We know that \[({a}+{b})^2={a}^2+2{a}{b}+{b}^2.\]
Using the above identity,we proceed to prove an **impossible** task:Proving \(0=1\)!

Try to figure out the fallacy in the below set of equations.Here we go:

\[\begin{equation} \begin{split} (n+1)^2=&n^2+2n+1 \quad & \ldots (1) \\ (n+1)^2-(2n+1)=&n^2 \quad & \ldots (2) \\ (n+1)^2-(2n+1)-n(2n+1)=&n^2-n(2n+1) \quad & \ldots (3) \\ (n+1)^2-(2n+1)(n+1)=& n^2-n(2n+1) \quad & \ldots(4) \\ (n+1)^2-(2n+1)(n+1)+\left(\dfrac{2n+1}{2}\right)^2=& n^2-n(2n+1)+\left(\dfrac{2n+1}{2}\right)^2 \quad & \ldots(5) \\ \left[(n+1)-\dfrac{2n+1}{2}\right]^2=&\left[n-\dfrac{2n+1}{2}\right]^2 \quad & \ldots(6) \\ (n+1)-\dfrac{2n+1}{2}=& n-\dfrac{2n+1}{2} \quad & \ldots(7) \\ n+1=& n \quad & \ldots (8) \\ n-n=& 1 \quad & \ldots(9) \\ 0=&1 \quad & \ldots(10) \end{split} \end{equation}\]

Could you figure it out? What do you think? Comment below!

## Comments

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TopNewest\(7^{th}\) step is wrong for the similar reason to Sandeep Bhardwaj. More ever it can be seen that LHS of the equality indicated in the \(7^{th}\) step is always positive while LHS is always negative. – Aditya Sky · 1 year ago

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Step (7) is wrong.In simpler terms it's like \((1)^2=(-1)^2 \Rightarrow 1=-1\) (taking square root on both the sides). But don't forget that \(\sqrt{x^2}=|x|\).

The same mistake is made in the journey from step (6) to step (7). – Sandeep Bhardwaj · 1 year ago

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– Rohit Udaiwal · 1 year ago

That' right :)Log in to reply

If you have any more doubts on the subject,see this. – Rohit Udaiwal · 1 year ago

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