Can you tell me what's wrong with it ?

With no doubts in your mind, i can write :


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

=> 11=11\sqrt{\frac{-1}{1}}=\sqrt{\frac{1}{-1}}

=> 11=11\frac{\sqrt{-1}}{\sqrt{1}}=\frac{\sqrt{1}}{\sqrt{-1}}

=> i1=1i\frac{i}{1}=\frac{1}{i}

cross multiplying

=> i2=1i^{2}=1

=> 1=1\boxed{-1=1} (Because we know i2=1i^{2}=-1)

Note by Sandeep Bhardwaj
6 years, 1 month ago

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1 vote

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you 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

ab=iaib=abandnot(1)(1)ab=ab\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

ab=iab=aib=abandthenucorrectlyget1=1allbackagain\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\\ \\

Mvs Saketh - 6 years, 1 month ago

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

Abhinav Raichur - 5 years, 11 months ago

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

Satvik Golechha - 6 years, 1 month ago

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If I say that

ab=ab\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

ab=i.ab\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 a.b=a.b\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

Sandeep Bhardwaj - 6 years, 1 month ago

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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.

Keshav Tiwari - 6 years ago

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@Keshav Tiwari I hope so that it's true. But i ain't 100% sure.!

Sandeep Bhardwaj - 6 years ago

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Isn'tab=ab\sqrt {\frac{a}{b}} =\frac {\sqrt {a}}{\sqrt {b}} possible when a and b are positive

Apoorv Srivastava - 6 years, 1 month ago

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

Sandeep Bhardwaj - 6 years, 1 month ago

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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

Apoorv Srivastava - 6 years, 1 month ago

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@Apoorv Srivastava Ok . So is there any RULE which says that ab=ab\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}} , if and only if both a and b are positive. It is true for both negatives too. And also true in one way :if one of them is positive and the other one is negative, for eg : 12=12=i.12\sqrt{\frac{-1}{2}}=\frac{\sqrt{-1}}{\sqrt{2}}=i.\frac{\sqrt{1}}{\sqrt{2}} . Here i denotes iota.

Sandeep Bhardwaj - 6 years, 1 month ago

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Roots with complex donot follow distributive law of algebra

Gauri shankar Mishra - 4 years, 9 months ago

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