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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Gaurav Tiwari
Gaurav Tiwari is a professional graphic & web designer from New Delhi, India. gauravtiwari.org is his personal space where he writes on blogging, digital marketing, content writing, learning and business growth. Gaurav has contributed in developing more than 325 brands worldwide and while you are reading this, he's busy building a couple more.

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