Science Magazine

Smart Fallacies: i Squared Equals to 1, 1 Equals to 2 and 3

Posted on the 11 August 2013 by Gaurav Tiwari @wpgaurav

This mathematical fallacy is due to a simple 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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