# Mathematical Fallacy 4:Is $0=1$?;A Discussion

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{aligned} (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{aligned}

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

Note by Rohit Udaiwal
3 years, 8 months 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).

- 3 years, 8 months ago

That' right :)

- 3 years, 8 months ago

$7^{th}$ 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.

- 3 years, 8 months ago

If you have any more doubts on the subject,see this.

- 3 years, 8 months ago