Two Plus Two Older Archives  

Go Back   Two Plus Two Older Archives > Other Topics > Science, Math, and Philosophy

Reply
 
Thread Tools Display Modes
  #1  
Old 10-13-2005, 11:00 PM
DrPublo DrPublo is offline
Member
 
Join Date: Jan 2004
Location: Princeton, NJ
Posts: 38
Default 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.

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
Reply With Quote
  #2  
Old 10-13-2005, 11:23 PM
Guest
 
Posts: n/a
Default 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.
Reply With Quote
  #3  
Old 10-13-2005, 11:28 PM
DrPublo DrPublo is offline
Member
 
Join Date: Jan 2004
Location: Princeton, NJ
Posts: 38
Default 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
Reply With Quote
  #4  
Old 10-13-2005, 11:30 PM
benkahuna benkahuna is offline
Junior Member
 
Join Date: Oct 2004
Posts: 4
Default 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.
Reply With Quote
  #5  
Old 10-13-2005, 11:37 PM
DrPublo DrPublo is offline
Member
 
Join Date: Jan 2004
Location: Princeton, NJ
Posts: 38
Default 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
Reply With Quote
  #6  
Old 10-13-2005, 11:38 PM
Guest
 
Posts: n/a
Default 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).
Reply With Quote
  #7  
Old 10-13-2005, 11:41 PM
Guest
 
Posts: n/a
Default 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).
Reply With Quote
  #8  
Old 10-13-2005, 11:49 PM
DrPublo DrPublo is offline
Member
 
Join Date: Jan 2004
Location: Princeton, NJ
Posts: 38
Default 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
Reply With Quote
  #9  
Old 10-13-2005, 11:52 PM
benkahuna benkahuna is offline
Junior Member
 
Join Date: Oct 2004
Posts: 4
Default 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.
Reply With Quote
  #10  
Old 10-13-2005, 11:56 PM
gumpzilla gumpzilla is offline
Senior Member
 
Join Date: Feb 2005
Posts: 1,401
Default 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.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off

Forum Jump


All times are GMT -4. The time now is 04:38 AM.


Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2024, vBulletin Solutions Inc.