# Is it actually incorrect to say $x/1 = x$?

The rational numbers $\mathbb{Q}$ are defined as the field of quotients of $\mathbb{Z}$ under the relation $(a, b) \sim (c , d) \iff$ $ad = bc$. There is an obvious isomorphism between the subring $\{[(a, 1)] : a \in \mathbb{Z}$} and $\mathbb{Z}$ . So technically we only pair every integer $a$ with $[(a, 1)]$. They're not equal though. Am I just making a big deal out of nothing?

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math.stackexchange.com/questions/14828/… related and probably a duplicate. – user17762 Jan 16 '13 at 7:08

You are absolutely correct. However, this embedding $\iota\colon\mathbb Z\to \mathbb Q$ is canonical and it is customary to view it as the inclusion. Note that the same holds for $\mathbb N\to \mathbb Z$, $\mathbb Q\to\mathbb R$ and $\mathbb R\to \mathbb C$. However, once you have constructed either of these number sets from the one below, you are hardly interested in the ugly construction below the surface. You can either replace $\mathbb Q$ with $(\mathbb Q\setminus \iota(\mathbb Z))\cup \mathbb Z$ or simply demand that e.g. $\mathbb Q$ is an arbitrary field that is a superset of $\mathbb Z$ and has no proper subfield (and the explicit construction shows the existence of such a field)
There are many instances where an equivalence class is not denoted as such. For example, elements of the Hilbert space $\mathcal{H} := {L^{2}}(X,\Sigma,\mu)$ are equivalence classes of square-integrable functions, where two functions are said to be equivalent if and only if they differ on a $\mu$-null subset of $X$. However, one rarely denotes an element of $\mathcal{H}$ by $[f]_{\sim}$. Indeed, one simply picks a function $f$ that represents a given equivalence class and pretends that it is the class itself.
For your problem, the choice of $x$ as a representative of $[(x,1)]$ is canonical, so although the two objects are not exactly equal from the set-theoretic point of view, the usage of such a shorthand should not cause much confusion. In fact, we have been doing arithmetic this way since elementary school without much trouble! :)