Ok so I just proved 1 = -1. Someone help me find my error.

[DrPublo]

Hi guys. First post in this forum.

Working on a problem set recently, a few friends and I accidentally discovered a proof of -1=1, and for the life of us we can't find out what we did wrong. And it's not like we're math slouches either; we're all graduate students in physical/theoretical chemistry.

From what I understand posting TeX doesn't work on 2+2, so you'll have to follow my algebra.

Start with the identity

(E-V)^(1/2) = (E-V)^(1/2)

Now multiply each side by -1, except on the RHS substitute i^2 for -1 (where i of course is the imaginary number).

(-1)(E-V)^(1/2) = (i^2)(E-V)^(1/2)

Now divide through by i

(-1/i)(E-V)^(1/2) = i*(E-V)^(1/2)

But since i is just the square root of -1, we can subsume it into the square root of E-V

(-1)[(E-V)/-1]^(1/2) = [(-1)(E-V)]^(1/2)

and then rearrange the interior of the square root to find

(-1)(V-E)^(1/2) = (V-E)^(1/2)

or

-1 = 1.

No dividing by zero in this proof either. Where did I make a mistake?

The Doc

I suspect the error occurs when you take the square root of -1. Normally the square root of a positive number can be positive or negative. eg sqrt(9) = +3 or -3
When you take the square root of -1 you are saying there is only one possible answer (i)
I might be out to lunch, but I suspect that is where the problem lies.

[DrPublo]

No question that -1 has both "positive" and "negative" square roots, as i^2 = (-i)^2 = -1. But I think taking the same root on both sides of the equation should maintain equality? For example, it would be odd to remark that 9 = 9 but then upon taking the square root of both sides of the equation conclude 3 = -3.

The Doc

[benkahuna]

You have performed an invalid operation between steps 3 and 4.
The 2 terms in step 3 equal each other. The 2 in step 4 do not.
Maybe you can't use -1/i instead of i algebraicly as you have.

[DrPublo]

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You have performed an invalid operation between steps 3 and 4.
The 2 terms in step 3 equal each other. The 2 in step 4 do not.

[/ QUOTE ]
Yes, I know that. The question is what principle of mathematics have I violated.
[ QUOTE ]

Maybe you can't use -1/i instead of i algebraicly as you have.

[/ QUOTE ]

-1/i = i. Try multiplying through by i.

The Doc

[benkahuna]

[ QUOTE ]

[ QUOTE ]

Maybe you can't use -1/i instead of i algebraicly as you have.

[/ QUOTE ]

-1/i = i. Try multiplying through by i.

The Doc

[/ QUOTE ]

I know they're equivalent, thanks. In other news 4/2=2.

I'm sure an elementary search of a place like wikipedia will tell you why you can't deal with i as you have.

I've told you what I know and I'm not going to search for the answer for you. I might know, but I just don't deal with a imaginary numbers very often in daily life.

I was thinking what the other poster was thinking about there being positive and negative square roots being the issue, but since no operations were performed taking a sqaure root of a number, I don't think it applies here. It's not like that lame proof that 1+1=1.

I'm not sure if this helps, but I agree with the previous poster that that particular step was incorrect.


sqrt(-1/1) = sqrt(-1)/sqrt(1), but sqrt(1/-1) is - sqrt(1)/sqrt(-1) not sqrt(1)/sqrt(-1).

(-1)(-1) = 1, so sqrt((-1)(-1)) = 1, but sqrt(-1)sqrt(-1) = i^2 = -1 (not 1).

[DrPublo]

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(-1)(-1) = 1, so sqrt((-1)(-1)) = 1, but sqrt(-1)sqrt(-1) = i^2 = -1 (not 1).

[/ QUOTE ]

You just blew my mind.

What is the technical reason this proof is incorrect?

The Doc

I'm not sure I can give you a technical reason, but certain mathematical laws or priciples that work for real numbers don't necessarily work for complex numbers.

sqrt(a/b) = sqrt(a) / sqrt(b) is only true for real numbers.

I have a hard enough time even understanding what a complex number actually is. [img]/images/graemlins/smile.gif[/img]