Let $R$ be a subring of $\mathbb{Q}$ containing $1$. We denote the absolute value of an integer $n$ by $|n|$.
Let $I$ denote a non-zero ideal of $R$. Let $0\ne b,a\in \mathbb{Z}$ be such that $\frac{a}{b}\in I$ and $|a| = \min \{ |c|: 0\ne \frac{c}{d} \in I\}$. Prove that $I$ is a principal ideal generated by $\frac{a}{b}$
My question is, is this a professional question? does it make sense? I think the norm is defined in a confusing way right?
There is nothing wrong with the norm. As already mentioned in the comments, it is not a definition of a norm but it is a condition on $a$. It just says that $a$ is an element with a particular property. Let me try to clarify. Let $X$ be the set of all $c \in \mathbb{Z}$ with $c \neq 0$ such that there exists a $d \in \mathbb{Z}$ with $\frac{c}{d} \in I$. In more formal terms: $$X:=\{c \in \mathbb{Z}:c\neq 0,\exists d \in \mathbb{Z}:\frac{c}{d} \in I\}$$ Now consider the set $$Y:=\{|c| : c \in X\}$$ where $|c|$ is the absolute value of $c$. Note that $Y$ is a non-empty set of natural numbers and therefore contains a smallest element. Let $n$ be the smallest element of $Y$. Now, the condition $$\frac{a}{b} \in I\phantom{aa}\text{ and }\phantom{aa}|a|=\text{min}\{|c|:0\neq \frac{c}{d} \in I\}$$ has the following meaning: $a$ is an element from $X$ with $|a|=n$. In short: $a$ is a non-zero integer for which there exists an integer $b$ with $\frac{a}{b} \in I$ and moreover, all other integers with this property have absolute value $\geq |a|$.