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, $1 = 0$

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Comments

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

Lu Chee Ket - 6 years, 3 months ago

At step5

Aishwarya Warke - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$