Question

I came across a math fallacy and i don't know why it is incorrect

Answer 1

alright show us what it is

Answer 2

\[\huge -1 = -1 \]
\[\huge \frac{ -1 }{ 1 } = \frac{ -1 }{ 1 }\]
\[\huge \frac{ -1 }{ 1 } = \frac{ 1 }{ -1 }\]
\[\huge \sqrt{\frac{ -1 }{ 1 }} = \sqrt{\frac{ 1 }{ -1 }}\]
\[\huge \frac{ i }{ 1 } = \frac{ 1 }{ i }\]
\[\huge i=\frac{ 1 }{ i }\]
\[\huge i ^{2} = 1\]
\[\huge -1 = 1\]

Answer 3

cool :)

Answer 4

yeah

Answer 5

i have no idea sorry

Answer 6

@sidsiddhartha I posted it finally

Answer 7

Cool !!

Answer 8

but one thing left ,
if i considerd that sqrt -1 = i
then sqrt 1would be i,-i,1 ,-1

Answer 9

Implying imaginary number exist.

Answer 10

I didn't get you @BSwan

Answer 11

haha its cool trick :)
but when you start from here

there are 2^4 possible solution :)

Answer 12

\(x^2 = y^2 \not \iff x = y\)

Answer 13

How does that apply here

Answer 14

nvm, i think bswan has it !

Answer 15

ill tell you :-
i^2=1
means
|i^2|=|1|
mmm @ganeshie8 lol u got the same thing :O

Answer 16

x^2=y^2
Apply that
|x|=|y|

Answer 17

was there some typo, i^2 = -1 right ?

Answer 18

yeah there was

Answer 19

i got that part

Answer 20

what typo ?
im talking about the trick it self step 7

Answer 21

there seems no trick in step 7
u square a iota u get
-1

Answer 22

-.-

Answer 23

What?

Answer 24

@BSwan in ur complex analysis course, is this a valid identity :
\(\sqrt{\dfrac{z_1}{z_2}} = \dfrac{\sqrt{z_1}}{\sqrt{z_2}}\)

?

Answer 25

nope :)
i said that early