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June 1st, 2014, 06:25 PM   #11
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Originally Posted by Rishabh View Post
If it diverges then it will be equal to infinity.
No. We use notation like $\lim f(x) = \infty$ to denote divergence, but it doesn't mean any equality in the usual sense of the word.

The fact that your continued fraction diverges means that it doesn't converge to any value, so it tells you nothing with the possible exception that you can't make $i$ by adding and dividing real numbers.
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Last edited by skipjack; June 7th, 2014 at 07:16 PM.
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June 2nd, 2014, 01:34 AM   #12
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Originally Posted by topsquark View Post
then there is a flaw in your original set of hypotheses.
I use only two hypotheses:
1) Division is closed for real numbers.
2) Subtraction is closed for real numbers.
I think the first one is wrong because of division by zero.
P.S. I suggest you read through my work. It has left senior mathematicians puzzled!!!
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June 2nd, 2014, 06:03 AM   #13
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You're clearly using more than that because you're manipulating a(n infinite) continued fraction which brings in all kinds of continuity and convergence issues.

You may wish to read about regularization before continuing further:
Euler summation - Wikipedia, the free encyclopedia
Abel's theorem - Wikipedia, the free encyclopedia
Borel summation - Wikipedia, the free encyclopedia
Cesàro summation - Wikipedia, the free encyclopedia
Ramanujan summation - Wikipedia, the free encyclopedia
Mittag-Leffler summation - Wikipedia, the free encyclopedia
1 + 2 + 3 + 4 + ⋯ - Wikipedia, the free encyclopedia
1 + 1 + 1 + 1 + ⋯ - Wikipedia, the free encyclopedia
etc.
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June 2nd, 2014, 07:51 AM   #14
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Originally Posted by Rishabh View Post
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!!!
You assumed it converges, because you wrote x = cf - 1. I other words, cf = x + 1, where x is a complex number. Since C is also closed under addition, you assumed cf is a complex number, i.e. that it converges. And then, from that you concluded that i is real, which it isn't. So, you proved, by contradiction that cf diverges. a proof by contradiction is when you want to prove something (in our case the divergence of cf), and assume the opposite, and then arrive at a lie (in our case i is real). Since i isn't real, the only possible answer it that the assumption is false (in our case cf converges is false, i.e. it diverges). You didn't even need to realize for yourself it diverges, you already proved it anyway! So, although you failed in proving i is real, which of course you did, because it's simply impossible, you succeeded, in a quite elegant way, to prove cf diverges. (maybe I'll dare to say with an infinitely small help from me).

Maybe just one more thing about convergence/divergence. If a sequence of real numbers $a_1$, $a_2$, ... (which your cf is, it's even a sequence of rationals) diverges, it doesn't converge, so there's no a € R such that for any $\epsilon$>0 there's $n_0$ € N such that for any n > $n_0$ the absolute value of $a_n$ - a_n_0 <$\epsilon$. There are 2 types of divergence. We can get a sequence like 1, 2,...for which we say it diverges to infinity, it will exceed any value. Or -1, -2, ... for which we say it diverges to - infinity, it will become smaller than any value. The other type is a bounded sequence, like 1, 0, 1, 0, ...

Last edited by skipjack; June 7th, 2014 at 07:19 PM.
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June 2nd, 2014, 08:38 AM   #15
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Disregard , i must have laid a brain fart.



It is interesting to see the Continued Fraction truncated at regular intervals

$$x + 1 \ = \ \frac{1}{1-1} \ = \ \frac{1}{0} \ \ $$ undefined

$$x + 1 \ = \ \frac{1}{1- \frac{1}{2 - 1}} \ = \ \frac{1}{1-1} \ = \ \frac{1}{0} \\ $$ undefined again


And the pattern continues , you get undefined at every truncation.

Nice trick BTW , I upvoted this thread to 5 bars.


Last edited by agentredlum; June 2nd, 2014 at 09:23 AM. Reason: More clear explanation + erased brain fart. :)
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June 2nd, 2014, 08:28 PM   #16
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Unhappy Sorry!

I just realized that I am a big unknowing fool standing between great mathematical geniuses.!!! Forgive me for this stupid thread.
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June 2nd, 2014, 08:31 PM   #17
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Cool

There may be a positive aspect though. Firstly, I learnt many new things. Secondly, I may write a book someday about faulty law-defying proofs and include this one.(among others which I have discovered such as the proof that 1=2, 1+2+4+8+16+... = -1, etc.) LOL
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June 2nd, 2014, 08:34 PM   #18
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Here's another proof that i is real. I know the mistake, you also try to find it.
(√(-1))^2 = √(-1) * √(-1);
= √(-1*-1);
= √1;
= 1;
=>√(-1) = √(1) = 1
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June 2nd, 2014, 08:36 PM   #19
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I just realized that I am a big unknowing fool standing between great mathematical geniuses.!!! Forgive me for this stupid thread.
Don't worry. You're a 14 year old with an inquisitive mind. There's nothing wrong with that. There's even less wrong with being wrong some of the time. It's the best way to learn.

All I would say is that at 14 you are unlikely to shake the foundations of any part of mathematics. If you think you have found something that does, you have probably made a mistake somewhere. So perhaps asking what is wrong would be better than claiming a new discovery.
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June 3rd, 2014, 05:24 AM   #20
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Originally Posted by v8archie View Post
All I would say is that at 14 you are unlikely to shake the foundations of any part of mathematics. If you think you have found something that does, you have probably made a mistake somewhere.
I was ~12 when I believed that I may have shaken the foundations of math with my discovery of triangular numbers (which I denoted $_{\scriptscriptstyle+}\!\!\!^\mid$ [as close as I can get in TeX]).

Anyone else have similar admissions to make?
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