I will show that 3=0.
Let x²+x+1=0. Clearly, x=1 is not a solution to the equation. Divide both sides by "x":
x+1+1/x=0 => x+1=-1/x
Thus, x²+x+1=0 => x²-1/x=0. Multiply by "x":
x³-1=0. Clearly, x=1 is a solution to this equation. So going back to the original equation:
1²+1+1=0 => 3=0
QED
#math
Aaron Toponce ⚛️:debian: /
npub14d…s5uyy
2023-09-09 03:55:33
Author Public Key
npub14dfr55yyvfw0leduur7e4af7fnkeegww6e2n64laxkkdyn4sufrqhs5uyyPublished at
2023-09-09 03:55:33Event JSON
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"content": "I will show that 3=0.\n\nLet x²+x+1=0. Clearly, x=1 is not a solution to the equation. Divide both sides by \"x\":\n\nx+1+1/x=0 =\u003e x+1=-1/x\n\nThus, x²+x+1=0 =\u003e x²-1/x=0. Multiply by \"x\":\n\nx³-1=0. Clearly, x=1 is a solution to this equation. So going back to the original equation:\n\n1²+1+1=0 =\u003e 3=0\n\nQED\n\n#math",
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