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title: '0.999...=1',
clickbait: 'No, it's not "infinitesimally far" from 1 or anything like that. 0.999... and 1 are literally the same number.',
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text: 'Although some people find it counterintuitive, the [4sl decimal expansions] $0.999\\dotsc$ and $1$ represent the same [-4bc].\n\n# Informal proofs\n\nThese "proofs" can help give insight, but be careful; a similar technique can "prove" that $1+2+4+8+\\dotsc=-1$. They work in this case because the [-series] corresponding to $0.999\\dotsc$ is [-absolutely_convergent].\n\n* \\begin{align}\nx &= 0.999\\dotsc \\newline\n10x &= 9.999\\dotsc \\newline\n10x-x &= 9.999\\dotsc-0.999\\dotsc \\newline\n9x &= 9 \\newline\nx &= 1 \\newline\n\\end{align}\n\n* \\begin{align}\n\\frac 1 9 &= 0.111\\dotsc \\newline\n1 &= \\frac 9 9 \\newline\n&= 9 \\times \\frac 1 9 \\newline\n&= 9 \\times 0.111\\dotsc \\newline\n&= 0.999\\dotsc\n\\end{align}\n\n* The real numbers are [-dense], which means that if $0.999\\dots\\neq1$, there must be some number in between. But there's no decimal expansion that could represent a number in between $0.999\\dots$ and $1$.\n\n# Formal proof\n\nThis is a more formal version of the first informal proof, using the definition of [-4sl].\n\n%%hidden(Show proof):\n$0.999\\dots$ is the decimal expansion where every digit after the decimal point is a $9$. By definition, it is the value of the series $\\sum_{k=1}^\\infty 9 \\cdot 10^{-k}$. This value is in turn defined as the [-limit] of the sequence $(\\sum_{k=1}^n 9 \\cdot 10^{-k})_{n\\in\\mathbb N}$. Let $a_n$ denote the $n$th term of this sequence. I claim the limit is $1$. To prove this, we have to show that for any $\\varepsilon>0$, there is some $N\\in\\mathbb N$ such that for every $n>N$, $|1-a_n|<\\varepsilon$. \n\nLet's prove by [5fz induction] that $1-a_n=10^{-n}$. Since $a_0$ is the sum of {$0$ terms, $a_0=0$, so $1-a_0=1=10^0$. If $1-a_i=10^{-i}$, then \n\n\\begin{align}\n1 - a_{i+1} &= 1 - (a_i + 9 \\cdot 10^{-(i+1)}) \\newline\n&= 1-a_i - 9 \\cdot 10^{-(i+1)} \\newline\n&= 10^{-i} - 9 \\cdot 10^{-(i+1)} \\newline\n&= 10 \\cdot 10^{-(i+1)} - 9 \\cdot 10^{-(i+1)} \\newline\n&= 10^{-(i+1)}\n\\end{align}\n\nSo $1-a_n=10^{-n}$ for all $n$. What remains to be shown is that $10^{-n}$ eventually gets (and stays) arbitrarily small; this is true by the [archimedean_property] and because $10^{-n}$ is monotonically decreasing.\n%%\n\n\n# Arguments against $0.999\\dotsc=1$\n\nThese arguments are used to try to refute the claim that $0.999\\dotsc=1$. They're flawed, since they claim to prove a false conclusion.\n\n* $0.999\\dotsc$ and $1$ have different digits, so they can't be the same. In particular, $0.999\\dotsc$ starts "$0.$," so it must be less than 1.\n\n%%hidden(Why is this wrong?):\nDecimal expansions and real numbers are different objects. Decimal expansions are a nice way to represent real numbers, but there's no reason different decimal expansions have to represent different real numbers.\n%%\n\n* If two numbers are the same, their difference must be $0$. But $1-0.999\\dotsc=0.000\\dotsc001\\neq0$.\n\n%%hidden(Why is this wrong?):\nDecimal expansions go on infinitely, but no farther. $0.000\\dotsc001$ doesn't represent a real number because the $1$ is supposed to be after infinitely many $0$s, but each digit has to be a finite distance from the decimal point. If you have to pick a real number to for $0.000\\dotsc001$ to represent, it would be $0$.\n%%\n\n* $0.999\\dotsc$ is the limit of the sequence $0.9, 0.99, 0.999, \\dotsc$. Since each term in this sequence is less than $1$, the limit must also be less than $1$. (Or "the sequence can never reach $1$.")\n\n%%hidden(Why is this wrong?):\nThe sequence gets arbitrarily close to $1$, so its limit is $1$. It doesn't matter that all of the terms are less than $1$.\n%%\n\n* In the first proof, when you subtract $0.999\\dotsc$ from $9.999\\dotsc$, you don't get $9$. There's an extra digit left over; just as $9.99-0.999=8.991$, $9.999\\dotsc-0.999\\dotsc=8.999\\dotsc991$.\n\n%%hidden(Why is this wrong?):\nThere are infinitely many $9$s in $0.999\\dotsc$, so when you shift it over a digit there are still the same amount. And the "decimal expansion" $8.999\\dotsc991$ doesn't make sense, because it has infinitely many digits and then a $1$.\n%%',
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