1-1-1-1-2-2-1-2-1-2-1-1-2-1-1-what-do-you-think-about-this-

Question Number 80084 by behi83417@gmail.com last updated on 30/Jan/20

- 1 = {(- 1)}^{1} = {(- 1)}^{\frac{2}{2}} = {({(- 1)}^{2})}^{\frac{1}{2}} = {(1)}^{\frac{1}{2}} =

= \sqrt{1} = 1

what do you think about this ?

Commented by key of knowledge last updated on 30/Jan/20

{(- 1)}^{\frac{2}{2}} = {({(- 1)}^{\frac{1}{2}})}^{2} \neq {({(- 1)}^{2})}^{\frac{1}{2}}

Commented by behi83417@gmail.com last updated on 31/Jan/20

thanks alot {sir}^{'} s .

perfect explanation .

Commented by MJS last updated on 31/Jan/20

{we}^{'} ve got a definition :

C = {z = e^{i θ} r ∣ - π < θ ⩽ π; r \in R^{+}}

we know that e^{i θ} = e^{i (θ + 2 π n)} with n \in Z but we

only use this for finding all possible solutions

of certain equations

we define

z^{q} = {(e^{i θ} r)}^{q} = e^{i (π - \mod (π - q θ, 2 π)} r^{q} for q \in Q, R, C

with \mod (x, y) ⩾ 0 i . e . \mod (- 3, 7) = 4

[{there}^{'} s only one exception for z \in R^{+} and

n \in Z : {(- z)}^{\frac{1}{2 n + 1}} = - (z^{\frac{1}{2 n + 1}}) This is useful in

elementary mathematics, like 3 d geometry

or solving polynomes of 3^{rd} degree using

Cardano .]

\Rightarrow {(z^{q})}^{\frac{1}{q}} \neq {(z^{\frac{1}{q}})}^{q} in many cases

{({(- 1)}^{2})}^{\frac{1}{2}} = {(1)}^{\frac{1}{2}} = 1

{({(- 1)}^{\frac{1}{2}})}^{2} = {(i)}^{2} = - 1

{({(3 + 4 i)}^{4})}^{\frac{1}{4}} = 4 - 3 i

{({(3 + 4 i)}^{\frac{1}{4}})}^{4} = 3 + 4 i

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