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text: 'The real numbers, when [50d constructed as equivalence classes of Cauchy sequences] of [4zq rationals], form a [540 totally ordered] [481 field], with the inherited field structure given by \n\n- $[a_n] + [b_n] = [a_n+b_n]$\n- $[a_n] \\times [b_n] = [a_n \\times b_n]$\n- $[a_n] \\leq [b_n]$ if and only if either $[a_n] = [b_n]$ or for sufficiently large $n$, $a_n \\leq b_n$.\n\n# Proof\n\nFirstly, we need to show that those operations are even well-defined: that is, if we pick two different representatives $(x_n)_{n=1}^{\\infty}$ and $(y_n)_{n=1}^{\\infty}$ of the same equivalence class $[x_n] = [y_n]$, we don't somehow get different answers.\n\n## Well-definedness of $+$\n\nWe wish to show that $[x_n]+[a_n] = [y_n] + [b_n]$ whenever $[x_n] = [y_n]$ and $[a_n] = [b_n]$; this is an exercise.\n%%hidden(Show solution):\nSince $[x_n] = [y_n]$, it must be the case that both $(x_n)$ and $(y_n)$ are Cauchy sequences such that $x_n - y_n \\to 0$ as $n \\to \\infty$.\nSimilarly, $a_n - b_n \\to 0$ as $n \\to \\infty$.\n\nWe require $[x_n+a_n] = [y_n+b_n]$; that is, we require $x_n+a_n - y_n-b_n \\to 0$ as $n \\to \\infty$.\n\nBut this is true: if we fix rational $\\epsilon > 0$, we can find $N_1$ such that for all $n > N_1$, we have $|x_n - y_n| < \\frac{\\epsilon}{2}$; and we can find $N_2$ such that for all $n > N_2$, we have $|a_n - b_n| < \\frac{\\epsilon}{2}$.\nLetting $N$ be the maximum of the two $N_1, N_2$, we have that for all $n > N$, $|x_n + a_n - y_n - b_n| \\leq |x_n - y_n| + |a_n - b_n|$ by the [-triangle_inequality], and hence $\\leq \\epsilon$.\n%% \n\n## Well-definedness of $\\times$\n\nWe wish to show that $[x_n] \\times [a_n] = [y_n] \\times [b_n]$ whenever $[x_n] = [y_n]$ and $[a_n] = [b_n]$; this is also an exercise.\n%%hidden(Show solution):\nWe require $[x_n a_n] = [y_n b_n]$; that is, $x_n a_n - y_n b_n \\to 0$ as $n \\to \\infty$.\n\nLet $\\epsilon > 0$ be rational.\nThen $$|x_n a_n - y_n b_n| = |x_n (a_n - b_n) + b_n (x_n - y_n)|$$ using the very handy trick of adding the expression $x_n b_n - x_n b_n$.\n\nBy the triangle inequality, this is $\\leq |x_n| |a_n - b_n| + |b_n| |x_n - y_n|$.\n\nWe now use the fact that [cauchy_sequences_are_bounded], to extract some $B$ such that $|x_n| < B$ and $|b_n| < B$ for all $n$;\nthen our expression is less than $B (|a_n - b_n| + |x_n - y_n|)$.\n\nFinally, for $n$ sufficiently large we have $|a_n - b_n| < \\frac{\\epsilon}{2 B}$, and similarly for $x_n$ and $y_n$, so the result follows that $|x_n a_n - y_n b_n| < \\epsilon$.\n%%\n\n## Well-definedness of $\\leq$\n\nWe wish to show that if $[a_n] = [c_n]$ and $[b_n] = [d_n]$, then $[a_n] \\leq [b_n]$ implies $[c_n] \\leq [d_n]$.\n\nSuppose $[a_n] \\leq [b_n]$, but suppose for [46z contradiction] that $[c_n]$ is not $\\leq [d_n]$: that is, $[c_n] \\not = [d_n]$ and there are arbitrarily large $n$ such that $c_n > d_n$.\nThen there are two cases.\n\n- If $[a_n] = [b_n]$ then $[d_n] = [b_n] = [a_n] = [c_n]$, so the result follows immediately. %%note:We didn't need the extra assumption that $[c_n] \\not \\leq [d_n]$ here.%%\n- If for all sufficiently large $n$ we have $a_n \\leq b_n$, then [todo: this part]\n\n## Additive [3jb commutative] [3gd group structure] on $\\mathbb{R}$\n\nThe additive identity is $[0]$ (formally, the equivalence class of the sequence $(0, 0, \\dots)$).\nIndeed, $[a_n] + [0] = [a_n+0] = [a_n]$.\n\nThe additive inverse of the element $[a_n]$ is $[-a_n]$, because $[a_n] + [-a_n] = [a_n-a_n] = [0]$.\n\nThe operation is commutative: $[a_n] + [b_n] = [a_n+b_n] = [b_n+a_n] = [b_n] + [a_n]$.\n\nThe operation is closed, because the sum of two Cauchy sequences is a Cauchy sequence (exercise).\n%%hidden(Show solution):\nIf $(a_n)$ and $(b_n)$ are Cauchy sequences, then let $\\epsilon > 0$.\nWe wish to show that there is $N$ such that for all $n, m > N$, we have $|a_n+b_n - a_m - b_m| < \\epsilon$.\n\nBut $|a_n+b_n - a_m - b_m| \\leq |a_n - a_m| + |b_n - b_m|$ by the triangle inequality; so picking $N$ so that $|a_n - a_m| < \\frac{\\epsilon}{2}$ and $|b_n - b_m| < \\frac{\\epsilon}{2}$ for all $n, m > N$, the result follows.\n%%\n\nThe operation is associative: $$[a_n] + ([b_n] + [c_n]) = [a_n] + [b_n+c_n] = [a_n+b_n+c_n] = [a_n+b_n] + [c_n] = ([a_n]+[b_n])+[c_n]$$\n\n## [3gq Ring structure]\n\nThe multiplicative identity is $[1]$ (formally, the equivalence class of the sequence $(1,1, \\dots)$).\nIndeed, $[a_n] \\times [1] = [a_n \\times 1] = [a_n]$.\n\n$\\times$ is closed, because the product of two Cauchy sequences is a Cauchy sequence (exercise).\n%%hidden(Show solution):\nIf $(a_n)$ and $(b_n)$ are Cauchy sequences, then let $\\epsilon > 0$.\nWe wish to show that there is $N$ such that for all $n, m > N$, we have $|a_n b_n - a_m b_m| < \\epsilon$.\n\nBut $$|a_n b_n - a_m b_m| = |a_n (b_n - b_m) + b_m (a_n - a_m)| \\leq |b_m| |a_n - a_m| + |a_n| |b_n - b_m|$$ by the triangle inequality.\n\nCauchy sequences are bounded, so there is $B$ such that $|a_n|$ and $|b_m|$ are both less than $B$ for all $n$ and $m$.\n\nSo picking $N$ so that $|a_n - a_m| < \\frac{\\epsilon}{2B}$ and $|b_n - b_m| < \\frac{\\epsilon}{2B}$ for all $n, m > N$, the result follows.\n%%\n\n$\\times$ is clearly commutative: $[a_n] \\times [b_n] = [a_n \\times b_n] = [b_n \\times a_n] = [b_n] \\times [a_n]$.\n\n$\\times$ is associative: $$[a_n] \\times ([b_n] \\times [c_n]) = [a_n] \\times [b_n \\times c_n] = [a_n \\times b_n \\times c_n] = [a_n \\times b_n] \\times [c_n] = ([a_n] \\times [b_n]) \\times [c_n]$$\n\n$\\times$ distributes over $+$: we need to show that $[x_n] \\times ([a_n]+[b_n]) = [x_n] \\times [a_n] + [x_n] \\times [b_n]$.\nBut this is true: $$[x_n] \\times ([a_n]+[b_n]) = [x_n] \\times [a_n+b_n] = [x_n \\times (a_n+b_n)] = [x_n \\times a_n + x_n \\times b_n] = [x_n \\times a_n] + [x_n \\times b_n] = [x_n] \\times [a_n] + [x_n] \\times [b_n]$$\n\n## Field structure\n\nTo get from a ring to a field, it is necessary and sufficient to find a multiplicative inverse for any $[a_n]$ not equal to $[0]$.\n\nSince $[a_n] \\not = 0$, there is some $N$ such that for all $n > N$, $a_n \\not = 0$.\nThen defining the sequence $b_i = 1$ for $i \\leq N$, and $b_i = \\frac{1}{a_i}$ for $i > N$, we obtain a sequence which induces an element $[b_n]$ of $\\mathbb{R}$; and it is easy to check that $[a_n] [b_n] = [1]$.\n%%hidden(Show solution):\n$[a_n] [b_n] = [a_n b_n]$; but the sequence $(a_n b_n)$ is $1$ for all $n > N$, and so it lies in the same equivalence class as the sequence $(1, 1, \\dots)$.\n%%\n\n## Ordering on the field\n\nWe need to show that:\n\n- if $[a_n] \\leq [b_n]$, then for every $[c_n]$ we have $[a_n] + [c_n] \\leq [b_n] + [c_n]$;\n- if $[0] \\leq [a_n]$ and $[0] \\leq [b_n]$, then $[0] \\leq [a_n] \\times[b_n]$.\n\nWe may assume that the inequalities are strict, because if equality holds in the assumption then everything is obvious.\n%%hidden(Show obvious bits):\nIf $[a_n] = [b_n]$, then for every $[c_n]$ we have $[a_n] + [c_n] = [b_n] + [c_n]$ by well-definedness of addition.\nTherefore $[a_n] + [c_n] \\leq [b_n] + [c_n]$.\n\nIf $[0] = [a_n]$ and $[0] \\leq [b_n]$, then $[0] = [0] \\times [b_n] = [a_n] \\times [b_n]$, so it is certainly true that $[0] \\leq [a_n] \\times [b_n]$.\n%%\n\nFor the former: suppose $[a_n] < [b_n]$, and let $[c_n]$ be an arbitrary equivalence class.\nThen $[a_n] + [c_n] = [a_n+c_n]$; $[b_n] + [c_n] = [b_n+c_n]$; but we have $a_n + c_n \\leq b_n + c_n$ for all sufficiently large $n$, because $a_n \\leq b_n$ for sufficiently large $n$.\nTherefore $[a_n] + [c_n] \\leq [b_n] + [c_n]$, as required.\n\nFor the latter: suppose $[0] < [a_n]$ and $[0] < [b_n]$.\nThen for sufficiently large $n$, we have both $a_n$ and $b_n$ are positive; so for sufficiently large $n$, we have $a_n b_n \\geq 0$.\nBut that is just saying that $[0] \\leq [a_n] \\times [b_n]$, as required.',
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