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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
1 year, 8 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. · 1 year, 8 months ago