vishal's doubt made me to write a note

\[\sqrt{-1}\times\sqrt{-1}\] \[case (1)=>\sqrt{-1}\times\sqrt{-1}=\sqrt{-1\times-1}=\sqrt{1}=1\] \[case(2)=>\sqrt{-1}\times\sqrt{-1}=(\sqrt{-1})^2=-1\] which one is correct either first case or the second case can anyone say please calvin lin can you say it

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Comments

$\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ if and only if $a,b> 0$

Siddhartha Srivastava - 6 years, 4 months ago

Case 1 is incorrect since you cannot split the square-root if both radicands are negative.

B.S.Bharath Sai Guhan - 6 years, 4 months ago

Case 2 is correct...

Since $\sqrt{-1} = i$ , then, the equation above is simply equal to $i^{2} = -1$ .. :)

Christian Daang - 6 years, 4 months ago

mr.subbu case 2 is correct

SARAN .P.S - 5 years, 7 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}`	$\boxed{123}$