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  "body": "\n## Proof\n\nWe use the [[construction from Cauchy sequences|0D8HE681SC]] and the [[order on the reals|0337CYB0VD]] defined there.\n\n1. A real number is a class $x = [(x_n)]$ of Cauchy sequences of rational numbers. Two sequences give the same real number exactly when their difference tends to $0$.\n2. The number $0$ is the class of the constant sequence $0$. Since $(x_n) + (-x_n)$ is the constant sequence $0$, we get $-x = [(-x_n)]$. So $y - x = [(y_n - x_n)]$.\n3. By definition, $x \u003c y$ means: there are a rational $\\varepsilon \u003e 0$ and an $N$ with $y_n - x_n \u003e \\varepsilon$ for all $n \\ge N$.\n4. This does not depend on the representatives. Let $(x'_n)$ and $(y'_n)$ be equivalent to $(x_n)$ and $(y_n)$. Choose $N' \\ge N$ with $\\abs{x'_n - x_n} \u003c \\varepsilon/4$ and $\\abs{y'_n - y_n} \u003c \\varepsilon/4$ for $n \\ge N'$.\n5. Then $y'_n - x'_n \u003e (y_n - x_n) - \\varepsilon/2 \u003e \\varepsilon/2$ for $n \\ge N'$. All computations here take place in the [[ordered field of rationals|09RFKMBQ63]].\n6. Lemma. Let $(z_n)$ be a Cauchy sequence of rationals that does not tend to $0$. Then there are a rational $r \u003e 0$ and a $P$ with $z_n \u003e r$ for all $n \\ge P$, or $z_n \u003c -r$ for all $n \\ge P$.\n7. Proof of the lemma. Since $(z_n)$ does not tend to $0$, there is a rational $s \u003e 0$ such that for each $N$ some $n \\ge N$ has $\\abs{z_n} \\ge s$.\n8. Since $(z_n)$ is Cauchy, there is $N$ with $\\abs{z_m - z_n} \u003c s/3$ for all $m, n \\ge N$. Pick $P \\ge N$ with $\\abs{z_P} \\ge s$.\n9. If $z_P \\ge s$, then $z_n \u003e z_P - s/3 \\ge 2s/3$ for each $n \\ge P$. If $z_P \\le -s$, then $z_n \u003c z_P + s/3 \\le -2s/3$ for each $n \\ge P$. Take $r = 2s/3$.\n10. Let $x \\ne y$ be real numbers. Then $(y_n - x_n)$ does not tend to $0$, by step 1. By the lemma, $y_n - x_n \u003e r$ eventually or $x_n - y_n \u003e r$ eventually. So $x \u003c y$ or $y \u003c x$.\n11. The relations $x \u003c y$ and $y \u003c x$ do not both hold. Otherwise, for all large $n$, $y_n - x_n \u003e \\varepsilon$ and $x_n - y_n \u003e \\varepsilon'$ with $\\varepsilon, \\varepsilon' \u003e 0$. Adding gives $0 \u003e \\varepsilon + \\varepsilon'$, which is false.\n12. The relation $x \u003c x$ never holds. By step 4 we may use the same representative twice, and then $x_n - x_n = 0$ is never larger than $\\varepsilon \u003e 0$.\n13. Recall that $x \\le y$ means $x \u003c y$ or $x = y$. It is reflexive. It is antisymmetric: if $x \\le y$, $y \\le x$ and $x \\ne y$, then $x \u003c y$ and $y \u003c x$, against step 11.\n14. It is transitive. Let $x \\le y \\le z$. If one of the two relations is an equality, then $x \\le z$ at once. Otherwise $z_n - y_n \u003e \\varepsilon$ and $y_n - x_n \u003e \\varepsilon'$ for large $n$. Then $z_n - x_n \u003e \\varepsilon + \\varepsilon'$, so $x \u003c z$.\n15. It is total by step 10. So $\\le$ is a [[total order|0316AYH1SC]] on $\\R$.\n16. Let $a \\le b$ and let $c$ be real. If $a = b$, then $a + c = b + c$. If $a \u003c b$, then $a + c$ and $b + c$ have the representatives $(a_n + c_n)$ and $(b_n + c_n)$. Their difference is $b_n - a_n$, which is eventually larger than some $\\varepsilon \u003e 0$. So $a + c \u003c b + c$.\n17. Let $0 \\le a$ and $0 \\le b$. If $a = 0$ or $b = 0$, then $a b = 0$ in the [[field of reals|0FGENPMPJX]], so $0 \\le a b$.\n18. Otherwise $0 \u003c a$ and $0 \u003c b$. By steps 2 and 3, there are rationals $\\varepsilon, \\varepsilon' \u003e 0$ and an $N$ with $a_n \u003e \\varepsilon$ and $b_n \u003e \\varepsilon'$ for $n \\ge N$.\n19. The product $a b$ is represented by $(a_n b_n)$, and $a_n b_n \u003e \\varepsilon \\varepsilon' \u003e 0$ for $n \\ge N$. So $0 \u003c a b$.\n20. The reals form a field, by the [[field property of the reals|0FGENPMPJX]]. With steps 15, 16, 17 and 19, $\\R$ is an [[ordered field|010JCV9J4E]].\n\n## Idea\n\nA real number that is not zero is represented by a sequence that eventually stays on one side of zero, at a fixed distance. This gives the trichotomy, so the order is total. The two axioms of an ordered field then hold term by term, because they hold in $\\Q$.\n",
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