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text: '[summary: An **ordered ring** is a [-3gq] that is [540 totally ordered], where the ordering agrees with the ring operations. In particular, adding something to two elements doesn't change which of them is bigger, and the product of two [-positive] elements is positive.]\n\nAn **ordered ring** is a [-3gq] $R=(X,\\oplus,\\otimes)$ with a [540 total order] $\\leq$ compatible with the ring structure. Specifically, it must satisfy these axioms for any $a,b,c \\in X$:\n\n- If $a \\leq b$, then $a \\oplus c \\leq b \\oplus c$.\n- If $0 \\leq a$ and $0 \\leq b$, then $0 \\leq a \\otimes b$\n\nAn element $a$ of the ring is called "[-positive]" if $0<a$ and "[-negative]" if $a<0$. The second axiom, then, says that the product of nonnegative elements is nonnegative. \n\nAn ordered ring that is also a [481 field] is an [-ordered_field].\n\n#Basic Properties\n\n- For any element $a$, $a \\leq 0$ if and only if $0 \\leq -a$.\n\n%%hidden(Show proof):\nFirst suppose $a \\leq 0$. Using the first axiom to add $-a$ to both sides, $a+(-a) = 0 \\leq -a$. For the other direction, suppose $0 \\leq -a$. Then $a \\leq -a+a = 0$.\n%%\n\n- The product of nonpositive elements is nonnegative.\n\n%%hidden(Show proof):\nSuppose $a$ and $b$ are nonpositive elements of $R$, that is $a,b \\leq 0$. From the first axiom, $a+(-a) = 0 \\leq -a$, and similarly $0 \\leq -b$. By the second axiom $0 \\leq -a \\otimes -b$. But $-a \\otimes -b = a \\otimes b$, so $0 \\leq a \\otimes b$.\n%%\n\n- The [-square] of any element is nonnegative.\n\n%%hidden(Show proof):\nLet $a$ be such an element. Since the ordering is total, either $0 \\leq a$ or $a \\leq 0$. In the first case, the second axiom gives $0 \\leq a^2$. In the second case, the previous property gives $0 \\leq a^2$, since $a$ is nonpositive. Either way we have $0 \\leq a^2$.\n%%\n\n- The additive [54p identity] $1 \\geq 0$. (Unless the ring is trivial, $1>0$.)\n\n%%hidden(Show proof):\nClearly $1 = 1 \\otimes 1$. So $1$ is a square, which means it's nonnegative.\n%%\n\n# Examples\n\nThe [4bc real numbers] are an ordered ring (in fact, an [-ordered_field]), as is any [-subring] of $\\mathbb R$, such as [4zq $\\mathbb Q$].\n\nThe [4zw complex numbers] are not an ordered ring, because there is no way to define the order between $0$ and $i$. Suppose that $0 \\le i$, then, we have $0 \\le i \\times i = -1$, which is false. Suppose that $i \\le 0$, then $0 = i + (-i) \\le 0 + (-i)$, but then we have $0 \\le (-i) \\times (-i) = -1$, which is again false. Alternatively, $i^2=-1$ is a square, so it must be nonnegative; that is, $0 \\leq -1$, which is a contradiction.',
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