With no doubts in your mind, i can write :

\(\sqrt{-1}=\sqrt{-1}\)

**[\(\sqrt{-1}=i\) i.e. iota]**

=> \(\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}\)

=> \(\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}\)

=> \(\frac{i}{1}=\frac{1}{i}\)

**cross multiplying**

=> \(i^{2}=1\)

=> \(\boxed{-1=1}\) **(Because we know** **\(i^{2}=-1\))**

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## Comments

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TopNewestyou cannot write root(a) root(b) as root(ab) if a and b are both negative numbers,,,

certain laws of exponents holds only for positive real numbers,, and why? specifically because it leads to contradictions as root(-1) root (-1) = root(-1*-1)=root(1)=1 which is not true,,,

instead we write it as

\(\sqrt { -a } \sqrt { -b } \quad =\quad i\sqrt { a } i\sqrt { b } =-\sqrt { ab } \quad \quad and\quad not\quad \sqrt { (-1)(-1)ab } =\sqrt { ab } \\\) you did the same contradiction in a rather elongated fashion and i can show that your way is same as the one i pointed above,,

Look at the 2nd step.... the result you obtained by exchanging signs among roots is a logical fallacy,

instead you should write

\(\frac { \sqrt { -a } }{ \sqrt { b } } =\frac { i\sqrt { a } }{ \sqrt { b } } =-\frac { \sqrt { a } }{ i\sqrt { b } } =\quad -\frac { \sqrt { a } }{ \sqrt { -b } } \\ \\ and\quad then\quad u\quad correctly\quad get\quad -1=-1\quad all\quad back\quad again\\ \\\)

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ure right ! ....... great mathematician LEONHARD EULER did the same mistake once! {of course in a slightly different manner}

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I'm searching for a rigorous fault, and I could find some possibilities. I don't like \(\sqrt{-1}\) being replaced by \(\iota\), cuz \(\iota = \pm \sqrt{-1}\), though I may be wrong. There may be something extraneous done while playing with the square roots and squares.

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If I say that

\(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\) , only if a and b are both positive or both negative...

And if exactly one of a and b is positive, then

\(\sqrt{\frac{a}{b}}=i. \frac{\sqrt{|a|}}{\sqrt{|b|}}\).

You can say that i am trying to make a new rule , which would give its best to control some functionings of such expressions.

One rule I know is \(\sqrt{a}.\sqrt{b}=\sqrt{a.b}\) ; if and only if at least one of a and b is non-negative. And the second rule may be which i discussed above :P

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So is the rule stated by you true or not ? I have read about the latter but not the first one.

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Isn't\[\sqrt {\frac{a}{b}} =\frac {\sqrt {a}}{\sqrt {b}}\] possible when a and b are positive

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Offcourse, you can write it ; if a and b are both positive. But what in case of above contradiction ?

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I am saying that you can split the term only if the statement i said is true then your 3 rd step will be wrong

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Roots with complex donot follow distributive law of algebra

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