Where-is-the-error-1-1-1-1-2-2-1-2-1-2-1-1-2-1-1-

Question Number 217772 by yamane last updated on 20/Mar/25

W h e r e i s t h e e r r o r

(- 1) = {(- 1)}^{1}

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

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

= {[1]}^{\frac{1}{2}}

= \sqrt{1}

= 1

Commented by Ghisom last updated on 21/Mar/25

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

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

{it}^{'} s similar to this mistake

6 / 3 = 6 / (2 + 1) \overset{? ? ?}{=} 6 / 2 + 6 / 1 = 3 + 6 = 9

Answered by Marzuk last updated on 20/Mar/25

\sqrt{1} = \pm 1 n o t o n l y 1

r_{1} = 1 a n d r_{2} = - 1

y o u c o u l d a l s o s a y

{[{(- 1)}^{\frac{1}{2}}]}^{2}

= i^{2}

= - 1

s o t h e m a i n e r r o r i s i n t h e u n d e r s t a n d i n g

o f a p p l i c a t i o n t h e p r o p e r t y {(a^{m})}^{n} = a^{m n}

Commented by Ghisom last updated on 21/Mar/25

\sqrt{1} = 1

\sqrt{1} = - 1 is wrong

this has been explained over and over again .

we have exactly one rule of exponentials :

z \in C \Rightarrow z = r e^{i θ} with r ⩾ 0 and - π < θ ⩽ π

x \in Q (also works for x \in R, x \in C)

z^{x} = {(r e^{i θ})}^{x} = r^{x} e^{i x θ}

{there}^{'} s no 2^{nd} result . never .

if z = 1 \land x = \frac{1}{2} :

z = {1 e}^{0 i}

z^{x} = {({1 e}^{0 i})}^{1 / 2} = 1^{1 / 2} e^{0 i / 2} = {1 e}^{0} = 1

how would you get - 1 ?^{(*)}

the mistake comes ftom solving equations

where we have a different rule .

x^{2} = 1 \Rightarrow x = \pm 1

because here {we}^{'} re looking for all values

of x which give 1 if we square x

{(+ 1)}^{2} = 1

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

(*)

the only exception is being made for

geometric problems to stay in R, i g n o r i n g

numbers \in C (because they had not been

invented yet in the age of geometry) :

\sqrt[3]{- 8} = {(- 8)}^{1 / 3} = - 2

we artificially take the root \in R

but for x \in C we have

{(- 8)}^{1 / 3} = {({8 e}^{i π})}^{1 / 3} = 8^{1 / 3} e^{i π / 3} = {2 e}^{i π / 3} \notin R

Commented by mr W last updated on 21/Mar/25

y e s .

i f x^{2} = 4, t h e n x = \pm \sqrt{4} .

b u t \sqrt{4} = 2, - \sqrt{4} = - 2, n o t \sqrt{4} = \pm 2 .

Commented by Marzuk last updated on 21/Mar/25

I a m a p o l o g i z i n g f o r m y m i s u n d e r s t a n d i n g .

I w i l l t r y m y b e s t i n f u t u r e . T h a n k s f o r

s h o w i n g t h e m i s t a k e!

Commented by Ghisom last updated on 21/Mar/25

no need to apologize

you live - you learn

I might be a fraction of a step further than

you are but still looking at the countless

backs of those before me

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