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 May 30th, 2014, 04:35 AM #1 Member     Joined: May 2014 From: India Posts: 87 Thanks: 5 Math Focus: Abstract maths! A discovery which could shake the foundation of complex numbers! I am a 14 year old (please continue reading despite my age) Math lover and explorer. I just discovered something I earnestly want to share with experienced Mathematicians. It is for this reason I have joined this site. What I discovered is that i(root of -1) can be expressed as a continued fraction. The proof is fairly simple but the discovery is important. I want to safely share this discovery because I believe that this discovery has the potential to prove that i is real(though I couldn't prove it myself since I could not find a good definition for real numbers.) I want it safe because even though I don't want to gain popularity, I don't want others to copy my proof without credit. I tried arXive but it requires you to be endorsed and I am only a 14-year old schoolboy. Is there another way to publish it "safely"? Please help me as the excitement is difficult to contain. Last edited by skipjack; June 7th, 2014 at 07:09 PM.
 May 30th, 2014, 05:34 AM #2 Senior Member     Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology It would be unbelievable if i was real, because it simply isn't! A real number is a member of the set of real numbers, which is defined as a set that satisfies certain 16 axioms. Most common way to prove that such a set exists is by so-called Dedekind cuts-intuitively you define a real as a convergent sequence of rationals (fractions). Well, that's not exactly it, but let's say it is. https://en.wikipedia.org/wiki/Real_number https://en.wikipedia.org/wiki/Dedekind_cut Complex numbers are defined as the set of all ordered pairs of reals, with addition "componentwise", and multiplication as follows: (a, b) (c, d) = (ac-bd, bc+ad) So i^2 = (0, 1) (0, 1) = (-1, 0). The reason for such a definition of multiplication is the fact that C is then an algebraically closed field-each polynomial in C has a zero. Then again, I'm not saying you're wrong, stranger things have happened. Last edited by raul21; May 30th, 2014 at 06:01 AM.
May 30th, 2014, 06:15 AM   #3
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Quote:
 Originally Posted by Rishabh I am a 14 year old (please continue reading despite my age) Math lover and explorer.
Welcome!

Quote:
 Originally Posted by Rishabh What I discovered is that i(root of -1) can be expressed as a continued fraction.
Yes. The standard form for the continued fraction is

i = [i; ]

but you could write it in alternate ways like

i = [i-1; 1]

if you like.

Quote:
 Originally Posted by Rishabh has the potential to prove that i is real
i is not a member of $\mathbb{R},$ the real numbers.

Quote:
 Originally Posted by Rishabh (though I couldn't prove it myself since I could not find a good definition for real numbers.)
The usual method is to construct them from the rationals via Cauchy sequences. An example of a Cauchy sequence would be taking successive decimal digits of a real number: 3, 3.1, 3.14, 3.141, etc. They get closer and closer to something, then you use a special property of the real numbers (the supremum property) to show that there is a number which is the limit of the sequence.

An alternate method is axiomatic: we list certain properties that the real numbers have which are jointly enough to define them.

Both approaches are discussed, for example, here:
Construction of the real numbers - Wikipedia, the free encyclopedia

Edit: Actually, Dedekind cuts (as Raul mentioned) are also quite popular. All three methods work, just pick the one you like best.

Quote:
 Originally Posted by Rishabh I want it safe because even though I don't want to gain popularity, I don't want others to copy my proof without credit. I tried arXive but it requires you to be endorsed and I am only a 14-year old schoolboy. Is there another way to publish it "safely"?
Unsolicited advice: No one is trying to steal your ideas, don't worry about it. But if they were hiding it would be the worst thing you could do, because then some nefarious person who managed to get a copy could spread it as their own. Better would be to have made some copy publicly available (with a timestamp!) so credit could be traced back to you.

(It's arXiv, by the way.)

Quote:
 Originally Posted by Rishabh Please help me as the excitement is difficult to contain.
I'm glad you find math so exciting! I hope you're able to keep that feeling.

Last edited by skipjack; June 7th, 2014 at 07:12 PM.

 May 30th, 2014, 06:42 AM #4 Senior Member   Joined: May 2014 From: Allentown PA USA Posts: 111 Thanks: 6 Math Focus: dynamical systen theory To Rishabh Don't be afraid, Rishabh. Most of the My Math Forum community are good people. I myself am somewhat foggy on some rather important concepts, given that I am middle-aged. You have probably have heard of Srinivasa Ramanujan, Sir Chandrasekhara Venkata Raman, and Subrehmanyan Chandrasekhar. There is a friend of mine, whom I haven't seen in many years, who has his master's degree in math from the University of Madras. Good luck! Sincerely yours, Carl J. Mesaros
 May 31st, 2014, 02:15 AM #5 Member     Joined: May 2014 From: India Posts: 87 Thanks: 5 Math Focus: Abstract maths! Thanks to all who have replied. I myself was very unsure that such a thing could happen, so I wanted to discuss it. But before that, I have already fallen in love with this site.! Here's the original discovery. To prove: âˆš(-1) = 1/(1-1/(2-1/(1-1/(2-...)))) - 1; Proof: Code: Let x = 1/(1-1/(2-1/(1-1/(2-...)))) - 1; â‡’x + 1 = 1/(1-1/(2-(x+1))); â‡’x + 1 = (1 - x)/(-x); â‡’x^2 + x - x +1 = 0; â‡’x^2 + 1 = 0; â‡’x = âˆš(-1); Hence proved. To prove: 1/(1-1/(2-1/(1-1/(2-...)))) - 1 is real. (Not very sure of this proof.) Proof: Axioms/theorems used in this proof: â€¢Division is closed under real numbers. â€¢Subtraction is closed under real numbers. Statement: Since this continued fraction involves only division and subtraction, I don't see why it shouldn't be real. Afterword: I don't forcefully assert this proof. I want to know what exactly is wrong with the proof. I will be very thankful to those who help.
 May 31st, 2014, 03:05 AM #6 Senior Member     Joined: Apr 2014 From: zagreb, croatia Posts: 234 Thanks: 33 Math Focus: philosophy/found of math, metamath, logic, set/category/order/number theory, algebra, topology Maybe the error lies in the fact you assumed in advance the continued fraction converges. Thanks from topsquark and Rishabh
 May 31st, 2014, 12:05 PM #7 Senior Member   Joined: May 2014 From: Allentown PA USA Posts: 111 Thanks: 6 Math Focus: dynamical systen theory Rishabh, Whether you realize it or not, complex variables are very useful in some areas, such as electrical engineering. You're one up on me, kiddo! I personally deal with complex numbers very rarely. Good luck and may you find mathematics very good! Sincerely yours, Carl J. Mesaros Thanks from Rishabh
June 1st, 2014, 01:40 AM   #8
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Quote:
 Originally Posted by raul21 Maybe the error lies in the fact you assumed in advance the continued fraction converges.
I just realized that it actually diverges as the denominator becomes lesser(due to subtraction) as we move down the ladder.
This however still poses a very important question.
If it diverges then it will be equal to infinity. But i is not equal to infinity!!!

June 1st, 2014, 01:43 AM   #9
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Quote:
 Originally Posted by Carl James Mesaros Good luck and may you find mathematics very good!
Thanks for the good luck. And I already find Math very interesting.

June 1st, 2014, 04:27 PM   #10
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Quote:
 Originally Posted by Rishabh If it diverges then it will be equal to infinity. But i is not equal to infinity!!!
I haven't looked at your work in any detail, but typically if your method is saying something is impossible (like i = infinity) then there is a flaw in your original set of hypotheses. Seeing that infinity is not a real number the method you are using might be giving you a big push to say that neither is i.

-Dan

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