We can construct $\mathbb Q$ from $\mathbb Z$ using a quotient. Topologically, $\mathbb R$ is construct as the space where all Cauchy-sequence of $\mathbb Q$ converge (and we can also construct $\mathbb R$ in a different way). But is there a way to construct algebraically $\mathbb R$ from $\mathbb Q$ ?
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$\begingroup$ $\Bbb Q$ is not "constructed" from $\Bbb Z$ using a quotient, to be pedantic. $\endgroup$– Kenny LauCommented Sep 9, 2017 at 7:49
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1$\begingroup$ You define $(a,b)\sim (c,d)\iff ad-bc=0$ and then $\mathbb Q=\mathbb Z/_\sim$. @KennyLau $\endgroup$– SurbCommented Sep 9, 2017 at 7:50
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$\begingroup$ Yes, and there is no quotient involved. Also, $\Bbb Q = (\Bbb Z \times (\Bbb Z \setminus \{0\}))/\sim$. $\endgroup$– Kenny LauCommented Sep 9, 2017 at 7:51
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$\begingroup$ Did you mean quotient in the sense of a set quotient an equivalence relation? $\endgroup$– Kenny LauCommented Sep 9, 2017 at 7:52
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1$\begingroup$ @KennyLau: Yes of course. $\endgroup$– SurbCommented Sep 9, 2017 at 7:52
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1 Answer
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We can construct the reals directly from the integers by an algebraic process (it's the quotient of the ring of quasi-homomorphisms of $\mathbb Z$ by a certain equivalence relation). See this question and, above all, the comments and the answer.
answered Sep 9, 2017 at 7:53
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$\begingroup$ It's exactly what I was looking for. Thank you. $\endgroup$– SurbCommented Sep 9, 2017 at 7:54
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$\begingroup$ @Surb I'm glad I could help. $\endgroup$ Commented Sep 9, 2017 at 7:55
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$\begingroup$ @Surb consider accepting his answer if you considered it helpful. $\endgroup$ Commented Sep 9, 2017 at 8:01