Can you tell me what's wrong with it ?

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 $i^{2}=-1$ )

#Algebra #ComplexNumbers #Contradiction #Mathematics

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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`\boxed{123}`	$\boxed{123}$

Comments

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

$\sqrt { -a } \sqrt { -b } \quad =\quad i\sqrt { a } i\sqrt { b } =-\sqrt { ab } \quad \quad and\quad not\quad \sqrt { (-1)(-1)ab } =\sqrt { ab } \\$ 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

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

Mvs Saketh - 6 years, 8 months ago

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

Abhinav Raichur - 6 years, 6 months ago

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.

Satvik Golechha - 6 years, 8 months ago

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

Sandeep Bhardwaj - 6 years, 8 months ago

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, 7 months ago

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

Sandeep Bhardwaj - 6 years, 7 months ago

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

Apoorv Srivastava - 6 years, 8 months ago

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

Sandeep Bhardwaj - 6 years, 8 months ago

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, 8 months ago

@Apoorv Srivastava – Ok . So is there any RULE which says that $\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ , 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 : $\sqrt{\frac{-1}{2}}=\frac{\sqrt{-1}}{\sqrt{2}}=i.\frac{\sqrt{1}}{\sqrt{2}}$ . Here i denotes iota.

Sandeep Bhardwaj - 6 years, 8 months ago

Roots with complex donot follow distributive law of algebra

Gauri shankar Mishra - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$