Proving the Set of Rational Numbers is the collection of equivalence classes of ratios of integers with...












-1














I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!



Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}

We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}



Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.










share|cite|improve this question









New contributor




UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
    – YiFan
    Jan 4 at 6:15
















-1














I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!



Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}

We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}



Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.










share|cite|improve this question









New contributor




UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
    – YiFan
    Jan 4 at 6:15














-1












-1








-1







I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!



Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}

We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}



Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.










share|cite|improve this question









New contributor




UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I read in a textbook that the set of Rational Numbers is the collection of equivalence classes of ratios of integers with nonzero denominators. I found it interesting and tried to prove it but ended coming up with a disproof. I'm just curious as to where my logic falls through!



Assume $mathbb{Q}$ is the union of disjoint equivalence classes over ratios of integers.
Let the set of rational numbers be defined as such.
begin{equation}
mathbb{Q} = bigg{ frac{p}{q}:p,q in mathbb{Z} , q neq 0 bigg}
end{equation}

We may equivalently define the following as a definition of the rationals.
begin{equation}
mathbb{Q} = bigg{ frac{a}{b}:(a,b) in mathbb{Z} times mathbb{Z} setminus { 0 } bigg}
end{equation}



Contradiction. Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible). Therefore $mathbb{Q}$ cannot be the union of disjoint equivalence classes over ratios of integers.







equivalence-relations rational-numbers






share|cite|improve this question









New contributor




UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 8:58









Asaf Karagila

302k32426757




302k32426757






New contributor




UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked Jan 4 at 4:32









UmamiBoyUmamiBoy

11




11




New contributor




UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






UmamiBoy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
    – YiFan
    Jan 4 at 6:15


















  • What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
    – YiFan
    Jan 4 at 6:15
















What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
– YiFan
Jan 4 at 6:15




What? Can you come up with a pair $(a,b)inmathbb Ztimesmathbb Zsetminus{0}$ where equivalence classes cannot be defined in the usual sense?
– YiFan
Jan 4 at 6:15










1 Answer
1






active

oldest

votes


















4















Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).




Says who?



You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.



Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.



More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.






share|cite|improve this answer





















  • Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
    – UmamiBoy
    Jan 4 at 5:09











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});






UmamiBoy is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061323%2fproving-the-set-of-rational-numbers-is-the-collection-of-equivalence-classes-of%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4















Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).




Says who?



You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.



Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.



More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.






share|cite|improve this answer





















  • Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
    – UmamiBoy
    Jan 4 at 5:09
















4















Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).




Says who?



You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.



Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.



More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.






share|cite|improve this answer





















  • Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
    – UmamiBoy
    Jan 4 at 5:09














4












4








4







Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).




Says who?



You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.



Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.



More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.






share|cite|improve this answer













Equivalence classes are not possible on Cartesian products whose sets are not equal (as, at the every least, reflexivity is not possible).




Says who?



You seem to be confusing the fact that an equivalence relation on $A$ is a certain subset of $Atimes A$ with the fact that in this case $A$ itself is a cartesian product.



Here the equivalence relation is a certain subset of
$$ (mathbb Ztimes(mathbb Zsetminus{0}))times(mathbb Ztimes(mathbb Zsetminus{0})) $$
and there's nothing that prevents such a relation from being reflexive.



More precisely, the relation is
$$ { ((a,b),(p,q)) mid aq=pb } $$
which is reflexive because $((a,b),(a,b))$ is in the relation for every $ainmathbb Z$, $binmathbb Zsetminus{0}$, because $ab=ab$ is always true.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 4:35









Henning MakholmHenning Makholm

238k16303540




238k16303540












  • Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
    – UmamiBoy
    Jan 4 at 5:09


















  • Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
    – UmamiBoy
    Jan 4 at 5:09
















Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
– UmamiBoy
Jan 4 at 5:09




Thank you, Having A as a cartesian product itself fixes my problem. Much appreciated.
– UmamiBoy
Jan 4 at 5:09










UmamiBoy is a new contributor. Be nice, and check out our Code of Conduct.










draft saved

draft discarded


















UmamiBoy is a new contributor. Be nice, and check out our Code of Conduct.













UmamiBoy is a new contributor. Be nice, and check out our Code of Conduct.












UmamiBoy is a new contributor. Be nice, and check out our Code of Conduct.
















Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061323%2fproving-the-set-of-rational-numbers-is-the-collection-of-equivalence-classes-of%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

1300-talet

1300-talet

Display a custom attribute below product name in the front-end Magento 1.9.3.8