Mathematical Fallacy 4:Is 0=10=1?;A Discussion

We know that (a+b)2=a2+2ab+b2.({a}+{b})^2={a}^2+2{a}{b}+{b}^2. Using the above identity,we proceed to prove an impossible task:Proving 0=10=1!

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

(n+1)2=n2+2n+1(1)(n+1)2(2n+1)=n2(2)(n+1)2(2n+1)n(2n+1)=n2n(2n+1)(3)(n+1)2(2n+1)(n+1)=n2n(2n+1)(4)(n+1)2(2n+1)(n+1)+(2n+12)2=n2n(2n+1)+(2n+12)2(5)[(n+1)2n+12]2=[n2n+12]2(6)(n+1)2n+12=n2n+12(7)n+1=n(8)nn=1(9)0=1(10) \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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1 vote

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Step (7) is wrong.

In simpler terms it's like (1)2=(1)21=1(1)^2=(-1)^2 \Rightarrow 1=-1 (taking square root on both the sides). But don't forget that x2=x\sqrt{x^2}=|x|.

The same mistake is made in the journey from step (6) to step (7).

Sandeep Bhardwaj - 3 years, 8 months ago

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That' right :)

Rohit Udaiwal - 3 years, 8 months ago

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7th7^{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 7th7^{th} step is always positive while LHS is always negative.

Aditya Sky - 3 years, 8 months ago

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

Rohit Udaiwal - 3 years, 8 months ago

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