Christmas Streak 69/88: Where is wrong? #2

Algebra Level 3

Picture is drawn by me.

$\text{Where is wrong?}$

Notation: $i$ is the imaginary unit.

This problem is a part of <Christmas Streak 2017> series .

i\times i=\sqrt{-1}\times\sqrt{-1}

i^2=i\times i

\sqrt{1}=1

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

-1=i^2

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

Nothing's wrong

1 solution

Boi (보이)
Dec 9, 2017

Generally, for reals $a,~b,$ the following holds:

$\sqrt{a}\sqrt{b}=\begin{cases} -\sqrt{ab} & (a<0,~b<0)\\\\ \sqrt{ab} & (\text{else})\end{cases}$

Since $-1<0,$ it should be: $\sqrt{-1}\times\sqrt{-1}={\color{#E81990}-}\sqrt{(-1)\times(-1)}.$

This is wrong. At the second line the imaginary unit i does NOT equal to sqrt(-1). It's a common mistake as in the complex field ever number has two square roots. And for your solution: generally we do NOT define sqrt(a) when a<0. I have never seen somebody saying that sqrt(-2) * sqrt(-3) = - sqrt(6).

A Former Brilliant Member - 1 year, 10 months ago

Yes a way of understanding i is sqrt(-1) and yes every number has two square roots but we use the term sqrt(4) for 2 don't we?

Refer: Wikipedia

Boi (보이) - 1 year, 10 months ago

Christmas Streak 69/88: Where is wrong? #2

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