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 $i= \sqrt{-1}=+\sqrt{-1}$ .

Updated: September 14th, 2014

#### Gaurav Tiwari

A designer by profession, a mathematician by education but a Blogger by hobby. Loves reading and writing. Just that.