Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
math.stackexchange.com/questions/14828/… related and probably a duplicate. –  user17762 Jan 16 '13 at 7:08

2 Answers 2

up vote 14 down vote accepted

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)

share|improve this answer
    
Note that the harmless looking work "canonical" has a very important (and impressive) impact. –  Hagen von Eitzen Jan 16 '13 at 15:59

I would like to mention a point that contributes slightly to Hagen's beautiful answer.

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! :)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.