This mathematical fallacy is due to a simple assumption, that $ -1=\dfrac{-1}{1}=\dfrac{1}{-1}$ .

Proceeding with $ \dfrac{-1}{1}=\dfrac{1}{-1}$ and taking square-roots of both sides, we get:

$ \dfrac{\sqrt{-1}}{\sqrt{1}}=\dfrac{\sqrt{1}}{\sqrt{-1}}$

Now, as the Euler’s constant $ i= \sqrt{-1}$ and $ \sqrt{1}=1$ , we can have

$ \dfrac{i}{1}=\dfrac{1}{i} \ldots \{1 \}$

$ \Rightarrow i^2=1 \ldots \{2 \}$ .

This is complete contradiction to the fact that $ i^2=-1$ .

Again, as $ \dfrac{i}{1}=\dfrac{1}{i}$

or, $ i^2=1$

or, $ i^2+2=1+2$

or, $ -1+2=3$

$ 1=3 \ldots \{3 \}$ .

or, in general $ a=a+2, \ \forall a \in \mathbb{C} $

Again using equation $ \{1 \}$ and dividing both sides by 2, we get

$ \dfrac{i}{2}=\dfrac{1}{2i}$

$ \Rightarrow \dfrac{i}{2}+\dfrac{3}{2i}=\dfrac{1}{2i}+\dfrac{3}{2i}$

$ \Rightarrow i \dfrac{i}{2}+i \dfrac{3}{2i}=i \dfrac{1}{2i}+i \dfrac{3}{2i}$

$ \Rightarrow \dfrac{i^2}{2}+\dfrac{3}{2}=\dfrac{1}{2}+\dfrac{3}{2}$

$ \dfrac{-1}{2} +\dfrac{3}{2}=\dfrac{1}{2}+\dfrac{3}{2}$

$ 1=2 \ldots \{4 \}$

or, in general $ b=b+1, \ \forall b \in \mathbb{C} $

$ \Box$

Where is the error?

These fallacies were derived since we ignored the negative ‘Square-roots’ of 1 & -1. If we put $ \sqrt{1}=\pm 1$ and $ \sqrt{-1}=\pm i$ , then the results would have been different. Also note that $ \sqrt{-1}=\pm i$ but $ i= \sqrt{-1}=+\sqrt{-1}$ .

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