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-   -   Ok so I just proved 1 = -1. Someone help me find my error. (http://archives2.twoplustwo.com/showthread.php?t=357250)

 DrPublo 10-13-2005 11:00 PM

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

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.

(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

 10-13-2005 11:23 PM

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

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 10-13-2005 11:28 PM

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

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 10-13-2005 11:30 PM

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

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 10-13-2005 11:37 PM

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

[ QUOTE ]
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

 10-13-2005 11:38 PM

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

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).

 10-13-2005 11:41 PM

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

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

 DrPublo 10-13-2005 11:49 PM

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

[ QUOTE ]
(-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

 benkahuna 10-13-2005 11:52 PM

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

[ 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.

 gumpzilla 10-13-2005 11:56 PM

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

You're picking the two different branches, since both i and 1 / i when squared yield -1, but themselves differ by a factor of -1, and this is where the problem comes in. So the answer to your question is that you are being inconsistent in how you define sqrt(-1), and if you're consistent in that regard then the problem goes away. Difficulties of this kind are extremely common when working with complex variables, which is part of what makes them confusing.

 10-13-2005 11:56 PM

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

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]

 theTourne 10-13-2005 11:58 PM

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

I could certainly be wrong, but I think the only problem is in the final division.

I don't see a problem with:

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

To use some concrete numbers, I think we all agree that sqrt(4) = -2

Then,

sqrt(4) = -2
sqrt(4) = (-1)*(2)
sqrt(4) = (-1)*(sqrt(4))

I don't believe you can divide both sides of the equation by a root, because it could be either positive or negative.

 gumpzilla 10-14-2005 12:00 AM

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

Here's another "proof" in this vein that I've always found enjoyable:

x^2 = x + x + . . . x (x copies of x)

Differentiate both sides:

2x = 1 + 1 + 1 . . . = x

Divide by x:

2 = 1.

 DrPublo 10-14-2005 01:33 AM

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

[ QUOTE ]
So the answer to your question is that you are being inconsistent in how you define sqrt(-1), and if you're consistent in that regard then the problem goes away.

[/ QUOTE ]

I believe you, and your answer makes a lot of sense. I've also tried to do the identical "proof" in terms of the generalized complex number r*exp(i*theta) and can't seem to make it work, indicating that my problem lies somewhere with the casual use of i.

One question though. I thought I am using sqrt(-1) = i on both sides of the equation, so wherein lies the inconsistency?

The Doc

 10-14-2005 01:57 AM

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

[ QUOTE ]
You're picking the two different branches, since both i and 1 / i when squared yield -1, but themselves differ by a factor of -1, and this is where the problem comes in. So the answer to your question is that you are being inconsistent in how you define sqrt(-1), and if you're consistent in that regard then the problem goes away. Difficulties of this kind are extremely common when working with complex variables, which is part of what makes them confusing.

[/ QUOTE ]
Right. To elaborate a bit (for the benefit of the others), you end up with this sort of thing any time you have a multi-valued function. For example, the function x^2 takes the number 1 to the number 1. It also takes the number -1 to the number 1. Thus the inverse of the function x^2 logically could take 1 to 1 or to -1. You must choose a consistent rule for making sense of this (we always choose the positive value without thinking much about it).

With real numbers, this is easy, and remains consistent once the rule is chosen. With complex numbers, however, this "choice" cannot be consistent over the entire complex plane if you are dealing with a multivalued function -- you need to specify where your choice is valid. If you try, for example, to integrate some multivalued function over some line in the complex plane, you must draw a branch cut line over which your line of integration can not cross. Crossing the branch cut would intuitively mean that part of the time you integrate using one rule, and after crossing the branch cut you use another rule. This is bad, since the choice of where you draw the line is somewhat arbitrary -- all that is important is that it must be drawn somewhere to ensure that things are well-defined.

 10-14-2005 02:12 AM

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

[ QUOTE ]
To elaborate a bit (for the benefit of the others)

[/ QUOTE ]

Thanks. I understand it perfectly now. [img]/images/graemlins/smile.gif[/img]

 BruceZ 10-15-2005 01:26 PM

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

[ QUOTE ]
Here's another "proof" in this vein that I've always found enjoyable:

x^2 = x + x + . . . x (x copies of x)

Differentiate both sides:

2x = 1 + 1 + 1 . . . = x

Divide by x:

2 = 1.

[/ QUOTE ]

Good one. The problem here is that

x^2 = x + x + . . . x (x copies of x)

is only true for integer x, and if this function is only defined for integer x, then the function is discreet and is not differentiable.

 ninjia3x 10-16-2005 11:54 AM

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

[ QUOTE ]
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.

(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

[/ QUOTE ]

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

for the (-1/i), times top and bottom by i

you get -i/i^2 = -i/-1 = i

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

...pretty obvious mistake u made having root(-1) = -i

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