1-1-1-1-1-1-i-i-i-2-1-where-is-the-mistake-

Question Number 171406 by floor(10²Eta[1]) last updated on 14/Jun/22

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

where is the mistake ?

Commented by infinityaction last updated on 14/Jun/22

i f a a n d b a r e p o s i t i v e n u m b e r

t h e n \sqrt{a b} = \sqrt{a} \sqrt{b}

a a n d b a r e n e g a t i v e n u m b e r

t h e n \sqrt{a b} = \sqrt{a} \sqrt{b}

Answered by MJS_new last updated on 14/Jun/22

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

\sqrt{a b} = \sqrt{(a b)} \Rightarrow 1 . calculate a b = c 2 . calculate \sqrt{c}

\sqrt{a} \sqrt{b} \Rightarrow 1 . calculate \sqrt{a} and \sqrt{b} 2 . multiply them

r, s > 0 \land 0 ⩽ α, β < 2 π

\sqrt{e^{i α} r \times e^{i β} s} = e^{i \frac{\mod (α + β, 2 π)}{2}} \sqrt{r s}

\sqrt{e^{i α} r} \times \sqrt{e^{i β} s} = e^{i \frac{α + β}{2}} \sqrt{r s}

\sqrt{(- 1) (- 1)} = \sqrt{e^{i π} \times e^{i π}} = e^{i \frac{\mod (2 π, 2 π)}{2}} = e^{i 0} = 1

\sqrt{(- 1)} \times \sqrt{(- 1)} = e^{i \frac{π + π}{2}} = e^{i π} = - 1

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