maybe you did state this argument already and I overlooked, but what about the following calculation.
Let \(x=0.\bar{9}\). Then \(10x=9.\bar{9}\) and \(9x=10x-x=9.\bar{9}-0.\bar{9}=9\). Hence, \(x=1\).
Diffgeometer1
npub1hd…k9se4
2025-08-07 04:17:59 UTC
in reply to nevent1q…3kq9
Author Public Key
npub1hdkuf2expv4gele847pq24raux2mp20dkrtvrvkvcrvpmfq3mqmq8k9se4Seen on
wss://relay.momostr.pinkPublished at
2025-08-07 04:17:59 UTCEvent JSON
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"content": "maybe you did state this argument already and I overlooked, but what about the following calculation. \n\nLet \\(x=0.\\bar{9}\\). Then \\(10x=9.\\bar{9}\\) and \\(9x=10x-x=9.\\bar{9}-0.\\bar{9}=9\\). Hence, \\(x=1\\).",
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