Thread: Proof that -1=1!

I have found an interesting "proof" that i would like to see someone here disprove. It runs as follows:

1=√1

1=-1*-1 then

1=√-1*-1

since √A*B=√A*√B

1=√-1*√-1

since √-1=i

1=i*i=i^2 so

1=i^2=-1 thus

1=-1

Q.E.D. 8)

I am not really sure how that can be disputed, lol.

uve made one fatal error
1=√(-1x-1)
1=√(-1x)√(-1 )
1=i√(x)i=√(x)i^2=-1*√(x)
then we square the whole shit
1=(-1)^2*(√(x))^2=x
all u have proven here is that x=+-1 nothing else. and one of those answers are wrong

so next time u try to prove something that goes against something so freasking obvius think it throu about a million times before posting it anywhere

hmmmmm....

You messed up your math. Ill show you yours and the corrected.

1=√(-1x-1)
1=√(-1x)√(-1 )
1=i√(x)i=√(x)i^2=-1*√(x)
1=(-1)^2*(√(x))^2=x

1=√(-1x-1)
1=√(-1)x√(-1 )
1=ixi=i^2=-1
1=(-1)^2

you tried to bring an x into there out of nowhere. Sorry, you cant do that unless you set x=1!

The problem lies in that the √(-1) as used here is not a well defined continuous function everywhere - it has two branches and thus the step
√(uv) = √u x √v cannot work in general for non-negative real numbers.

We have instead that √(uv) = (+-) √u x √v where you have to pick the right sign.

That is what i felt was the problem with the "proof." Doesnt really surprise me with the same thing, considering all of the other steps were relatively legitamate.

i didnt bring x from nowhere, it says 1=-1x-1
i see a x there, and i uses a x
if u mean times u should use * not x, it might get confused else

sorry about that. I changed it.

good, cause if i use x as times i could write xxx wich either means x^3 or x times x wich is x^2

Is there no way for the powers that be on this forum to include a latex translator - its a lot easier to just write in latex code and then for it to be rendered into mathematical notation!

couldnt the leader of this forum add like in MS Words equation program? that would make equations alot easier, both mathematical/physical/chemical

I think I'm gonna go and find a program that makes lines of mathematics to post on forums like this.

Vroomfondel - if you are interested in real mathematics and physics then go and learn latex. It is the langauge used in the real academic world and is quite handy to know!

Anything but ms equation editor Zelos!

check your 4th line. that equation is true if and only if a and b are non-negative.

Originally Posted by river_rat

Is there no way for the powers that be on this forum to include a latex translator - its a lot easier to just write in latex code and then for it to be rendered into mathematical notation!

i have no idea what a latex translator is but for powers you can use the < sup> < /sup> tags
note: i have a space in between the first bracket so that the tags will be displayed, you don't want that space.

E=MC²

Log₂2³

god dammed logarithms.

river_rat i dont care if its MS or anything, we need somekinda equation editor ehre

how u assume 1=root(1)=1;
aactually it is eq to +_1 na

Originally Posted by Vroomfondel

I have found an interesting "proof" that i would like to see someone here disprove. It runs as follows:

1=√1

1=-1*-1 then

1=√-1*-1

since √A*B=√A*√B

1=√-1*√-1

since √-1=i

1=i*i=i^2 so

1=i^2=-1 thus

1=-1

Q.E.D. 8)

I am not really sure how that can be disputed, lol.

Two things have been forgotten here. First, review your algebra theorems (line 4-- only works with positive integers I believe). Second, root 1 is +/- 1, and you must remember the issue of extraneous solutions.

I think elusive neutrino is right - √(AB) = √A*√B is only valid for positive numbers

i squared can be 1 or -1, but since -1 makes more sense graphically we accept it as being -1

Originally Posted by Neutrino

I think elusive neutrino is right - √(AB) = √A*√B is only valid for positive numbers

√(AB) = √A*√B is true for all real numbers, the problem is that this does not hold for complex numbers because the square root cannot be continuously extended around origin.

Originally Posted by Nima Rahnemoon

i squared can be 1 or -1, but since -1 makes more sense graphically we accept it as being -1

Sorry thats wrong, i^2 = -1 and nothing else. The map x -> x^2 is well defined so you can only have one answer. The ambiquity comes from its inverse, √(-1) = +/- i