1=0
(n+1)2 = n2+2n+1
//Expansion
(n+1)2-(2n+1) = n2
//Subtract from both sides
(n+1)2-(2n+1)-n(2n+1) = n2-n(2n+1)
//Add to both sides
(n+1)2-(n+1)(2n+1) = n2-n(2n+1)
//Factor
(n+1)2-(n+1)(2n+1)+(2n+1)2/4 = n2-n(2n+1)+(2n+1)2/4
//Add to both sides
[(n+1)-(2n+1)/2]2 = [n-(2n+1)/2]2
//Factor
(n+1)-(2n+1)/2 = n-(2n+1)/2
//Take square roots of both sides
n+1 = n
//Subtract from both sides
1 = 0
Impossible!
The operations listed above utilize basic arithmetics to arrive at the false conclusion. Starting by simply expanding a squared equation, we can subtract 2n+1 from both sides to isolate n2. Subtracting n(2n+1) from both sides now allows the left side to be factored. Adding (2n+1)2/4 to both sides once again will enable both sides of the equation to be factored down to squared forms. By taking the square roots and then subtracting the n-(2n+1)/2, the proof is complete and 1=0.
If two numbers are equal, their squares are also equal. However, the reverse form of such a statement does not hold. In short, u = v does not imply square root of u equals square root of v due to the fact that the result of a square root is not unique. Without this fact, the above proof becomes actually legitimate.
