Mathematical Fallacy: Is 1=21=2?

Algebra Level 1

11=11(1)11=11(2)11=11(3)i1=1i(4)i2=12i(5)i2+32i=12i+32i(6)i(i2+32i)=i(12i+32i)(7)i22+3i2i=i2i+3i2i(8)12+32=12+32(9)1=2\begin{aligned} \dfrac{-1}{1}=&\dfrac{1}{-1} \quad & \ldots(1) \\ \sqrt{\dfrac{-1}{1}}=&\sqrt{\dfrac{1}{-1}} \quad & \ldots (2) \\ \dfrac{\sqrt{-1}}{\sqrt{1}}=& \dfrac{\sqrt{1}}{\sqrt{-1}} \quad &\ldots (3) \\ \dfrac{i}{1}=&\dfrac{1}{i} \quad & \ldots(4) \\ \dfrac{i}{2}=&\dfrac{1}{2i} \quad & \ldots(5) \\ \dfrac{i}{2}+\dfrac{3}{2i}=&\dfrac{1}{2i}+\dfrac{3}{2i} \quad &\ldots(6) \\ i\cdot\left(\dfrac{i}{2}+\dfrac{3}{2i}\right)=& i \cdot\left( \dfrac{1}{2i}+\dfrac{3}{2i}\right) \quad & \ldots(7) \\ \dfrac{i^2}{2}+\dfrac{3i}{2i}=& \dfrac{i}{2i}+\dfrac{3i}{2i} \quad & \ldots(8) \\ \dfrac{-1}{2}+\dfrac{3}{2}=& \dfrac{1}{2}+\dfrac{3}{2} \quad & \ldots (9) \\ 1=&2 \end{aligned}

Consider the steps above.

In which step is the mistake committed?

Note: i=1i=\sqrt{-1} in equation (9)(9) and the last one we assume to work out the L.H.S. and the R.H.S. separately.


Source:Spencer,1998

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