BlameThePixel!

go on then, i dare ya - find an erro in this little beauty.

-1 squared is 1... and u cant get a route of a negative, so 2 does not equal 0

you're joking, right? lol

This is true. You [I]can[/I] find the cube route of a negative, but that would just make it a plain 1 anyway.

Also, you cannot cancel a - into two equivelent fractions (well, you can, it's just not a good idea).

Regarding the sqrt of 1/-1 and -1/1 (if it were possible), these will be constantly equal. This is due to the fact that you may continually multiply either fraction by sqrt of -1.

Finally, since 1/-1 and -1/1 would be equal, no matter how many things you do to it, they will ALWAYS BE EQUAL. Remember, it doesn't matter where you put the negative in a fraction, it is still the same number.

in the step from the fractions to the step w/o fractions, you've made an error.

Solving anything with the fractions produces a true answer, anythign below does not. Let me try to remember why though.

ps: i = rad(-1), and i^2 IS -1 so that is not the problem

AH, when introduced to imaginaries i remember this.

My prof called it a collision of two rules, and said that one rule is "more correct" then the other.

I dont remember the exact reasons or errors in the above, as i did some math which leads to the exact above answer. I'll get my math genius friend to disprove it and actually explain why, that that might take until monday

Anywhere, heres some proof for the invalidness:


(btw, i = rad(-1) and i^2 = -1. the math shown is inherently wrong, but I dont remember why)

messed up in saying that 1/sqr(-1)=sqr(1/-1). the laws that made that true with real numbers don't apply with imaginary numbers. sqr(-1)=i. 1/i= i^(-1)=-i. also, the squareroot isn't a function, there are 2 different answers. sqr(-1)=i and sqr(-1)=-i.

for those of you people in calculus, try this one

ladder leaning against a wall, someone pulls at the bottom of the ladder so the top moves straight down. the length of the ladder is L. the height of the ladder on the wall is h. the distance of the ladder from the bottom of the wall is x. pythagoreas tod us that x^2+h^2=L^2. the rate that you pull away from the wall is dx. then taking the derivitave of the whole equation, we get dx+2y*dy=2L*dL. the length of the ladder is a constant, so dl=0. dx+2y*dy=0. solving for dy(that's the speed that the ladder is moving down the wall) we get dy= dy=dx/(2y). so as the top of the ladder approaches the ground, it's moving at some number divided by 0. which approaches infinity, easily surpassing the speed of light. now to turn this into a time machine. . . .

told you you couldn't use complex numbers to prove something that isn't true is ... beats me as to why, we haven't touched complex numbers and probably never will...

and lol at the ladder time machine, stand on thge ladder and see what happens...

shut up with the math.. it hurts my head.. + it doesnt look to pretty either

well the way i look at it is that

√x² = x

therefore

√(-1)² = -1

however if you want to use complex numbers, then i =√(-1) and i² = -1

so:

i/√1 = √1/i
i² = √1²
-1 = 1

you still get the same result - the flaw is actually in the √1√1 since a root of a positive number can either be +ve or -ve then √1√1 can be equal to 1 OR -1 giving:

i² = √1√1
-1 = (-1)(1)
-1 = -1

it was flawed , i just wanted to see if anyone would find the mistake.

er, (-1)² = 1, as i know
and √ of 1 can be 1 or -1

it's (√(-1))(√(-1)) really, so thats (√(-1))²

root then square cancel out so it's left as -1

Readme 0 : 1 everyone else

gg

lol



we've all heard that before

well no-one else noticed the actual flaw, you were all worrying about the root -1 when thats not the bit thats mathematically wrong.