Question

THINK ABOUT IT :::::::::::::::::::;
see commments

Answer 1

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

Answer 2

did any 1 think about it ?

Answer 3

nice trick, but each statement must stand on its own, the last one does not no matter what train of thought you force us down!

Answer 4

so finally -1=1 or not ? if not then why ?

Answer 5

nope
i think something should be in 4 or 5 step

Answer 6

1/i = -i

Answer 7

think about this one, then:

if a=b, then

a - b = 0.

a - b = a - b.

a - b = 2(a - b).

a - b = 2
-----
a - b

therefore, 1 = 2.

what is my trick????? :D

Answer 8

cant divide by 0

Answer 9

2/0 dne

Answer 10

where is the 2/0?????

Answer 11

that was from that second riddle...

Answer 12

nonononon, thats not what i'm asking. there is no step which says 2/0 :P

Answer 13

oh, a-b was defined as 0

Answer 14

by dividing the left side by a-b you implicitly divide the right side by it also...

Answer 15

2/0 is complex infinity and 0/0 is undefined. The last statement is simply incorrect. If a-b had not been defined as 0, this would be different.

Answer 16

i is imaginary therefore it is a "what if" statement that could not be solved so when you take the square root of -1

Answer 17

i was correct it is that step

there is no rule that gurantees the\[\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\]

Answer 18

it gurantees that this is true only when a and b are postive

Answer 19

i^2 = 1 bcuz i = -1 so it is correct

Answer 20

last step is wrong it should be 1=1

Answer 21

nope that is correct i^2 = -1

Answer 22

how

Answer 23

-1 x -1 = 1

Answer 24

good one, snaef999.

Answer 25

i= sqrt (-1) jay

Answer 26

so it's sqrt(-1)^2 = -1

Answer 27

the squre root of -1 is not a real root... try using a calc

Answer 28

@tanvidais13
in your 4th equation, you have multiplied one side of the equation by 2 and not the other

Answer 29

Because of the discontinuous nature of the square root function in the complex plane, the law √zw = √z√w is in general not true.--- straight from wiki because i believed that the square root was the problem

Answer 30

so values must be >=0 for the distributing of the square root

Answer 31

good work friends

Answer 32

lolz =]

Answer 33

*continuosly proving math is correct..... most of the time*