Two Plus Two Older Archives Ok so I just proved 1 = -1. Someone help me find my error.
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#11
10-13-2005, 11:56 PM
 Guest Posts: n/a
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]
#12
10-13-2005, 11:58 PM
 theTourne Junior Member Join Date: Sep 2004 Posts: 23
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.
#13
10-14-2005, 12:00 AM
 gumpzilla Senior Member Join Date: Feb 2005 Posts: 1,401
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.
#14
10-14-2005, 01:33 AM
 DrPublo Member Join Date: Jan 2004 Location: Princeton, NJ Posts: 38
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
#15
10-14-2005, 01:57 AM
 Guest Posts: n/a
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.
#16
10-14-2005, 02:12 AM
 Guest Posts: n/a
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]
#17
10-15-2005, 01:26 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
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.
#18
10-16-2005, 11:54 AM
 ninjia3x Junior Member Join Date: Mar 2005 Posts: 7
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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