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Proof of 1 = 0

At what step did the error started? \[(x+1)^2 = x^2+2x+1\] Step 1: \[(x+1)^2 - (2x+1) = x^2\] Step 2: \[(x+1)^2 - (2x+1)-x(2x+1)=x^2-x(2x+1)\] Step 3: \[(x+1)^2 -(x+1)(2x+1)+ \frac{1}{4}(2x+1)^2 = x^2 -x(2x+1) + \frac{1}{4}(2x+1)^2\] Step 4: \[ [(x+1) - \frac{1}{2}(2x+1)]^2=[x- \frac{1}{2}(2x+1)]^2\] Step 5: \[(x+1) - \frac{1}{2}(2x+1) = x - \frac{1}{2}(2x+1)\] Step 6: \[x+1 = x\] Therefore, \(1 = 0\)

Note by Paul Ryan Longhas
2 years, 10 months ago

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Step 5 corrected to: (x + 1) - (1/ 2) (2 x + 1) = (1/ 2) (2 x + 1) - x would be all right.

Lu Chee Ket - 2 years, 10 months ago

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At step5

Aishwarya Warke - 2 years, 10 months ago

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