 

if you can step from $a$ to $b$ and from $b$ to $c$, then you can step directly from $a$ to $c$.

If you can’t step directly from $a$ to $b$ and from $b$ to $c$, this property tells you nothing about the possibility of stepping directly from $a$ to $c$: maybe you can, and maybe you can’t, and either possibility is consistent with $R$ being transitive.
Suppose, for instance, that the underlying set is $A=\{a,b,c,d,e\}$, and the relation $R$ contains the pairs $\langle a,b\rangle,\langle a,c\rangle,\langle c,d\rangle$, and $\langle a,d\rangle$. You can step from $a$ to $b,c$, or $d$, and you can step from $c$ to $d$. The only two steps that link up are the ones from $a$ to $c$ and from $c$ to $d$: $\langle a,c\rangle\in R$ and $\langle c,d\rangle\in R$. 
Is there a shortcut step directly from $a$ to $d$? Yes: $\langle a,d\rangle\in R$. Therefore $R$ is transitive. There’s only one pair of ‘steps’ for which transitivity of $R$ actually requires a shortcut, and the shortcut is present. The fact that $R$ doesn’t even mention $e$ is irrelevant.
A: EDIT: This question is about microeconomic theory, not set theory. Directly following is an answer related to microeconomics. Further below is a set-theoretic answer, which I'm going to leave there in case someone searches for similar keywords.
Microeconomic Answer:
In economic preference, we say that a relation is complete if and only if for all $A,B$, we have $A \succsim B$ or $B \succsim A$ or both. So consider $A,B,C$ such that $A\succsim B$ and $B \succsim C$. Then the relation is not complete because we do not have an explicit relationship $A \succsim C$. However, we can infer one by transitivity: if $A\succsim B$ and $B \succsim C$, then by transitivity $A \succsim C$, though technically the relation is not `complete'.

Set-Theoretic Answer:
I don't understand your terms (what do you mean with completeness?). So I am somewhat guessing at what your question actually asks. Please comment with clarifications so I can provide a better answer.
A relation $R$ on a set $X$ is a subset of $X \times X$.
An equivalence relation on $X$ is a relation $R$ on $X$ such that:


*

*Reflexivity: For all $a \in X, (a,a) \in R$

*Symmetry: For all $a,b \in X$, if $(a,b) \in R$, then $(b,a) \in R$.

*Transitivity: For all $a,b,c\in X$, if $(a,b) \in R$ and $(b,c) \in R$, then $(a,c) \in R$.


Your notion of completeness seems to be a restatement of trichotomy: in an ordered set, one of the three is true: $a=b$, or $a>b$, or $a<b$.
Notably, we can have transitivity in terms of an equivalence relation on a set, while the set itself is unordered and not transitive. Is this what you mean?
A: For a great example of a relation that is not complete but is transitive, consider the relation of whether a shape fits into a another shape.  You can imagine you have three shapes, one a large triangle, one a smaller triangle that fits into the larger triangle and a circle that fits into the smallest triangle, it's easy to see the smallest circle will fit in the largest triangle.  On the other hand if you had a larger circle, we can imagine the circle not fitting into the larger triangle, but the smaller triangle still fits in both.
Hence the relation is not complete (because we don't have the relation between every element), but transitivity is satisfied.
You can find more from these lecture notes by Joseph Guse. 
