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# 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:

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

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

Note by Rohit Udaiwal
1 year, 5 months ago

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$$7^{th}$$ 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. · 1 year, 5 months ago

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). · 1 year, 5 months ago

That' right :) · 1 year, 5 months ago