Expert answers to "Why −1⋅−1≠−12?"

Answered question

2022-01-17

Why

\sqrt{- 1 \cdot - 1} \neq {\sqrt{- 1}}^{2} ?

Answer & Explanation

nick1337

Expert2022-01-19Added 777 answers

Step 1 Isaacs

Vasquez

Expert2022-01-19Added 669 answers

Step 1
Given $a, b \in R$
Rule 1:
$\sqrt{a b} \Leftrightarrow \sqrt{a} \sqrt{b}$ is valid if $a, b \geq 0$
Similarly,
$\sqrt{\frac{a}{b}} \Leftrightarrow \frac{\sqrt{a}}{\sqrt{b}}$
is valid only if
$a \geq 0, b > 0$
Rule 2:
And, if $a > 0$
Case 1: $\sqrt{a}$ will give only one positive value (more precise: you will get purely real complex number with positive real part) eg.
$\sqrt{4} = 2$
Case 2: $\sqrt{- a} = i \sqrt{a}$
will give only one complex number with positive imaginary part. (more precise: you will get purely imaginary complex number with positive imaginary part) eg.
$\sqrt{- 4} = 2 i$
So, you have done wrong in fourth line
Lets take a example, shows if you do not follows above rules you can create blunder. Take
$a > 0$
$\sqrt{- a} = i \sqrt{a}$
But if do not follow above rule you can do blunder as
$\sqrt{- a} = \sqrt{\frac{a}{- 1}}$
$\Rightarrow \sqrt{- a} = \frac{\sqrt{a}}{- 1}$
$\Rightarrow \sqrt{- a} = - i \sqrt{a}$
All the above taking and discussion is done under the consideration that in general the notion of
$\sqrt[n]{x}, x \in C$
talking about only about principal root and principal root and it is unique
Step 2
But for all the roots of $\sqrt[n]{x}, x \in C$ Some Mathematician generally uses $x^{\frac{1}{n}}; x \in C$
For example:
$(4)^{\frac{1}{2}} = 2, - 2 {more precisely: (4)^{\frac{1}{2}} \Rightarrow 2, - 2 but 2 ⧸ \Rightarrow (4)^{\frac{1}{2}} or (- 2) ⧸ \Rightarrow (4)^{\frac{1}{2}}}$
Similarly, $(- 1)^{\frac{1}{2} = i}, - i$ because
$(- i) \times (- i) = \sqrt{- 1 = i \times i}$

alenahelenash

Expert2022-01-24Added 556 answers

There is always a danger with dealing with roots of any kind, that one might not be dealing with the same number all the time. This comes in part from both

1^{2} = 1

and

(- 1)^{2} = 1

Writing these into an equation as

\sqrt{1} = 1

and

\sqrt{1} = - 1,

gives a result that

a = - a

. Such might be true in some mantissa-space (mantissa here is a multiplication form of modulus: ie just as

a + b n = a (\mod n)

, so would

a \times n^{b} m a n n)

By taking square roots, one is effectively dealing in a potential mantissa-space where

+ x = - x

, and some subtly is needed to distingiush the two. This is one of the reasons that

\sqrt{x} \geq 0

is taken as convention. The actual mistake in the calculations, is that

\sqrt{x}

is let to vary by sign.

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