The set of rational numbers is defined as $\mathbb{Q} = \left\lbrace \frac{a}{b} \mid a, b \in \mathbb{Z} \land b \neq 0 \right\rbrace$.

This apparently means that $\frac{1}{2}$ and $\frac{2}{4}$ are distinct two elements of the set $\mathbb{Q}$. And similarly, every $\frac{0}{n}$ for all $n \in \mathbb{Z} \setminus \left\lbrace 0 \right\rbrace$ are also distinct elements of $\mathbb{Q}$. Is this right?

And if that is right, for a function $f : \mathbb{Z} \to \mathbb{Q}$ to be a bijection, there has to exists distinct $n, m \in \mathbb{Z}$ which satisfy $f(n) = \frac{0}{1}$ and $f(m) = \frac{0}{2}$, and of course $n \neq m$. Is this right as well?

  • 5
    $\begingroup$ The rational numbers are formed from equivalence classes of pairs where $\frac{a}{b} = \frac{c}{d}$ if $ad - bc = 0$. So $\frac{1}{2}$ and $\frac{2}{4}$ are the same element of $\mathbb{Q}$. $\endgroup$
    – ndruiven
    Commented Dec 7, 2014 at 17:47
  • 1
    $\begingroup$ Hi, no it's not the case (although it's quite fair to be confused by the set notation above). A more complete description: it is the set of pairs of integers (a,b) (excepting those where b=0) quotiented by the equivalence relation that (a,b) ~ (c,d) iff ad = bc. $\endgroup$
    – amomin
    Commented Dec 7, 2014 at 17:49
  • $\begingroup$ The set of rational numbers is defined as $\displaystyle \mathbb{Q}=\{\frac{a}{b}| a,b \in \mathbb{Z}, b\neq 0, $ gcd $\{a,b\}=1\}$. So we cannot have $\frac{1}{2}$ and $\frac{2}{4}$ as two distinct elements in $\mathbb{Q}$, it is just $\frac{1}{2}$. $\endgroup$
    – Extremal
    Commented Dec 7, 2014 at 17:54
  • $\begingroup$ @Mathi: "is defined": can be defined. This is not the only definition (indeed, it is not the usual one). $\endgroup$
    – TonyK
    Commented Dec 7, 2014 at 17:57
  • $\begingroup$ Hello everyone. I have never thought that I would receive a "no" for the first part of my question. I had been told by our instructor that equivalent elements $\frac{1}{2}$ and $\frac{2}{4}$ of $\mathbb{Q}$ are indeed distinct, and merely just equivalent. If you insist that they aren't, then I have one more to ask: For $f : \mathbb{Z} \to \mathbb{Q}$ to be a bijection, should it be the case that $m = n$, given $f(m) = \frac{1}{2}$ and $f(n) = \frac{2}{4}$? $\endgroup$ Commented Dec 7, 2014 at 18:00

1 Answer 1


By virtue of the fact that the definition is not in terms of ordered pairs but rather in terms of fractions, all of the rules of fractions apply. In particular, for any $a\in\mathbb{Z}$, $b\in\mathbb{Z}-\{0\}$, and $c\in\mathbb{Z}-\{0\}$ we have $$\frac{ac}{bc}=\frac{a}{b}$$ and, as mentioned in the comments, $$\frac{a}{b}=\frac{c}{d}$$ if and only if $ad-bc=0$.


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