Question

Find the wrong step

${i}^{2} = { (\sqrt{ - 1} )}^{2}$
$= \sqrt{ - 1 } \times \sqrt{ - 1}$
$= \sqrt{( - 1) \times ( - 1)}$
$= \sqrt{1}$
$= 1$




​

Answer 1

Answer:

Wrong step : $\sqrt{(-1)\times(-1)}$

If we consider $\sqrt{-1}$ as $(-1)^{1/2}$,

then clearly,

$\sqrt{-1} \times \sqrt{-1}$ = $(-1)^{1/2} \times (-1)^{1/2}$

When $x^a$ is multiplied by $x^b$, the final answer is $x^{a+b}$;

Therefore,

it is equal to $(-1)^{(1/2+1/2)} = (-1)^{(1)} = -1$

Hence,

The wrong step is $\sqrt{(-1)\times(-1)}$

Answer 2

Answer:

Answer:

Wrong step : \sqrt{(-1)\times(-1)}

(−1)×(−1)

If we consider \sqrt{-1}

−1

as (-1)^{1/2}(−1)

1/2

,

then clearly,

\sqrt{-1} \times \sqrt{-1}

−1

×

−1

= (-1)^{1/2} \times (-1)^{1/2}(−1)

1/2

×(−1)

1/2

When x^ax

a

is multiplied by x^bx

b

, the final answer is x^{a+b}x

a+b

;

Therefore,

it is equal to (-1)^{(1/2+1/2)} = (-1)^{(1)} = -1(−1)

(1/2+1/2)

=(−1)

(1)

=−1

Hence,

The wrong step is \sqrt{(-1)\times(-1)}

(−1)×(−1)