Integral domain is unique factorization domain if every non zero , non unit element can be written uniquely as finite product of irreducible elements . why in this definition non unit element is required?
what happens if i will take unit element ?
eg. in $\mathbb{Q}$ , 6 is unit he nce $\lt6\gt=\mathbb{Q}$and we can write 6=2.3 where 2 and 3 are both unit in $\mathbb{Q}$ so what i can conclude? can i say 6 is irreducible? that is every unit element is irreducible?
Because you lose the uniqueness property if you also take in units into the definition. Say $a,b,c,d$ in some ring $R$ are units such that $a=bc$. But you can also write $a=dbcd^{-1}$, then $e=db$ and $f=cd^{-1}$ are units again. All in all we would have $a=bc=ef$, and none of the factorisations are more "right".
In your example $6=2*3$, but also $6=\frac{5}{1}\frac{6}{5}$. You have to distinct here between $6$ as an element in the integral numbers and as an element in the rational numbers. In the former $6$ is not a unit and you have a unique factorisation, but in the latter $6$ is a unit and you would have many decompositions technically.
The rational numbers are a fraction field of a UFD, that is why you could argue that such decomposition would work and which might lead to some confusion, but in general you can't assume this.