December 22nd, 2016, 09:45 AM  #1 
Member Joined: Jan 2014 Posts: 86 Thanks: 4  Proof that i is poth positive and negative
Where i satisfies i^2 = 1, i = 1/(i) QED 
December 22nd, 2016, 09:53 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,390 Thanks: 2100 Math Focus: Mainly analysis and algebra 
$\displaystyle i$ is neither positive nor negative. The ordering relations $\displaystyle \lt$ and $\displaystyle \gt$ do not apply to the complex numbers.

December 22nd, 2016, 10:00 AM  #3 
Member Joined: Jan 2014 Posts: 86 Thanks: 4 
I know that but how does that attack directly my above derivation? It seems to me that that proves that if we accept (or imagine) that a number squared gives a negative, that number HAS to be both positive and negative. How do you escape that conclusion?

December 22nd, 2016, 10:44 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,390 Thanks: 2100 Math Focus: Mainly analysis and algebra 
"Positive" means "greater than zero". "Negative" means "less than zero". $i$ is neither greater than zero nor less than zero. $\frac{1}{i}$ (which is actually equal to $i$ as demonstrated by your derivation) is neither greater than zero nor less than zero. We can say that $i \gt 0 = 0$, but $\left\frac1i\right \gt 0$ too (as we'd hope, since the two numbers are equal). 
December 22nd, 2016, 10:56 AM  #5 
Member Joined: Jan 2014 Posts: 86 Thanks: 4 
It's not zero neither, so we "proved" it's not a real number hahah. But how do you prove that it has no sign? Or is it an axiom? Furthermore how is the expression "neither negative nor positive" any different than "both positive and negative"? Or alternatively how is "neither yes or no" different than "both yes and no"? 
December 22nd, 2016, 11:44 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,390 Thanks: 2100 Math Focus: Mainly analysis and algebra 
You can generate orderings on the complex plane, but they are not the total ordering implied by your use of the terms positive and negative. I suggest to do some reading about ordering on the complex plane to understand the subject. Regarding your second paragraph, the difference is quite stark. How do "mammal" and "fourfooted" apply to an elephant? How do they apply to a snake? The main reason I use "neither" is that the ordering you are using either does not exist or is not defined. For an ordering to make sense "both" is not an option. Your logic says that because, in the real numbers, $0=0$ zero must be both positive and negative. However, by the usual definition of those terms it is neither. 
December 22nd, 2016, 09:47 PM  #7  
Member Joined: Sep 2016 From: USA Posts: 98 Thanks: 36 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
Cleary $i \neq 0$. Now suppose $0 < i$ so we can take $0 < a < i, b = i = c$ and we have a contradiction to the total ordering on the reals. Repeat this by assuming $i < 0$ and applying the argument for $i$. In any case, you conclude that trichotomy fails for $i$ so it can't fit into the total ordering on the reals.  
December 23rd, 2016, 06:36 AM  #8 
Senior Member Joined: May 2016 From: USA Posts: 466 Thanks: 198  You have not defined what you mean by positive and negative. Consequently, your "proof" has a giant hole. It is of course correct that $\ i^2 = (\ 1)(i^2) = (\ 1)(\ 1) = 1.$ $\therefore \ i^2 =(\ i)(i) = 1 \implies i = \dfrac{1}{\ i}.$ And that relates to definitions of positive and negative how? You skip over that part entirely. I strongly suspect that the highly confusing use of the $$ symbol in mathematics has led you astray. The symbol is used to indicate the operation of subtraction. With respect to a numeral representing a real number, the same symbol is ALSO used to indicate that the real number represented is negative, meaning less than zero. Finally, with respect to a pronumeral, which represents an unspecified number, the $$ symbol represents an additive inverse and says nothing about whether the unspecified number is positive or negative. $x = \ a \iff a + x = 0$, which does not imply that $\ a$ is negative, meaning less than zero. It is terrible notation that invites confusion. Last edited by JeffM1; December 23rd, 2016 at 06:39 AM. 
December 23rd, 2016, 08:06 AM  #9 
Senior Member Joined: May 2016 From: USA Posts: 466 Thanks: 198 
Here is another approach. $Assume\ i\ is\ both\ positive\ and\ negative \implies $ $(\ 1) * i = \ i\ is\ both\ negative\ and\ positive \implies$ $\ i\ is\ both\ positive\ and\ negative \implies$ $i + (\ i) > 0\ and\ i + (\ i) < 0\ and\ i + (\ i) = 0 \implies$ $both\ 0 \ne 0\ and\ 0 = 0,$ $which\ is\ a\ contradiction.$ $\therefore i\ is\ both\ positive\ and\ negative\ is\ FALSE.$ 
December 24th, 2016, 09:53 AM  #10  
Member Joined: Jan 2014 Posts: 86 Thanks: 4  Quote:
Quote:
i  i = i + 1/i, the only way to make that zero is if i is 1 in one instance and 1 at the other instant. So i = 1 = 1, 2 = 0, 1 = 0, something is nothing. The only way to escape this is if we declare that i^2 = 1 is false. Last edited by Tau; December 24th, 2016 at 10:12 AM.  

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